Topic 4: Genetics - Biology Notes Form 6

GENETICS

Genetics is the study of heredity and variation

Heredity- is the passage of character from one generation to another.

Variation – these are differences among individuals of the same species.

GENETICS & VARIATION

HEREDITARY MATERIALS:-

Hereditary or genetic materials are chemical substances or units on the chromosome that are responsible for the passage of genetic information from one generation to another.

Characteristics of hereditary materials:-

The features that characterize hereditary materials include the following:-

1. Metabolic stability. Hereditary materials are metabolically very stable or chemically inert. If it were altered to any extent, impercfect copies would be made.

2. Mutation: There is a close correlation between hereditary materials and mutation agents that is, when the hereditary materials are exposed to mutagens undergo mutations.

3. Self replication – Hereditary materials are capable of reproducing themselves.

4. Constancy within the cell – The amount of hereditary materials remains constant within a cell or in the cells of organisms of the same species.

5. Carriage of information – The hereditary materials are capable of carrying genetic information from one generation to another.

6. Linearity – The information or the genetic materials if always arranged in a linear array.They are macromolecules

SPECIES CONCEPT:

There are several ways if defining what a species is:-

(a) According to genetics: A species is defined as a group of organisms that share a common gene pool and have the same number of chromosomes. Gene pool is the total of all genetical make up in a given population.

(b) According to ecology: A species is defined as a group of organisms that share a common ecological niche no two species can share the same genetic niche.

(c) According to plant and animal breeding: A species is a group of organisms as that can freely interbreed and produce fertile offspring.

Question;-How does a breeder define a species?

By the above definition is a horse and donkey of the same species? Give reasons.

SOLN:-

According to definition of species given by a breeder as horse and a donkey are of different species. This is because although they interbreed freely producing a mule but a mule is non – fertile and therefore it cannot produce another mule.

Question:-In a certain research programme at Kwamsisi Rodent research centre, cage of 159 rats from Usambara mountains and a cage of 162 rats from Pugu forest; reserve were researched. If you were one of the researchers how would you identify those rats of the same species?

SOLN:-

To identify those of same species the following should be done:-

(a) To allow interbreeding:

Those rats of the same species will interbreed freely and produce fertile offspring.
Those rats of different species will either fail to interbreed or if the will inter breed the product so produced will be non – fertile.

(b) Chromosomes analysis:

Those rats of the same species will have the same number of chromosome.

EXTRA OF HEREDITARY MATERIALS

Macromolecules

They are universal but restricted within species.
All are made due to phosphoric acid.
All are comprised of pentose sugar, nitrogen base and phosphate.

CHROMOSOMES AND THEIR STRUCTURE

Chromosomes carry the hereditary – material DNA. In addition they are made up of protein and RNA. Individual chromosomes are not visible in a non – dividing (resting) but the chromosomal material can be seen especially if stained. This material called Chromosomes become visible only during onset of cell division.

Each chromosome is seen to consist of two threads called chromatids joined to point called centromere. Chromosomes vary in shape and size both with and between species.

Homologous chromosomes are similar in structure.

Arrangement of homologous chromosomes in pair is known as Karyograi and the set of chromosomes is known as Karyotype.

Structure of chromosome:-

TOPIC 4: GENETICS|BIOLOGY FORM 5 & 6

THE NUCLEIC AIDS, TYPES OF HEREDITARY MATERIALS

There are two types of nucleic acids:-

(a) Ribonucleic acid, RNA.

(b) Deoxyribonucleic acid, DNA.

Chemical nature of nucleic acids:-

Chemically nucleic acids are composed of the following:-

1. Pentose sugar – This is a five carbon sugar.

In RNA, there is ribose sugar where as in DNA, there is deoxyribose sugar.

2.Nitrogenous (organic) base

There are two groups of organic bases:

(a)Purine bases-

These include:

(i) adenine (A)

(ii) guanine (G)

(b)Pyrimidine bases-

These include

(i) Thymine(T)

(ii) Cytosine (C)

(iii) Uracil (U)

Note that;

Thymine is a DNA pyrimidine while Uracil is an RNA pyrimidine. No uracil in DNA nor is there thymine in RNA

3. Phosphate group:-

This is derived from phosphoric acid and it is this group that makes compounds (DNA and RNA) acidic in nature.

The three components are combined by condensation reactions to give a nucleotide. By a similar condensation reaction a dinucleotide is formed and continued condensation reaction leads to the

formation of a polypeptide.The main function of nucleotides is the formation of nucleic materials RNA and DNA which have vital roles in protein synthesis and heredity.

Structure of a typical nucleotide:-

1. Chemical bonds:-

There are two types of chemical bonds:-

Phosphodiester bonds – These hold the nucleotides together.

Hydrogen bonds – These hold together the complementary base pair in DNA as well as RNA.

2. Protein cost:-

The DNA of the eukaryotes has a history protein coat over its surface.

(A) RIBONUCLEIC ACID (RNA)

Chemical nature:-

Ribonucleic acid is chemically composed of the following substances:-

(a) Pentose sugar – This is a 5 carbon sugar called ribose.

(b) Phosphate group– derived from phosphoric acid.

(c) Organic (nitrogenous) bases – These are of two types.

(i) Purines – These are Adenine (A) and Guanine (G).

(ii) Pyrimidines – These are Uracil (U) and Cytosine.

(d) Chemical bonds: These are of two types:-

(i) Phosphodiester bonds – Which hold the nucleotides together.

(ii) Hydrogen bonds – Which hold together the complementary base parts in tRNA molecule.

Diagram to show structure of RNA:

Role of RNA.

The role of RNA is situational:-

1. In the presence of DNA, RNA in collaboration with DNA.

Controls heredity.

Controls protein synthesis.

2. In the absence of DNA, RNA alone.

Controls heredity.

Controls protein synthesis.

Types of RNA

According to function and location in the cells, there are three types of RNA:-

(a) Messenger RNA (mRNA).

This is the type of RNA formed from one of the strands of DNA in the process called transcription.

Role of mRNA:-

Messenger RNA carries the genetic code from DNA in the nucleus to the ribosome in the cytoplasm. This genetic code contains the information about the types of amino acids that should be joined together to form a protein molecule.

(b) Ribosomal RNA (rRNA).

Ribosomal RNA (rRNA) or soluble RNA constitutes about 80% of the total RNA in the cell.
Ribosomal RNA is synthesized by a special DNA found in the nucleolus at a special region called a nucleolar organizer.
It makes a bulk of the ribosome.

Role of rRNA

(i) It is an integral part of the ribosome i.e large proportion of the ribosome is made up on rRNA.

(ii) It attracts other types of RNA i.e mRNA and tRNA towards the ribosome during protein synthesis.

(c) Transfer RNA (tRNA)

This constitutes about 15% of the total RNA in the cell.
Structurally, tRNA is a clover – lead shaped molecule with a folded loop – like chain.
The looping of the chain, results into pairing of the folded of organic bases. Hence the formation of hydrogen bonds.
The molecule has got four active / recognition sites.
The upper site recognizes an amino acid, where as the lower side (Anticodon) recognizes the mRNA. One of the sides recognizes the ribosome where as the other one recognizes in enzymes, amino – acyl tRNA synthetase.

Role of tRNA

The role of tRNA is to carry the activated amino acids from various parts of the cytoplasm to their binding site, the ribosome.

(B) DEOXYRIBONUCLEIC ACID (DNA)

Chemical nature:-

DNA is chemically composed of the following substances:

(i) Deoxyribose sugar– This is a pentose (5 – carbon) sugar.

(ii) Organic or nitrogenous bases – These are of two categories.

(a) Purine bases – These are Adenine (A) and Guanine (G).

(b) Pyrimidine bases – These are Cytosine (C) and Thymine (T).

Base pairing rules:-

Since DNA is double stranded molecule, the bases on the two strands appear in pairs being held together by the hydrogen bonds.
The strands run in opposite directions, that is are Antiparallel.
The base pairing rules make the chains, Complementary.
According to Watson – Crick modal of DNA structure, a purine pairs with a pyrimidine. The rules are that:

(a) Adenine pairs with thymine and the two bases are held together by two hydrogen bonds.

(b) Cytosine pairs with guanine and the two bases are held together by three hydrogen bonds.

(iii) Phosphate group – dividend from phosphoric acid.

(iv) Protein – Over the surface of DNA, there is a histone protein coat.

(v) Chemical bonds – There are two types of chemical bonds.

(a) Phosphodiester bonds – These hold the nucleotides together.

(b) Hydrogen bonds – These hold the complementary base parts together.

Diagrammatic structure of DNA:-

Role of DNA in protein synthesis.

This role of DNA is that, it instructs the cell of the types of amino acid that should be initiated to form a protein molecule. That is the message contains the information about the types of amino acids that should be joined up forming the protein molecules

Qn:-One of the characteristics of DNA as a hereditary material; is that it is metabolically very stable. State the features of DNA that account for this metabolic stabilit

Answer;-

The features of DNA account for this metabolic stability include the following:-

(a) Possession of a histone protein coat

(b) The helical nature, increases mechanical strength.

Evidence for the role of DNA in inheritance

It took many years to clarify whether genetic material was the DNA or the protein the chromosomes. It was suspected that protein might be the only molecule with staff verify of structures to act as genetic material.

Evidence from bacteria:

In the days before development of antibiotics pneumonia was often a fatal disease. It was intended in developing a vaccine against the bacterium Pneumococcus which was one form of pneumonia.

Two forms of pneumococcus where known, one covered with a gelalinious capsule one violent (disease producing) and the other non – capsulated and non – violent. The capsule protected the bacterium in some way from attack by immune system of the host.

Griffith hoped that by injecting the patients with either the non – capsulated the heat – killed capsulated forms, their bodies would produce antibodies which would give protection against pneumonia. In a series of experiments he injected with both forms of pneumococcus and obtained the results shown in a table below. The dead mice revealed the presence of live capsulated forms their bodies. On the basis of these results Griffith concluded that something must be passing from the heat – killed capsulated forms to the live non – capsulated forms which caused them to develop capsule and become virulent.

However the nature of this transforming principle, as it was known was not isolated and identified until 1994.

Results of Griffth’s experiments:-

Injected form of pneumococcus	Effect
Live non – capsulated	Mice live
Live capsulated	Mice live
Heat – killed capsulated	Mice live
Heat – killed capsulated	Mice live
Heat – killed capsulated + live non capsulated	Mice live

Later on analysis on the constituent molecule of heat – killed capsulated pneumococcal cells and testing their ability to bring about transformation in live non – capsulated cells. Removal of the polysaccharide capsule and the protein much from the cell extracts had no effect on the transformation, but the addition of the enzyme deoxyribonuclease (DNase), which hydrolyses (break down) DNA prevented transformation. Hence, demostration of the Griffth transforming principle basing on the fact of DNA.

Evidence from viruses:-

Experiment on bacteriophage which attacks the bacterium, it concluded that DNA physical and not the protein which is the hereditary materials.

DNA REPLICATION

DNA replication is a process whereby the exact copies of DNA (replicable) are produced by the old DNA molecules.

Significance of DNA replication:-

(i) Since it occurs prior to the nuclear division, DNA replication ensures that all newly formed cells have the same amount of DNA.

(ii) It ensures sameness and constancy of hereditary materials of the cells.

(iii) Occasional mistakes during DNA replication, results into genetic variations hence evolution.

(iv) If evidence mistake attracts uracil instead of thymine. RNA is constructed not DNA. This occurs when the enzyme fails to recognize the methyl group of uracil.

Mechanism of DNA replication:-

The two strands of a DNA unwind and separating thus acting as temperature to which a complementary set of nucleotides would attach by base pairing. In this way each original molecule of DNA give rise

to two copies with identical structures. In the presence of ATP an enzyme DNA polymerase links free DNA to form complementary bases.The unwinding of DNA, double helix is controlled by the enzyme helicase. DNA polymerase then move along the strand resulting formation of complementary bases and hence a free nucleotide and finally extending new stand of DNA. As the enzymes continue to move

along one base at a time, the new DNA strand grows. This is called continuous replication in which one strand is copied before another strand.

The formation (copying) of another strand involves movement of DNA polymerase away from unwinding enzyme. This results in the small gaps being left at some points along the newly constructed DNA stand. These gaps are then sealed by an enzymes DNA ligase. This is called Discontinuous replication.

Semi – conservative replication

In this method of replication, each newly formed double helix retains (conserves) of the two strands of the original DNA double helix.
That is in each of the newly constructed DNA molecules, there is an old and new strand.

Illustration:-

A representative portion of DNA, which is about to undergo replication is shown.
DNA polymerase causes the two strands of the DNA to separate.
The DNA polymerase completes the splitting of the strand. Meanwhile free nucleotides are attracted to their complementary bases.
Once the nucleotides are lined up joined together. The remaining unwinded bases continue to attract these complementary nucleotides.
Finally the nucleotides are joined to form a complete polynudeotide chain. In this way two identical strands of DNA are formed. As each strand retains half of the original DNA material, this method of replication is called Semi – conservative method.

The tree theories of DNA replication illustrated

Differences between DNA and RNA:-

	DNA	RNA
	Double stranded polynucleotide molecule	– Single stranded polynucleotide molecule.
	The pentose sugar is deoxyribose	– The pentose sugar is ribose
	The pyramidine base is Thymine	– The pyramidine base is uracil.
	It is found in molecules	– It is found in the cytoplasm
	It is constant in the cell	– The amount of RNA is variable
	It is more stable	– It is less stable.
	It has high molecular mass	– It has low molecular mass.
	The ratio of ‘A’ to ‘T’ and ‘G’ to ‘C’ is always 1.	– The ratio of ‘A’ to ‘U’ and ‘G’ to ‘C’ is variable
	Only one basic form, but with an infinity variety within that form.	– Three basic forms, messenger, transferred and ribosomal RNA
	Treatment	Stay exist temporary for short period

Study Questions:-

1. (a) What is DNA replication?

(b) Describe the mechanism of the process by which a DNA molecule is produced and explain why this is called a semi-conservative process

2. Summarize the structure differences between DNA and RNA

The nature of genes:-

What are genes?

Mendel defined gene as a unit of inheritance. This is an acceptable definition of gene but it does not tell us anything about the physical nature of gene.

Below are ways of overcoming this objection.

(i) A unit of recombination

It was shown that a gene was the shortest segment of a chromosome which is separated from adjacent segments by crossing over.

This definition regards gene as the specific region of chromosome determining a district chromosome in the organism.

(ii) A unit of function

It is known that genes are codes for proteins;

Therefore a gene is the DNA code for polypeptide.

Since some proteins are made up of more than one polypeptide chain and are coded by more than one gene.

The genetic code

The genetic code is the relationship between nitrogenous bases on the DNA and the acids.

It was suggested that the genetic information which passed from generation to and which controlled the activities of the cell, might be stores in the sequence for the production of protein molecules it become clear that these sequence of in the DNA must be a code for the sequence of amino acids in protein molecules relationship between bases and amino acids is known as the genetic code.

In other words the genetic code is a means by which the genetic information.

DNA controls the manufacture of specific proteins, by the cells.

The problems remained were to demonstrate that a base code consisted to break the code and to determine how the code is translated in to the amino acid sequence of a protein molecule.

The code is triplet code.

There are four bases in the DNA molecules, Adenine (A), Guanine (G), Thymine (T) and Cytosin.

Each base is a part of nucleotide and the nucleotides are arranged as a polynucleotide chain (strand). The sequence of base indicated by their first letters (alphabets) are responsible for carrying the code that

results in the synthesis of potentially infinite number of different protein molecules.

There are 20 common amino acids used to make protein and that the base in the DNA must code for. If one base determined the position of a single amino acid in the primary structure of a protein, the

protein could have four different amino acids. If a combination of base pairs coded for each amino acid then 16 acids could be specified into the protein molecule.

Only a code composed of three bases could incorporate all 20 amino acids into the structure of protein molecules.

It was therefore proved that the code is indeed a triplet code, meaning that three bases is the code for one amino acid.

Problems.

1. Using different pairs of the bases A, G, T and C list the 16 possible combinations of bases that can be produced.

Answer:-

Base	A	G	T	C
A	AA	AG	AT	AC
G	GA	GG	GT	GC
T	TA	TG	TT	TC
C	CA	CG	CT	CC

2. If four bases used singly would code for four amino acids, pairs of bases code for the 16 amino acids and triplets of bases code for 64 amino acids, deduce a material to expression to explain this.

Answer:-

4 bases used once = 4 x 1 = 4

4 bases used twice = 4 x 4 = 4² = 16

4 bases used thrice = 4 x 4 x = 64

The mathematical expression is X^y

Where: X = Number of bases and

Y = Number of bases used.

– It is thus a combination of three nitrogenous bases a three lettered ward of AGC, AUA, GCA etc.

Features (Characteristics) of the genetic code:-

1. It is a triplet of bases in the polynucleotide chain codes for an amino in the polypeptide chain.

2. The genetic code is degenerate i.e A given amino acid can be coded for by more to one code and (Codons-complementary triplets in the mRNA).

Example:

3. The genetic code is universal i.e. the same triplet codes for the same amino acids all organisms.

4. The genetic code can be punctuated i.e. It has got the ‘start’ and ‘end’ signals.

5. The genetic code is non-over lapping. E.g. If the base sequence is ACAGAGUCGGAC, then this will be read as ACA/GAG/UCG/GAC and not ACA / CAG / AGA.

6. The genetic code sequence has got no camma e.g. AAU, GCG, GAC, etc. This is because the bases are continuously sequenced on the DNA or RNA strand.

Note: The type of code where the number of amino acids is less than the number of codons is termed as degenerate.

Nonsense codons – These codons do not code for amino acids, they pregimably mark the end point of 2 chains. They act as stop signals for the termination of polypeptide chains during translation.

PROTEIN BIOSYNTHESIS. ‘DNA makes RNA and RNA
makes Protein’

Protein synthesis is a mechanism by which protein molecule is constructed by joining the amino acids with the peptide bonds according to the instruction in the mRNA coded from DNA.

1. Synthesis of amino acids.

2. Transcription (Formation of mRNA).

3. Amino acid activation.

4. Translation.

The site for protein synthesis is the ribosome.

These protein synthesized may have structural role such as Keratin and collagen, or a functional role such as insulin, fibrinogen and mostly important enzymes which are responsible for controlling all metabolism. It is the particular range of enzymes that determines what type of cell it becomes. This is the way in which DNA controls the activities of a cell.

The instructions and information for the manufactures of enzymes and all other proteins are located in the DNA. However, the actual synthesis of protein occurs in the ribosomes in the cytoplasm. Therefore a mechanism had to exist for carrying the genetic information’s from the nucleus to the cytoplasm. This link was from messenger RNA.

Adaptations of the ribosome to protein synthesis

1. Presence of appropriate enzymes that catalyze the synthesis of polypeptide bonds between the amino acids.

2. Presence of receptor site for messenger RNA attachment.

3. Presence of rRNA for attracting other types of tRNA towards the ribosome.

4. Ability to read and ‘translate’ the message contained in the codes of mRNA.

Mechanism of protein synthesis:-

There are four main stages in the synthesis of protein:-

1. Synthesis of amino acids:-

In plants, the formation of amino acids occurs in mitochondria and chloroplast in a series of stages:

(a) Absorption of nitrates from the soil.

(b) Reduction of those nitrates to the amino group (NH₂).

(d) Transfer of the amino group from one carbohydrate skeleton to another by a process called transamination.

Animals usually obtain their acids from the food they ingest, although they have capacity to synthesize their own non- essential amino acids.

2. Transcription (formation of mRNA).

This is a mechanism by which the base sequence of a section of DNA representing gene, is converted into a complementary base sequence of mRNA.
In this process a complementary mRNA copy is made from a specific region of the molecule which codes for a polypeptide.

Mechanism of transcription

A specific region of the DNA molecule, called Cistron, unwinds. This unwinding is a result of hydrogen bonds between base pairs in the DNA double helix being broken. This exposes the bases along each strand and one of these strands is selected as a template against which mRNA is constructed.

This mRNA molecule is formed by linking free nucleotides under the influence of RNA polymerase and according to the rules of base pairing between DNA and R

Table to show the RNA bases which are complementary to those of DNA:-

When the mRNA molecule has been synthesized they leave the nucleus via the nuclear pore and carry the genetic code to the ribosomes. Along the mRNA is sequence of triplet codes which have been determined by the DNA. Each triple called a codon.

When sufficient numbers of mRNA molecules have been formed from the gene the RNA polymerase molecule leave the DNA and the two strands ‘Zip up” reforming the double helix.

Illustration:-

3. Amino acid activation

Activation is the process by which amino acids combine with tRNA using energy from ATP. Each type of tRNA binds with the specific amino acid which means there must be at least 20 types of tRNA.

Each type differs among other things in the composition of a triplet of bases called terminates in the CCA. It is to the free end that the individual amino is not known. The tRNA molecules with attached amino acids form an amino acid tRNA complex known as aminoacyl-tRNA and their formation is under the enzyme aminoacyl-tRNA synthetase. The combination now moves towards the ribosome.

Illustration:-

4. Translation.

Translation is the mechanism by which the sequence of bases in an mRNA molecule converted into a sequence of amino acids in a polypeptide chain.

It occurs on ribosomes.
Several ribosomes may become attached to a molecule of mRNA like bodies on string end a whole structure is known as polyribosome or polysome.
The advantage of such an arrangement is that it allows several polypeptides to be synthesized at the same time.
The first two mRNA codons (a total of 6 bases) enter the ribosome. The first codon binds the aminoacyl–tRNA molecule having the complemetary ‘anticodon’ and which is carrying the first amino acid (Usually – methionine) of the polypeptide being synthesized.
The second codon then also attracts an amino acyl-tRNA molecule showing the complementary anticodon.
The function of the ribosome is to hold in position the mRNA, tRNA and the association enzymes controlling the process until a peptide bond form between the adjacent amino acids.

Once the new amino acid has been added to the growing polypeptide chain, the ribosome moves one codon along the mRNA. The tRNA molecule which was previously attached to the polypeplide chain now leaves the ribosome and passes back to the cytoplasm to be reconverted into a new aminoacyl – tRNA molecule.

This sequence of ribosome ‘reading’ and ‘translating’ the mRNA code continues until it comes to a codon signaling ‘stop’. These terminating codons are UAA,UAG, and UGA. At this point the polypeptide chain, now with its primary structure as determined by DNA, leaves the ribosome and translation is complete. The main steps involved in translation may be summarized under the following headings;-

Binding of mRNA to ribosome.
Amino acid activation and attachment to tRNA.
Polypeptide chain initiation.
Chain elongation.
Chain termination.
Fate of mRNA.

The polypeptides so formed must now be assembled into proteins. This may involve the spiralling of the polypeptides to give a secondary structure, its folding to give a tertiary structure and its combination with other polypeptides and or prosthetic group to give a quatenary structure.

If the ribosome is attached to ER (rough ER) the protein enters the ER to be transported.

Question.

(a) Describe how a single stand of mRNA is being constructed from one of the strands of DNA.

(b) If the base sequence on the portion of DNA strand is AGTCCACCATAA,

(i) What is the base sequence on the portion of mRNA constructed by this portion?

(ii) How many amino acid molecules are there in the base sequence given above?

SOLN

Thus the base sequence on the mRNA will be UCAGGUGGUAAU

Since there are four triplets each responding a single amino acid, then there will be four amino acids.

Introns and exons.

It was discovred that the DNA of eukaryotic gene is longer than its corresponding mRNA. It should be the same length because the messenger RNA is a direct copy discovered that immediately after the mRNA is made, certain sections of the molecule out before it is used in transaction. The sections of the gene that code for the unused pieces of RNA are called Introns. The remaining sections of the gene the code for the protein and are called exons.

Summary:

Eukaryotic genes contain regions called Introns which do not code for the amino. The parts of the genes that code for amino acids are called exons.

MENDELIAN GENETICS

Gregor Johan Mendel did studies of genetics using the Pisum sativan (garden peaces).

He was trying to find the laws that govern the passage of characters from one generation to another.

He established that Pisum sativum had the following advantages over other species:-

1. They were several varieties available which had quite district characteristics.

2. The plants were easy to cultivate

3. The reproductive structures were enclosed by the petals, this made the plant self pollinating and hence producing varieties of the some characteristics (pure breading).

4. Artificial cross – breeding between varieties was possible and resulting hybrids were confertile.

Mono hybrid inheritance and the principle of segregation:-

Monohybrid inheritance is a pattern of inheritance which involves two contrasting variations of only one characteristic.

Example:

Tall Vs short (height).

Red Vs White (colour).

Rough Vs Smooth (texture).

Glossary of common genetic terms:-

1. Gene -The basic unit of inheritance for a given characteristic.

2. Allele – One of number of alternative forms of the same gene responsible for determining contrasting characteristics e.g. A or a (pared genes).

3. Locus – Position of an allele within a DNA molecule. Alleles of one gene are on one locus.

4. Homozygous – The diploid condition in which the alleles at a given locus are identical e.g. AA or aa.

5. Heterozygous – The diploid condition in which the alleles at a given locus are different e.g. Aa.

6. Phenotype – The observable characteristics of an individual usually resulting from the interaction between the genotype and the environment in which development occurs e.g. Red, blue.

7. Genotype – The genetic constitution of an organism with respect to the allele under consideration e.g. AA2, A2, or did.

8. Dominant – The allele which influence the appearance of the phenotype even in the presence of an alternative allele e.g. A

9. Recessive – The allele which influence the appearance of the phonotype only be or in the presence of another identical allele e.g. a

10. F1 generation – The generation produced by crossing homozygous parents.

11. F2 generation – The generation produced by crossing two F1 organisms.

Basic Monohybrid ratio

This is the phenotypic ratio contained in the F2 generation of the original pure parents.

The ratio is always 3:1

Mendel’s experiment and the Monohybrid ratio

In one of his experiments, Mendel crossed a red flowered plant from a pure line with a white flowered plant also from a pure line. Al the resulting F1 plants had red flowers.
When the F1 members were selfed, the resulting F2 were a mixture of red and white phenotypes in the approximate ratio of 3:4.
This is the basic monohybrid ratio obtained from a cross between two heterozygous individuals.

Illustration:-

Non coding DNA.

Though human DNA contains large number of genes, the problem is about 95% of the DNA appears to have no obvious function because it is non – coding. In other words does not code for proteins or RNA.

(i) The factor for redness was dominant over that for whiteness which was red.

(ii) The factor for whiteness was present in the F1 though not expressed effect was obscured by the factor for redness.

(iii) The characteristic red and the characteristic white remained unchanged. I.e.: There was no an intermediate colour.

(iv) Each characteristic is controlled by a pair of factors that segregate during gamete formation.

This observation, suggested to Mendel the formulation of his first law “the law of segregation.”

Assumptions:–

(i) Let ‘R’ be factor redness and ‘r’ factor for whiteness.

(ii) Let ‘R’ dominate ‘r’ so that when the two are together, only R is expressed.

(iii)Let each character be controlled by a pair of factors that segregate gametes formation.

Consider the following cross:-

Parental phenotypes Pure breeding Red flower x Pure bleeding White flower

Phenotypic ratio 3Red : 1White

Mendel’s 1^st law of inheritance (Law of segregation)

The law states that:-

“The characteristics of an organism are determined by internal factors which occur in pairs. Only one of a pair of such factors can be represented in a single gamete.

Meiotic explanation of Mendel’s first law.

Although Mendel knew nothing about Meiosis, but his first law is explained by Meiosis as follows:-
During Meiosis, the paired homologous chromosomes, separate from each other as a result of which the gametes receive only one type of chromosome instead of the normal two.
Alleles also occur in pairs at the homologous chromosomes, thus separation of homologous chromosomes occurs currently with the separation of alleles.
Thus, there is similarity between separation of homologues chromosomes in Meiosis and segregation of Mendelian factors.

We know that Mendel’s factors are specific portion of a chromosome called genes. We also know that the process which produces gametes with only one of each pairs of factors is Meiosis. On the basis of his results, Mendel had effectively predicted the existence of genes and Meiosis.

Methods used to solve Mendelian problems:-

(a) Algebraic method.

(b) Punnet square/chequer board method.

(A) Algebraic method

– Consider a cross between two tall plants both heterozygous for height.

Symbols used in genetics:-

In genetics any symbol can be used to represent any characteristics provided it is defined.
However, it is common that a dominant characteristic is represented by the first latter of its name. Eg. R for red, T for tall, G for green etc. The characteristics will take the lower case letter of the dominant one e.g r white where R red is dominant to white. The symbol P1 stands for parents and F1 and F2 are filial generations 1 and 2respectively

Example: One of the causes of dwafirsm in man is the inheritance of dominant gene D. The allele for a normal height is d, Given that the genotype for Kijeba a man suffering from dwafirsm is Dd, work out the genotype and phenotype rations of the offspring if he marries.

(a) A normal woman

(b) A dwarf woman

Solution:-

Given: D – allele for dwarfness

d – allele for tallness

(b) If he marries, the genotypes and phenotypes of there child will depend on the genotype of the woman.

If she is homozygous dwarf, then the products will be.

P. phenotype: Dwarf x Dwarf

Genotype: D d D d

The genotype ratio will be 1 DD: 1 Dd

If she is heterozygous dwarf, then the products will be.

Genotype ratio is 1DD : 2Dd: 1dd

Phenotype ratio is 3 Dwarf : 1 tall

BACK CROSS AND TEST CROSS

Back cross – This is a cross between an organism and either of its parents.
Test cross – This is a cross between an experimental organism with a dominant phenotype and that of a recessive phenotype, of its parent so as to determine the genotype of that experimental organism.

Explanations:-

One common genetic problem is that an organism which shows a dominant character has two possible genotypes.

Example

A plant producing seeds with round coats could either be homozygous dominant (RR) heterozygous (Rr). The appearance of the seeds (phenotype) is identical in both cases. However it is often necessary to determine the genotype accurately.

This involves the use of a technique known as Test mass in which an organisms is unknown genotype is crossed with the one whose genotype is accurately known.

A genotype which can positively be identified from its phenotype alone is one which shows recessive features.

In the case of the seed coast, any pea seed with a linked coast must have the genotype “rr”. By crossing the dominant character, the unknown genotype can be identified.

Let R = allele for round seeds

R = allele for wrinkled seeds

If the plant producing round seeds have the genotype RR.

Conclusion: The only possible offspring are plants which produce round seeds, thus the unknown genotype is RR.

If the plant producing round seeds have the genotype Rr.

This 1:1 ratio is the monohybrid test cross ratio obtained from a cross investigation between heterozygous dominant and a homozygous recessive.

Questions

1. If a pure strain of mice with brown-coloured fur are allowed to breed with a pure of mice with grey-coloured fur, they produce offspring with brown-coloured fur. If F1 mice are allowed to interbreed

they produce an F2 generation with fur coloured in proportional of three brown-coloured to one grey.

Explain their result fully.

What would be the results of meeting a brown – coloured heterozygote from the generation with the original grey – coloured parent?

Answer:-

Let: B represents brown fur (dominant)

b represents grey fur (recessive)

F1 phenotype. All brown fur

NON – MONOHYBRID INHERISTANCE

This is a pattern of inheritance which involves more than one character. These may be two three etc.

Dihybrid inheritance and Mendel’s Law of Independent assortment.

Dihybrid inheritance is the pattern of inheritance which involves inheritance of two characters simultaneously.

In one of his experiments Mandel investigated the inheritance of the seed shape (size Vs Wrinkled) and seed colour (Yellow Vs green) at the same time. He knew from the monohybrid crosses that the round seeds were dominant to wrinkled ones and yellow seeds were dominant to green. He chose to cross plants with both dominant seed (round and yellow) with one that were recessive for both (Wrinkled and green).

The F1 generation yield plants all of which produced round, yellow seeds – hard surprising as these are two dominant features.

F1 seeds were planted and then allowed to self pollinate. The resulting members were a mixture of phenotype in the following proportions:

315 Round yellow (Two dominant features).

701 Wrinkled yellow (recessive and Dominant).

108 Round green (Dominant and recessive).

32 Wrinkled green (Two recessive features).

Those numbers represent an appropriate ratio of 9:3:3:1. This is the basic dihybrid ratio.
In a dihybrid cross, characteristic behaves independently of the other i.e Each characteristics behaves as if it is in the monohybrid cross.

Now, treating each characteristic separately we have:-

(a) Considering seed texture (Ignore colour)

Round Winkled

315 + 108 101 + 32

423 133

133 133

3 : 1

(b) Considering colour (Ignore seed texture)

Yellow Green

315 + 101 108 + 32

416 140

140 140

= 3 : 1

Thus, in the F2 generation of a dihybrid cross each characteristics, has a phenotype ratio of 3:1
The binomial of the two ratios renders.

(3:1) (3:1) = 9:3:3:1

Thus, the dihybrid ratio is a binomial expression of two bases monohybrid rations.

Genetic representation of the dihybrid cross:-

Let R = allele for round seed

r = allele for wrinkled seed

G = allele for yellow seed

g = allele for green seed.

Parents Phenotype: round yellow seed Vs wrinkled green seed

Genotype: RRGG rrgg

Punnet square to show the fusion of gametes:-

Gamets	RG	Rg	rG	rg
RG	RRGG	RRGg	RrGG	RrGg
Rg	RRGg	RRgg	RrGg	Rrgg
rG	RrGG	RrGg	rrGG	rrGg
rg	RrGg	Rrgg	rrGg	rrgg

rrGg = Wrinkled, Yellow seed – 2

rrgg = Wrinkled, green seed – 1

rrGG = Wrinkled, Yellow seed – 1

Hence the ratio 9:3:3:1

How to calculate the genotype and phenotype ratio of a dihybrid cross.

There are two alternative ways:-

(a) By counting the number of boxes on the punnet square containing the genotype and phenotype of interest.

(b) Using a method based on the probability principle that:-

“The chances that a number of independent events will occur together, is square to the product of the chances that each event occur separately.”

From above example, there is a1 in 4 chance of any gamete containing any of the F2 allele combination shown above.

From a consideration of monohybrid inheritance where ¾ of the F2 phenotypes show the dominant allele and ¼ the recessive allele, the probability of the four alleles appearing in any F2 phenotype as

follows:

Round (dominant) ¾.

Yellow (dominant) ¾.

Wrinkled (Recessive) ¼.

Green (Recessive) ¼.

Hence the probability of the following combinations of alleles appearing in the F2 phenotypes is as follows:-

Round and Yellow = ¾ x ¾ = 9/16.

Round and green = ¾ x ¼ = 3/16.

Round and yellow = ¼ x ¾ = 3/ 16.

Wrinkled and green = ¼ x ¼ = 1/16.

Mendel’s 2nd law of Inheritance (Low of Independent assortment)

In the dihybrid inheritance, Mendel realized that during gametes formation in each sex either one or another pair of factors may enter the same gametes cell (random combination) with either one or another

cell. The law states that:-“Any one of a pair of characteristics may combine with either one of another pair”

Meiotic explanation of Mendel’s second low

Mendel’s second law is explained by Meiosis as follows:-

During gametes formation, the distribution of each allele from a homologous chromosome pair, is entirely independent of the distribution of alleles of another pair. It is the random alignment of the homologous chromosomes on the equator spindle in “Metaphase 1” and their subsequent separation in “Anaphase I” that leads to a variety of alleles in the gametes.

Examples

1. In the guinea pig (cavia), there are two alleles for hair colour, black and white, and two alleles for hair length short and long. In a breeding experiment the F1 phenotypes produced from a cross between pure – breeding short black haired and pure – breeding, long white – haired parents had short black hair. Explain;

(a) Which alleles are dominant, and

(b) The expected proportions of F2 phenotypes.

Answer

(a) If short black hair appeared in the F1 phenotypes, then short hair must be dominant to long hair and black hair must be dominant to white.

(b) Let B represent black hair

b represent white hair

S represent short hair

s represent long hair.

F1 phenotypes Short black hair x short black hair

F1 genotypes (2n) SbBb SsBb

gametes SB Sb sB sb.

Gametes	SB	Sb	sB	sb
SB	SSBB	SSBb	SsBB	SsBb
Sb	SSBb	SSbb	SsBb	Ssbb
sB	SsBB	SsBb	ssBB	ssBb
sb	SsBb	Ssbb	ssBb	ssbb

9 short black hair : 3 short white hair : 3 long black hair : 1 long white hair

2. Flower colour in sweet pea plants is determined by two allelomorphic pairs of gene (R,r and S,s). If at least one dominant gene from each allelomoorphic is present in the flowers are purple. All other

genotypes are white. If two purple plants, each having the genotype RrSs, are crossed, what will be the phenotypic ratio of the offspring?

Parental phenotype: Purple x Purple

Parental genotype: RrSs x RrSs

gametes RS Rs rS rs

Gamete	RS	Rs	rS	rs
RS	RRSS	RRSs	RrSS	RrSs
Rs	RRSs	RRss	RrSs	Rrss
Rs	RrSS	RrSs	rrSs	rrSs
Rs	RrSs	Rrss	rrSs	rrss

Offspring phenotype: 9 purple : 7 white

3. Consider a pea plant with round yellow seeds of the genotype Rr Yy. This means there are two pairs of homologous chromosomes. One pair carrying the allele for the colour and another pair carrying the allele for the seed form (texture). Thus chromosomes carrying the alleles for seed colour are homologous with another as those for seed form.

– At Meiosis, the homologous chromosomes come together (assort), but they carry themselves on the spindle independently of each other. They may arrange themselves in one of the following way

RrYy: Ry, RY, rY, ry

Question:-

(i) State Mendel’s laws of inheritance.

(ii) State the observations made by Mendel that led him to formulate his laws of inheritance.

(iii) Discuss in fully as you can how the behaviour and movement of chromosomes during meiosis, explain Mendel’s laws of inheritance.

NECTA 1973

In guinea pig, rough coat is dominant over smooth coat and black coat is dominant over white coat. When a rough black guinea pig was crossed with a rough white guinea pig the offspring obtained were.

328 rough black

311 rough black

111 smooth black

110 smooth white

What were the genotypes of the parents?

Soln

Let – R- rough coat

r- smooth coat

B- black coat

b- white coat

– In the dihybrid cross, each character behaves independently of the other. Thus considering coat texture we have:-

Rough Smooth

328 + 311 110 + 111

639 221

221 221

= 3 : 1

This is a basic monohybrid ratio obtained from a cross involving two heterozygous individuals. Thus, the genotype of the rough coat with respect to this gene was Rr.

Considering coat colour.

Black White

328 + 111 311 + 110

439 421

421 421

= 1 : 1

This (1:1) is a monohybrid test cross ratio obtained from a cross of a heterozygous dominant and a homozygous recessive.
Therefore the genotype of a black coat was Bb and that of a white coat
Therefore, the genotypes of the parents were:-

Rough black : RrBb

Rough white : Rrbb

5. A tall plant with red flowers, form a true breeding line was crossed with a short plant with white flowers. One of the resulting plants was crossed in short red flowered plant unknown parentage.

This cross gave the following results:-

109 – short white

38 – tall red

29 – tall white

100 – short red

(i) Interprete the results.

(ii) What was the phenotype of the plants produced by cross I?.

Solution

According to Mendel’s 2^nd law, in a dihybrid cross, each characteristic behaves independently of the other.

– Thus, treating each characteristic separately we have:

Short Tall

109 + 100 38 + 29

209 67

67 67

= 3 : 1

– This is a basic monohybrid ratio obtained from a cross between two heterozygous plants

– From this ratio, short is dominant over tall.

Colour

White Red

109 + 29 38 + 100

138 188

138 138

1 : 1

– This is a monohybrid test cross ratio obtained when a homozygous recessive is crossed with a heterozygous dominant.

Consider the two crosses for colour only

Heterozygous dominant x Red (Unknown parentage)

1 White : 1 red

From the above cross, the red colour was recessive to white.

Defn: of symbols

Let: W = White

w = red

S = short

s = tall

Punett square to show the fusion of gametes.

Gametes	SW	Sw	sW	Sw
Sw	SSWw	SSww	SsWw	Ssww
sw	SsWw	SSww	ssWw	ssww

The phenotypes are:-

3 short white

3 short red

1 Tall white

1 Tall red

(ii) From cross 1 above, the phenotypes of the product was short white.

6. Two form IV students Sophia and Issa were eager to put into practice their genetic knowledge. They carried out the following crosses:-

CROSS I

A pure breed plant for terminal purple flowers was crossed with a home plant for axial white flowers.

CROSS II

A plant with axial purple flowers of unknown percentage was crossed with one of the products of the first cross. This cross produced the following results.

338 axial white flowers.

109 terminal purple flowers.

84 terminal white flowers.

304 axial purple flowers.

– Due to their elementary knowledge in genetics, Sophia and Issa failed to interprete their results.

– Using your advanced biology knowledge, show how Issa and Sophia could;

(i) Interprete their results

(ii) Identify the genotypes and phenotypes of the plants produced in the first cross.

Solution;-

(i) According to Mendel’s second law, each characteristic in a dihybrid cross behaves independently of the other. Thus, treating each characteristic separately we have.

– Considering position of the flowers, we have:-

Axial Terminal

338 + 304 109 + 34

193

= 3 : 1

This is a basic monohybrid ratio obtained given a cross between two heterozygous individuals.

From this ratio, axial flowers are dominant over terminal flowers.

Considering colour of the flowers:-

Purple White

304 + 109 338 + 84

= 1 : 1

This is a monohybrid test cross ratio obtained from a cross between a heterozygous dominant and homozygous recessive.

Considering the two crosses for flower colours only.

TOPIC 4: GENETICS | BIOLOGY FORM 6

Since this ratio is obtained when a heterozygous dominant is crossed with homozygous recessive, then purple was recessive and white was dominant.

Definition of symbols:-

Let:- A – axial

a – terminal

W – white

w – purple

Punett square to show the fusion of gametes:-

Gametes	AW	Aw	aW	aw
Aw	AAWw	AAww	AaWw	Aaww
Aw	AaWw	Aaww	aaWw	aaww

The phenotypes are in the following proportions-

3 Axial white

3 Axial purple

1 Terminal white

1 Terminal purple

The results Issa and Sophia have been interpreted since the ratio obtained corresponds with the figures given.

From Gross 1 above, the genotype and phenotype of the products of the flowers cross are AaWw and axial white respectively.

MERITS AND DEMERITIS OF MENDEL

MERITS:-

Mendel was successful in his work where others had failed.

He was very systematic and scientific in his researches and data analysis and for this reason he managed to come out with the laws of inheritance.

He realized the role of gametes in the transfer of genetic information from parents to the offspring.

The secret behind Mendel’s success is within the following facts:-

Preliminary investigations were carried out to obtain familiarity with experimental organisms.

He paid attention to one characteristic at a time.

He used organisms with limited continuous variations

Meticulous care was taken during data collection and analysis so as to avoid introduction of contaminating variables.

He collected sufficient data to have statistical significance

DEMERITS:

The shortfalls of Mendel include the following:-

His gametes describe only the diploid sexually reproducing organism. The haploid organisms such as Bryophylum are not explained.

(b) His gametes is only based on the dominating- recessiveness principle’s but not all the time that one characteristic is dominant over the other.

(d) Mendel did not consider gene interaction such as epistesis collaboration, lethal genes etc all of which interfere with his basic ratio.

More examples:

The position of starch in pollen grains in maize is controlled by the presence of one allele of certain gene. The other allele of that gene results in starch being deposited. Explain in terms of reasons why half the pollen grains produced by a heterozygous plant contain starch.

Solution:

The two alleles segregate during metaphase I and anaphase I.

Calculate the number of different combination of chromosomes in the pollen grains of the cross (cross balance) which has a diploid number of six (2n = 6).

Solution

The number of different combination of chromosomes in the pollen grants cell is calculated, using the formula 2n, whore n is the haploid number of chromosomes.

Since 2n = 6, n = 3

Therefore, combination = 23 = 8

NON – MENDELIAN GENETICS

This is simply a pattern of inheritance in which the basic Mendelian ratios are modified.

Examples:

INCOMPLETE DOMINANCE (under gene interactions)

Incomplete dominance/ Blending – Is a type of inheritance in which there occur the apparent failure of one allelic gene to dominate the other that when the two genes are together, they produce a character between them.

Example

A cross between a white Andalusian (fowl) and a black Andalusian produce a blue variety in the F1 generation. When the F1 members are selfed, the F2 individuals are a mixture of phenotypes ie: black, blue and white in the ratio of 1:2:1

Illustration:

Genotype ratio: 1 BB : 2 BW : 1 WW

Phenotype ratio: 1 black : 2 blue : 1 white

QN. NECTA 1993

A genetist who was verifying Mendel’s first law and second law crossed 45 homozygous red flowered plants with 45 homozygous white flowered plants. The resulting F1 were 530 plants all with pink flowered plants, the seeds obtained were planted and F2 offspring with the following phenotypes were obtained.

1292 red flowered.

2570 pink flowered.

1290 white flowered.

(a) Illustrate using symbols the crosses made and the results obtained in the experiment described above.

(b)(i) What is the name above experiment?

(ii) How do the above observations differ from the results of Mendelian work which led him to formulate his laws of inheritance?

(c) Describe the genetical test you would carry out to prove weather not the appearance of the pink flower in the above experiment is true deviation from Mendel’s principles of inheritance (20marks).

Solution:

(a) Let R – allele for red colour.

W – allele for white colour.

RW – genotype for pink colour (flower).

The F2 members are 1292 red flowered, 2570 Pink flowered, 1290 White flowered

(b) (i) The name of the mode of inheritance is Incomplete dominance inheritance.

(ii) The observation differ from Mendelian principles in that, the inheritance the flower colour does not follow the dominance – recessiveness principle.

All the F1 offsprings are pink flowered instead of them to show the dominant colour from either of the experiment is fully dominance or fully recessive. In the Mendelian experiment, when F1 are selfed

the resulting one in the dominance-recessive phenotypes ratio is of 3:1. But in this experiment, selfing the F1 individuals give 1:2:1.

(c) The genetical test to be carried out is back test cross in which the pink flowers plant will be test crossed with either of the homozygous plants say red flowered plant RR.

The result of the test cross above will give products of red (RR) and pink (RW) in the ratio of 1:1.

The above results show a siltation of RW (Pink) and RR (red). This can prove that there has been no true blending, has occurred in F1 generation then we could expect offspring which were again all pink flowers.
The appearance of pink colour trait in the F1 generation is not a true deviation from Mendel’s laws; otherwise the gene for red could not be reviewed unchanged.
It was just observed when in the presence of WW (White) while its identity was being retained.

Thus, the inheritance of the flower colour in the experiments precisely obeys Mendel’s principle of inheritance only the phenotype ratios are different.

Partial dominance (Co dominance & Incomplete dominance)

Sometimes both alleles express themselves in the phenotype, but one more so than another. This is an intermediate stage between complete dominance and co dominance.

LINKAGE

For just 23 pairs of chromosomes to determine the many thousands of different human characteristics, it follows that each chromosome must possess many different genes.

Any two genes which occur in the same chromosome are said to be linked. All the genes on a single chromosome form a linkage group.

Under normal circumstance, all the linked genes remain together during cell division and so pass into the gamete, and hence the offspring, together. They not therefore segregate in accordance with Mendel’s law of Independent Assortment.

The figure below shows the different gametes produced if a pair of genes A and B are linked rather than on separate chromosomes.

If gene A and B occur on separate chromosomes ie. are not linked

NOTE: Linked genes do not confirm to Mendel’s principle of independent assortment, therefore they fail to produce the expected 9:3:3:1 ratio in a breeding situation involving the inheritance of two pairs

of contrasted characters (dihybrid inheritance).

Crossing over and crossover values (COV)

During cross over (in chiasmata formation), the alleles of parent linked group separate and new associations of alleles are formed in the gamete cells, a process known as genetic recombination.

Offspring formed from these genes showing ‘new’ combinations of characteristics are known as recombinants. Hence crossing over is a source of variations.

The recombination frequency (COV) is calculated using the formula

no. if individual showing recombination x 100

no. of offspring

Gene mapping

Calculation of COV enables geneticists to produce maps showing the relative positions of genes on chromosome. Chromosome maps are constructed by directly converting the COV between genes into

hypothetical distances along the chromosomes.

Sex determination

Sex is a state of being male or female.

In human there are 23 pairs of chromosomes of these 22 pairs are identical in both sexes. The 23^rd pair, however is different in the male from the female.

The 22 identical pairs are called autosomes, the 23^rd pairs are referred to as sex chromosomes or heterosomes.

In females the two sex chromosome are identical (X chromosomes) are said to be Homogametic, while in males the two chromosomes are non- identical (Y – chromosome is smaller in size than X –

chromosome) and are said to be Heterogametic.

Unlike other features of an organization, sex is determined by chromosomes rather than genes.

Humans of genotype XXY are phenotypically male, while genotypes with just one X chromosome (XO) are phenotypically female. This suggests that the presence of the Y chromosome which makes the human male, in its absence the sex is female. How does the Y chromosome determine maleness?

The Y chromosome possesses several copies of a testicular differentiating gene which codes for the production of a substance that causes the undifferentiating gonads to become testes. In the absence of the gene and hence this substance, the gonads develop into ovaries.

In humans, sex is determined by the type of sex chromosome contained in the spermatozoa that fuses with an X – chromosome of the egg cell.
If the X – chromosome bearing sperm fuses with an X- chromosome of the egg cell, the resulting zygote will develop into a female (XX).
If the Y- chromosome bearing sperm fuses with an X – chromosomes of the egg cell, the resulting zygote will develop into a male (XY).

Consider a cross below

From this cross it is evident that;

(i) The chances of a zygote develop into a male or female are 50%

(ii) The sex of the individual before to be born is determined by the father.

Sex determination differs in other organisms. In birds, most reptiles and fish and all butterflies, the male is the homogametic sex (XX) and the female is from the cross, all the resulting individuals are phenotypically normal with all females beings carriers.

Questions.1.A certain species of flies has, the following genetic attributions

i. Female flies have two X- chromosomes (XX).

ii. Male flies have one X- and one Y – chromosome.

iii. Y chromosome does not bear extra genes.

iv. Eye colour is sex linked and red colour is dominant over white eye colour trait.

(b) What will be the genotype and phenotype of the female and males in the F1 and F2 when;

2. Siku and her brother Juma have their elder brother who is haemophilic. They and their parents are normal but they are worries that they may have haemophilic children in future. If they approach you for help how would you advice them?

3. A homozygous purple flowered short-stemmed plant was crossed with a homozygous stemmed-flowered long stemmed plant and the F1 phenotypes had purple flowers and short stems. When the F1

4. generation was test crossed with a double homozygous recessive plant the following progency were produced.

52 Purple flower, short stem.

47 Purple flower, long stem.

49 red flower, short stem.

45 red flower, long stem.

Explain these results fully.

The F1 phenotypes show that purple flower and short stem are dominant and red flower and long stem are recessive. The approximate ratio of 1:1:1:1 in a dihybrid cross suggests that the two genes controlling the characteristics of flower colour and stem length are not linked and the four alleles are situated on different pairs of chromosomes (see below).

Let P- Purple flower

p – Red flower

S – Short stem

s – Long stem

Since the parental sticks were both homozygous for both characteristics the genotypes must be PpSs.

Test cross phenotypes: Purple flower short stem x red flower, long stem

Test cross genotype: PpSs ppss

gametes	PS	Ps	ps
ps	PpSs	Ppss	ppss

Meiosis:

Gametes (n):

Random fertilization

(shown inpunett square):

Offspring genotypes (2n)

(Listed in each square):

Offspring phenotypes: 1 purple flower, short stem : 1 purple flower, long stem.

1 red flower, short stem : 1 red flower long stem.

MULTIPLE ALLELISM

Multiple alleles are those alleles of a single locus when there are more than alternatives in a population.

In humans, the inheritance of the ABO blood groups is determined by a gene which has different alleles. Any two of these can occur at a single locus at one time.

Allele A causes production of antigen A on red blood cells.

Allele B causes production of antigen B on red blood cells.

Allele O causes no production of antigens on red blood cells.

Alleles A and B are codominant and allele O is receive to both.

The transmission of these alleles occurs in normal Mendelian fashion.

A cross between an individual of group AB and one of group O therefore gives rise to individuals non of whom possess either parents blood group.

Phenotype: blood group AB x blood group O

		gametes
		I^A	I^B
gametes	I^O	I^AI^O	I^BI^O

Paternity suits.

Although blood group cannot prove who a father of a child is, it is possible to use inheritance to show that an individual could not possibly be the father.

Imagine a mother who is blood group B having a child of blood group O. She claims the father is man whose blood group is found to be AB. As the child is group O, its only possible genotype is I^OI^O. It

must therefore herited one I^Oallele from each parent. The mother, if I^BI^O could donate such an allele. The man with blood group AB can only have the genotype I^AI^B. He is unable to donate an I^O allele

and cannot therefore be the father.

Dominance series

Coat colour in rabbits is determined by a gene C which has four possible alleles.

Alleles C^F determines full coat colour and is dominant to

Allele C^CH which determines chinchill a coat and is in turn dominant to

Allele C^H which determines Himalayan coat and is in turn dominant to

Allele C^A which determines albino coat colour.

There is therefore a dominance series and each type has a range of possible genotypes.

Inheritance is once again in normal Mendelian fashion.

Other characteristics controlled by multiple alleles are coat colour in mice, eye colour.

Worked examples

1. A woman of blood group A claims that a father of blood group B is the father of her child whose blood group is O. How far are the woman’s claims valid?

Answer.

If both are heterozygous for their blood groups, then the woman’s claims are valid, but if either of them is homozygous for the blood group then women’s claims are invalid.

2. Anna is a woman married to John. This couple once had a child, Kitto was one day discovered that his parents were in bad terms. John claims that Kitto is an illegitimate child but Anna is opposing the case. Blood tests rewards that John is of type A and Kitto is of blood group O. Anna’s mother blood type is B and Anna’s father blood group type is AB. Using this information alone

(i) Suggest the possible genotype for Anna, show how you determine genotype.

(ii) Show dearly whether Kitto is or he is not an illegitimate child of the said family.

Solution:

1. Since Kitto is of blood group O, then his genotype is of no doubt.

This implies that the genotype for Anna should have an allele I^O considering Anna’s parents we have:

Mother: Blood group B whose possible genotypes are I^BI^B or I^B I^O

Father: Blood group AB whose possible genotype is I^A I^B

But for Anna to have an allele I^O in her genotypes, the mother should bear the allele I^O. Thus, the genotypes should be I^BI^B

A cross between Anna’s parent reveals the following;

2. Since Anna is heterozygous for her blood group and John is of blood group A, then the legitimacy of Kitto will depend on John’s genotype. If he is heterozygous for his blood group, then Kitto is a legitimate child as the cross below reveals.

If John is homozygous for his blood group, then for sure, Kitto is an illegitimate child of the family.

3. Write an essay on the statement that “the knowledge on the inheritance of blood group can be used to tell for sure that “the baby is not Yours” rather than the baby is yours”

4. Mama is a form six student with blood group A. She recently has a baby whose father she insisted was her fellow student Kashesha. Kashesha refused paternity and this paternity case was taken to court where the following facts were established. Kashesha’s mother blood group A; Kashesha’s father blood group B. Baby blood group O. Based on this factor explain whether the law will accuse or excuse Kashesha.

Sex Limited and sex influenced characters:

Sex limited characters are those characters that are concerned to only one sex eg. baldness, beards and Adam’s apple in males and enlarged breasts and hips in females.

The development of such character is controlled by sex hormones, they are thus said to be sex influenced characters.

PEDIGREE ANALYSIS

Pedigree is a sequential arrangement of individual in a given family to show the passage of certain character from one generation to another.

In analyzing a pedigree the first individual to show the characters of interest is called a propositous

Features of a pedigree.

(i) Circles represent females, squares represent males.

(ii) Shaded figures show a phenotypic expression of the character, open figures represent a normal phenotype.

(iii) Parents are connected by a horizontal line as children are connected to parents by vertical line.

Worked examples

1. Study the pedigree shown below, circles represent females, squares represent males, shaded figures represent colour blindness, open figures represent normal phenotype

(i) What is the probable genotype for 1?

(ii) What are the possible genotypes for 5 and 9?

(iii) If 8 marries a normal man, what are the chance that she will have a colour blind son?

Solution

Consider the following genetic attribute;

X^NX^N – Normal female.

X^NXⁿ – Carrier female.

XⁿXⁿ – Colour blind female.

X^NY – Normal male.

XⁿY- Colour blind male.

(i) The possible genotype for 1 is X^NX^N ie: homozygous normal.

(ii) The possible genotypes for, 5 is X^NXⁿ ie: heterozygous normal

(iii) Since 8 was born from a certain mother, she has 50% chance of receiving an allele for colour blindness.

There are also 50% chances that the allele will pass to the son.

Thus, the chances of 8 to have a colour blind son are:

½ x ½ x ½ = 1/8

(a) State Mendel’s laws of inheritance.

(b) In dogs coat colour is determined by a series of multiple alleles. The allele A^s produces a uniformly dark coat, the allele a^y produces a tan coat and the allele a^t produces a spotted coat. The dominance heredity achy is A^s> a^y> a^t which means A^s is dominant to both a^y and a^t where a^y is dominant to a^t only.

A family tree for dogs showing their coat colours is given below;

State the genotype of each of the individuals 1 – 5
By means of genetic diagram deduce the possible genotypes and phenotypes of the puppies which could be produced by mating between individual 4 and 6

Solution

(b) (i) Given dominance hierachy A^s> a^y> a^t

Phenotype: Possible genotypes

Dark: A^sA,^sA^sa^y,A^sa^t

Tan: a^ya^y,a^ya^t

Spotted: a^ta^t

Thus, the genotypes are:

Individual	Genotype
1	A^sa^y
2	A^sa^t
3	a^ya^y
4	a^ta^t
5	a^ya^t

Crossing between individual 4 and 6

More examples:

1. In cats, the genes controlling the coat colour are carried on the X chromosome and are codominant. A black coated female mated with a ginger coat produced a litter consisting of black male and tortoise shell female kittens. What is the expected F2 phenotypic ratio? Explain the results.

Soln:

Let B represent black coat colour.

G represent ginger coat colour.

XX represent female cat.

XY represent male coat.

Colour female colour female colour male colour male

(The parental female must be homozygous for black coat colour since this is the only condition to produce a black coat phenotype).

F2 phenotypes: Tortoise shell coat colour female : black coat colour female : ginger coat colour male : black coat colour male

1. (a) Explain using appropriate genetic symbols, the possible blood groups of children whose parents are both heterozygous, the father being blood group A and mother B.

(b) If these parents have non – identical twins, what is the probability that both twins will have blood group A?

Answer

(a)Let: I represent the gene for blood group

A Represent the allele A (equally dominant)

B Represent the allele B

O represent the allele O (receive)

(b) There is a probability of ¼ (25%) that each child will have blood group A. So the probability that both will have blood group A is ¼ x ¼ = 1/16 (6.25%)

GENE INTERACTION

In dihybrid crosses, two or more genes interact to determine a single phenotype. Such an interaction may modify the basic ratios.

Examples of gene interaction are:

(a) Lethal genes.

(b) Epistasis.

(d) Multiple gene interaction.

(e) Complementary genes.

(a) Lethal genes.

A lethal gene is that dominant or recessive gene which when occur in the homozygous state causes death to its bearer. eg : Sickle cell anaemia in humans.

Lethal genes may affect several characteristics including mortality.

Example

Consider the inheritance of fur colour in mice. Wild mice have grey coloured fur, a condition known as agouti. Some mice have yellow fur. Cross breeding yellow mice produces offspring in the ratio 2 yellow fur : 1 agouti fur. These results can only be explained on the basis that yellow is dominant to agouti and that all the yellow coat mice are heterozygous. The typical Mendelin ratio is explained by the fetal death of homozygous yellow coat mice.

Explanation of this rests on the fact that examination of the uteri of pregnant yellow mice from the above crosses revealed dead yellow features. Similar examination of the uteri of crosses between yellow fur and agouti fur mice revealed no dead yellow features. The explanation is that this cross would not provide homozygous yellow (YY) mice.

Let Y represent yellow fur (dominant).

y represent agouti fur (recessive).

NB: the ratio 2:1 talks of lethal. The gene for yellow is dominant for fur colour of the cat, the genotype Yy produce yellow cost but it is for viability. Hence gene YY represents lethal combination.

(b) Epistasis.

Epistasis arise when the allele of one gene suppress or marks the action of another.

Definition;

Epistasis is the type of gene interaction in which one gene (epistatic gene) effect the phenotype expression of the other (hypostatic gene).

An example occurs in mice where three genes determine the coat colour. However the absence of a dominant allele at one of the loci results in no pigment being produced and the coat being albino. This occurs regardless of the genes present at the other loci, even if these produce normal coat colour. The gene at third locus clearly suppresses the action of the others.

Example

In white leghorn fowl, plumage colour is controlled by two sets of genes, including the following: W (white) dominant over w (colour).

B (black) dominant over b (brown).

The heterozygous F1 genotype WwBb is white. Account for this type of gene interaction and show the phenotypic ratio of the F2 generation.

Answer.

Since both dominant alleles W, white and B, black are present in the heterozygous F1 genotype and the phenotype is white, it may be concluded that the alleles show an epistatic interaction where the white allele represents the epistatic gene.

The F2 generation is shown below:-

Using the symbols given in the question.

F1 phenotypes: White cock x White hen

F1 genotypes: WwBb WwBb

Gamete	WB	Wb	wB	wb
WB	WWBB	WWBb	WwBB	WwBb
Wb	WWBb	WWbb	WwBb	Wwbb
Wb	WWBb	WWbb	WwBb	Wwbb
wb	WwBb	Wwbb	wwBb	wwbb

F2 phenotypes: 12 white colour : 3 black colour : 1 brown colour

(Symbols)

Definition: Epistatic/Inhibiting gene are recessive genes which when occur in a genotype it surfers the showing up of another gene.

(c) Collaboration (complex gene/simple Gene Intervention).

Collaboration occurs when two genes controlling the same character, interacts to produce a single character phenotype which neither could produce alone.
An example of each collaboration genes is seen in the inheritance of the shape of the comb in domestic cowl (chicks).
One gene, P – produces a rose comb where as its recessive allele, p produces a single comb.
Another gene R produces a rose comb where as its recessive allele, r also produces a single comb.
When P and R occurs together (codominance), they collaborate to produce a walnut comb.Consider the case below:-

Phenotype	Possible genotype
Pea	PPrr or Pprr
Rose	ppRR or ppRr
Single	pprr
walnut	PPRR, PPRr, PpRR or PpRr

Consider the cross below:-

Selfing PpRr gives 9 Walnut comb : 3 pea comb : 3 rose comb : 1 single comb

(d) Complementary genes.

These are genes which are mutually dependent. Neither of them produces a given phenotype in the absence of the other.

Example

-In sweet peas, purple colour of the flowers is controlled by two genes C and P. In the absence of either the flowers are white one gene (C) regulates the production of raw materials for formation of a purple pigment where as another gene (P) regulates the conversion of raw materials into a purple pigment.

Consider a dihybid cross between a purple flowered plant and a white flowered plant

From the above cross, the F2 phenotype ratio is 9:7 instead of the normal 9:3:3:1. The last three phenotype classes have been combined.

(e) Polygenic inheritance (Multiple gene interaction).

Multiple gene interaction (polygenic inheritance) is a type of gene interaction in which a single character is controlled by a series of genes each exerting its effect on the present phenotype in an additive fashion.

Many genes acting together are referred to as polygenes.

Polygenes give rise to continuous variation.

VARIATION

Variations are differences among the individuals of the same species.

Those variations which can be inherited are determined by genes. These are called genetic or inheritable variations.

Some variations are determined by the individual’s environment and are known as acquired characteristics. Acquired characteristics such as big muscles developed from training and exercise are not inherited. Inheritable variations may be caused by mutation or by new combination of genes in the zygote. Non inheritable variations arise and disappear from a species when the individuals die.

In genetics we are concerned with inheritable variations. Many variations are controlled by genes. There are two types of inheritable phenotypic variations.

(i) Continuous variation and

(ii) Discontinuous variation.

Discontinuous Variation

This occurs when an organism must either have or not have a certain character.

There is no gradual change between the two extreme. This case of variation produces organisms with a clear cut differences between them and with no intermediate between them.

Such characteristics include sex where an individual is male or female, eye colour, blood group, finger prints, tongue rollers, non-tongue rollers.

Characteristics showing discontinuous variation are usually controlled by one major gene which may have two or more allelic forms.

Discontinuous variation cannot be altered by environment. For example you cannot change your blood group by altering your diet.

Continuous Variation

Continuous variation occurs when every member of species shows a certain characteristic but not to the same extent.

Some examples of such characteristics are hand span in humans, length of tail in other animals, number of leaves per plant, body weight and height of the people of the same age. These characteristics vary continuously in the population.

Characteristics which show continuous variation are controlled by the combining effect of a number of genes called polygenes and any character which results from the interaction of many genes is called polygenic characters.

HISTOGRAMS

Characteristic

Environmental influence

One of the reasons for continuous variation is that all phenotypic characters are influenced by the effects of the environment.

Many continuously variable characteristics are affected by environment or by what happens during individual’s life time. For example a genotypically tall organism may be dwarfed by not getting enough food or balanced diet and therefore appear similar to a child whose genotype is for shortness.

No character of any organism can be said to be completely due to effects of heredity (nature), or due to environment (nurture). Environment and heredity always interact in producing the phenotype.

Origins of variation:

Variation may be due to,

(i) Environment effect e.g.: Diseases and Nutritional standards.

For example: The action of sunlight on a light coloured skin may result in it becoming darker. Such changes have little evolutionary significance as they are not passed from one generation to the next.

(ii) Genetic factors.

These are much more important to evolution as they are inherited. These genetic changes may be the result of the normal and frequent reshuffling of genes which occur during sexual reproduction, or as a consequence of mutations.

Reshuffling of genes:

The sexual process in organisms has three inbuilt methods of creating variety.

The mixing of two different parental genotypes where cross-fertilization occurs (Fertilization)
The random distribution of chromosomes during metaphase I of meiosis (Independent assortment).
The crossing over between homologous chromosomes during prophase I of meiosis.

Mutations and deleterious genes

Definition

Mutations are sudden unpredictable changes that occur in the chromosome or genes and they may alter the phenotype expression of an organism.

In other words, mutations are defined as changes in the amount or structure of DNA of an organism as well as arrangement of DNA.

Mutation is a sudden inheritable change of the genotypes.

Significance of mutations

1. Mutations are rare events because DNA and chromosomes are stable structures.

When they occur they provide a source of new variability which is necessary for organisms to adapt to constantly changing physical and biological environment.

2. Mutation is the only means by which new genes arise there by acting as raw material for organic evolution.

NOTE: In diploid types the mutant gene, may be dominant, receive or intermediate to its effect. The most common mutants are recessive.

Only mutations occurring in the gametic cell can be inherited from one generation to another where as those mutations of the somatic cells are not inherited by the dominant cells. These are thus called SOMATIC MUTATIONS.

Question: Why somatic (autosomal) mutations are not inherited?

Causes of mutations:

The substances that cause mutation are called Mutagens or Mutagenic agents. The organism that has undergone mutation is called Mutant.
The mutagenic substances thus include,

Electromagnetic radiations such as X – rays, UV – radiations, É¤- ray etc.
High energy particles such as α – particles, β – particles, cosmic particles.
Chemicals such as caffeine, formaldehyde, some constituents of tobacco, for preservatives and pesticides.
Abrupt temperature changes.

Types of mutations:

Gene mutations

This is a type of mutation in which the DNA structure or chemistry of gene on a single locus is changed.
Since a single locus is affected then this type of mutation is also called Point mutation.
In protein Biosynthesis, we saw that the genetic code, which ultimately determines an organism’s characteristics, is made up of a specific sequence of nucleotide on the DNA molecule. Any change to one or more of those nucleotides will produce the wrong sequence of amino acids in the protein it makes. This protein is often an enzyme, which may have a different molecular shape and be unable to catalyze its reaction. The results will be that the end product of the reaction cannot be formed. This may have an effect to an organism.
Gene mutations are not easily detected by the microscope and they can be passed over several generations without being expressed in the phenotype.

Forms of gene mutation:

Dedication – a portion of a nucleotide obtain becomes repeated.
Addition (insertion) – an extra nucleotide sequence becomes inserted in the chain.
Deletion – a portion of the nucleotide chain is removed from the sequence.
Inversion – a nucleotide sequence becomes separated from the chain.
Substitution – one of the nucleotide is replaced by another which has different organic base.

Example of gene mutations:

1. Albinism

This is a type of gene mutation resulting from base substitution in which the correct base sequence is substituted for by incorrect base sequence.

The result of this substitution is the failure of the enzyme tyrosinase to convert an amino acid tyrosine into melanin. Hence absence of pigment melanin and hence development of a light coloured skin (albinism).

Albinism is caused by a recessive allele (a) whose dominant allele (A) produces normal skin colours pigment.

Consider a cross between two normal individuals producing an albino.

Mutation occurs during meiosis where a chromosome/gene may be Deleted, Duplicated, inverted or substituted, in the presence of mutagens.

2. Sickle cell anaemia

This is a bases substitution type of gene mutation.
It results into development of an individual with abnormal haemoglobin which causes sickling of the red blood cells.
In sickle cell anaemia, the replacement of a base in the DNA molecule results in the wrong amino acid being incorporated into two of the polypeptide chains which make up the haemoglobin molecule. The abnormal haemoglobin makes the red blood cells to become sickle – shaped, resulting in anemia and possible death.
The synthesis of normal haemoglobin in the body is controlled by a pair of gene with the genotype AA (Hb^AHb^A).
The mutated gene known as haemoglobin S (Hb^sHb^s) is recessive and is the one which causes sickle cell anaemia.
An individual with the genotype Hb^AHb^s(heterozygous stickler, a condition known as sickle cell trait) has no effect, rather the genotype gives an advantage such as a genotype produces normal shaped red blood cells: However, they are likely to lose their shape when the tension of O₂ get lowered.
When plasmodium enters such a cell, the tension of O₂in the cell get lowered consequently the cell sickles up.
These mis-shaped cells are cleared from the blood system by the spleen together with the parasites contained in them. In this way the person is said to be resistant to Malaria and the situation is referred to as heterozygous advantage.

Sickle cell anaemia is characterized by the following features;

Sickling of the red blood cells.
Severe and eventually lethal anaemia as the Hb is inefficient at carrying O₂.
Abnormal joint pains.
Enlarged spleen.
Resistance to Malaria for sickle cell trait.
Blocked blood vessels depriving organs of O₂ and permanently damaging.

Consider a cross below:

(b) Chromosome Mutations

These are results of changes in number or gross structure of the chromosomes. Such mutations are called chromosomal abberations.

Changes in whole sets of chromosomes

Sometimes organisms occur that have an additional whole sets of chromosomes.

Instead of having haploid set in the sex cells and a diploid set in the cells, they have several complete sets. This is known as polyploidy.

Where three sets of chromosomes are present, the organism is said to be triploid, with four sets, it is said to be tetraploid (4n).

If gametes are produced which are diploid and these self fertilizer, a tetraploid is produced. If instead the diploid gametes fuses with a normal haploid gamete, a triploid results. Polyploidy can also occur when whole sets of chromosome doubled after fertilization.

Tetraploids have two complete sets of homologous chromosomes which can undergo pairing during gamete production in meiosis. Triploids are thus sterile and hence propagation by asexual means.

Significance of polyploidy:

It is associated with advantageous features eg: Increased size, hardiness and resistance to disease. This is called hybrid vigour.

Most of our domestic plants are polyploids producing large fruits, storage organ flowers or leaves.

Forms of polyploidy

There are two forms namely;

(i) Auto polyploidy.

(ii) Allopolyploidy.

Auto polyploidy

This condition may arise naturally or artificially as a result of an increase in number of chromosomes within the same species.

Allopolyploidy

This condition arises when the chromosome number in a sterile hybrid become doubled and produces fertile hybrids.

F1 organisms are sterile as they cannot form homologous chromosome pairs during meiosis. This is called hybrid sterility. However, the multiples of the original number of chromosomes are fertile.

Changes in chromosome number (Aneuploidy)

Aneuploidy – A condition in which changes may involve the loss or gain of single chromosomes.

In this condition, half the daughter cells produced have an extra chromosome (n + 1),

(2n + 1) etc, whilst the other half have a chromosome missing (n – 1), (2n – 1)

and so on.

Anaeuploidy arise from the failure of a pair of chromosomes to separate during gamete formation. This may lead to formation of gamete cells containing more or few chromosome. This is known as non- disjunction.

Fusion of either of these gametes with a normal haploid gamete produces a zygote with an odd number of chromosomes. They are usually abnormal.

Non disjunction in gamete cell formation

Consequence of non- disjunction in humans (Genetic disorders)

Down’s syndrome (Mongolism) *** sex symptoms on next pages ***

In this case 21^st chromosome fails to segregate and the gamete produced possesses chromosomes. The fusion of this gamete with a normal one with 23 chromosomes results in the offspring having 47 (2n + 1) chromosomes. This leads to a presentation of three copies of chromosomes, a condition known as trisomy, hence down syndrome is also known as trisomy 21.

A Mongol is characterized by the following feature:
Big head, protruding tongue, flat facial features, puffy eyes, mental retardation, sterility and short life expectancy.
Non disjunction in the case of Down’s syndrome appears in the production of ova rather than sperm.
The chances to have a mongol child increases with an increased age of the mother. At the teenage, the chance is one in many thousands at age of 40 – 45, the chances are 35% and above.

3. Kilinefelter’s syndrome (in feminized males)

This is a male genetic disorder in which the victim has got 47 (2n + 1) chromosomes instead of 46.

It is due to an extra X chromosome. The genotype is therefore XXY instead of normal XY. It is like Mongolism, an example of trisomy.
It may occur during spermatogenesis or during oogenesis.

Symptoms:

Infertility – sperm are never produced.
Usually taller than average.
Enlarged breasts.
Enlarged hipbone.
Very small testes.
Low intelligency.
Little facial hair.
Smooth skin texture.
Voice pitched higher than normal.

Treatment – Male hormones can be given. Breast then returns to normal size and the condition is diagnosed only after puberty.

Turner’s syndrome (XO)

This is a female genetic disorder in which there are only 45 chromosomes. Patients can be described as incompletely developed females.
The sex constitution is said to be XO.

Symptoms:

Infertility – ovaries as are absent.
Small uterus.
Shortness of stature.
Broad chest with widely spaced nipples.
Under developed breasts.

Treatment:

From the age of puberty, a woman is given female sex hormones to make female develop breasts & have periods. Though this does not cure infertility.

Explanation of Klinefelter’s syndrome and Turner’s syndrome as a result

(a) Non – disjunction of the father’s sex chromosomes

(b) Non – disjunction of the mother’s sex chromosomes

Changes in the chromosome structure

There are four types:

Deletion – a portion of a chromosome is lost. As it involves the loss of genes, it can have a significant effect on an organism’s development, often proving lethal.
Inversion – a portion of chromosome become deleted, but becomes reattached in an inverted position. The sequence of genes on this portion is reversed and therefore the phonotype is changed although the overall genotype is unchanged.

This indicates that the sequence of genes on the chromosome is important.

Translocation – a portion of chromosome becomes deleted and rejoins at a different point on the same chromosome or with a different chromosome.
Duplication – a portion of chromosome is doubled resulting in repetition of a gene sequence.

Qn Discuss the genetic disorders that may result due to non – disjunction of somatic and sex chromosome during meiosis.

GENETIC ENGINEERING (Recombinant DNA technology)

Definition: Genetic engineering or recombinant DNA technology is defined as the manipulation of DNA of one organism (donor) and its transfer into another organism (the host) where its combines with that of the host organism.

Where it combines with that of the host organism

To create as new gene combination, genetic engineers must be able to

Located a specific gene in the donor cell.
Modify the donor DNA in a highly selective way.
Isolate the located gene.
Transfer the modified DNA into the host cell in such away the gene will be expressed strongly enough to be practical use.

Techniques used to manipulate DNA

The manipulation of DNA involves three techniques each of which uses specific enzyme or group of enzymes.
The molecules to modify DNA (the enzyme used) are;

1. Reverse transcriptase

This catalyses the synthesis of DNA from RNA.

2. Restriction endonucleases

These are used to cut DNA at specific sections.

3. DNA Ligase

This joins the donor and vector DNA section so as to form a recombinant DNA molecule.
The techniques of recombinant DNA technology are:-

Splitting the DNA molecule into smaller portions using restriction endonuclear which are specific to particular base sequences on the DNA.
Copying the required DNA section using the enzyme reverse transcriptase which controls the synthesis of DNA from the RNA. The resulting DNA is called copy or complementary DNA

(cDNA).

Adding the gene the vector DNA.
Formation of recombinant DNA molecule with vector.
Joining the DNA portions together using DNA ligase enzyme.

Illustration: Consider the diagram below showing the synthesis of insulin

Question

Define the terms:-

(b) Gene cloning.

Solution

A clone is a group of cells of similar characteristics that are able to replicate and produce more cells.
Gene cloning is a process whereby multiple copies of a given gene are produced which may then be used to manufacture larger quantities of valuable products.
A transgenic organism is a genetically modified organism (GMO) ie: the organism formed as a result of genetic engineering. New genes are added into embryo of an organism.

MERITS AND DEMERITS OF GENETIC ENGINEERING

Merits of genetic engineering

Synthesis of hormones such as insulin, growth hormones etc.
Production of vaccine and antibiotics. Already interferon has been synthesized by genetic engineering.
Increases plant resistance to pests eg: cotton and potatoes.
Improves quality and quantity of animal products such as milk.
It offers endless opportunities to manipulate DNA.

Demerits

The materials contained in the manipulated DNA, are likely to undergo mutation.
The use of genetic engineering in the manufacture of biological weapons is a mis-application of genetics.
The use of GMO’s for human consumption is dangerous as it increases cancer chances.
Some practices of Genetic engineering may not be in line with ethical and moral values.

Phenlyketonuria (PKU)

PKU is a recessive autosomal condition.
The disease is due to inability to convert the amino acid phenylalanine to another amino acid, tyrosine.

Phenylalanine

Hydrolase.

Phenylalanine (PAH) Tyrosine.

The enzyme PAH is normally there in the liver.

As a result of phenylalanine builds up in the body and the excess is converted into toxins which affect mental development.

Affected children appear normal at birth because, while in their mother’s uterus during pregnancy, excess phenylalanine moves across the placenta and is removed by the mother’s liver. If not treated soon, harmful effects are noted.

Identifying PKU in new born babies

Few days later after birth, blood test is carried on and higher level than normal of phenylalanine is detected.

Question: Why the baby not tested when is first born?

Answer

Its excess phenlylalanine is removed by the mother while it is in the uterus. It takes a few days for the levels of phenylalanine to build up.

Genetic screening and parental diagnosis:

Genetic screening is the detection of mutant genes in an individual.

There are three situations where genetic screening is of particular relevance namely;-

Prenatal diagnosis.
Carrier diagnosis.
Predictive diagnosis.

Prenatal diagnosis

This is the use of modern medical techniques to identify any health problems of the unborn baby.

It includes the detection of genetic disease. If such a disease is detected, it is usually possible to provide counseling about the quality of life the child can expect and other potential problems. The parents are usually also given the option to abort.

Carrier diagnosis:

This is the identification of people who carry a particular genetic disease, usually with no visible symptoms or harm to themselves.

Predictive diagnosis:

This is the prediction of a future disease which you are likely to suffer as a result of your genes but not yet produced any symptoms.

Chorionic villus sampling (CVS)

A small sample of chorion is withdrawn for examination.
The cells of the chorion are derived from the zygote, so are genetically identical to the embryo. The chromosomes are examined (Karyotype analysis) and the sex of the child can also be seen.
The rise of miscarriage is higher than Amniocentesis.

Amniocentesis

This involves analysis of amniotic fluid.
Amniotic fluid is genetically identical to chorion and zygote.

Mixed concepts

Successfulness of Mendel in his experiment was due to the following reasons.

He chose pea plants (P. sativum) in his crossing. Why?

Reasons:

P. sativum has short life span.
P. sativum has many contrasting characters (about 32 contrasting characteristics).
A plant is Bisexual.

He was Imaginative in ratio determination and careful, he covered the stigma with muslin cloth which allowed respiration but prevented pollens of one plant from reaching another plant.
He did his experiments in a large scale to avoid (eliminate) the effects of mutation.
He was intelligent as he used mathematical concepts in explaining the ratio of the results he got.

Modern Mendelian laws (1^st Law & 2^nd Law)

The characteristics of an organism occur in pairs of chromosomes which divide during meiosis such that only one of each chromosome appears in gametes.
Alleles occur in pairs, they divide by meiosis and during fertilization, zygote is formed from random fertilization in which any of the allele from each pair may combine with the other.

Application of Genetic engineering

(a) Medicine

The gene in man that codes for insulin is transferred in the bacterium, Escheriachia coli to produce pure insulin in large quantities.
Human growth hormone, somatotrophic hormone can be extracted from the pituitary glands of dead bodies.
Blood clothing factors such as fibrinogen needed by haemophiliacs are produced.
Vaccines are produced from viruses.

(b) Biological Warfare

Micro – organisms that cause diseases have been used in wars.

The micro organisms are cloned and thrown into the territory of the enzyme.

Infections of this bacteria (micro-organism) causes death within few days.

(c) Agriculture

Research is being done to produce plants that are capable of fixing nitrogen without relying on nitrogen fixing bacteria.
Genetically modified organisms are used to break down wastes from homes and industries.
Frozen embryos may be separated into cells, which can be made to grow into new embryos if implanted into the uterus.
Some plants such as pyrethrum are being propagated through tissue culture.

Genetic disorders

Pregnant woman can be told whether the fetus had deformities or not and hence prepare for its birth or terminate it.
Normal genes can be introduced into the embryos to curb or cure genetic disorder such as sickle – cell anemia.
Genetic engineers are designing to produce human like creatures as sources of human spare parts in surgeries and transplants.

Examples of monohybrid inheritance:

Rhesus factor – In which an organism is either positive or negative.
Albinism – If dominant for skin pigmentation (AA or Aa), then an individual is normal though in the later case, he/she is a carrier.

But if recessive homozygous, then an individual is albino.

In maize seeds – The seeds are either dominant white or recessive.

How genetic engineering is done

A section of DNA, extracted from an organism or synthesized artificially, is usually translocated to a bacterium or virus. The bacterium or virus used in genetic engineering is called transgenics. Inside the bacterial cell is a structure called plasmid.

The plasmid is split open by some enzymes called restriction endonucleases so as to allow the foreign DNA to enter.

A given restriction enzyme cuts the bacterial plasmid open at specific sites where is determined by the sequence of base in that region. This same enzyme cuts foreign DNA wherever an identical base sequence occurs.

This procedure of splitting open the bacterial plasmid and inserting the foreign DNA is called gene splicing.

The foreign DNA and the plasmid join up, and so the foreign DNA gets incorporated into the plasmid. The enzyme DNA ligase is responsible for joining the foreign DNA and plasmid. The result of the combination is called recombinant DNA.

The foreign DNA replicates along with the rest of the plasmid every time the bacterial cell divides.

The bacterium is selected because it replicates quickly and the offsprings resemble parents.

Once the bacterium has taken up a piece of foreign DNA successfully, it may divide repeatedly into a population of bacterial cell all of which contain replicas the foreign DNA.

This production of large quantities of identical genes by means of genetic engineering is called gene cloning. This technology is currently being used in production of human insulin to save diabetes.

Application of Genetics

1. Plant and animal breeding

It has been observed that crossing two genetically dissimilar organisms of the same species produces organisms that possess beneficial characteristics not shown by either of the parents.

The individual that results from crossing two individuals with contrasting characters is called a hybrid. In cattle, milk production, quick maturation and beef production can be obtained through hybridization.

For example: The Hereford, English breed shows high beef production and quick maturation.

The Boran from Tanzania shows disease resistance and grows on dry pasture.

A cross between the Herefore bull and a Boran cow produces a hybrid of all these qualities.

2. Blood transfusion

Blood transfusion is the transfer of blood from one person, the donor, into the stream of another person, the recipient.

Before blood transfusion, blood is tested to determine the blood group and Rhesus factor. If the blood of the donor is not compatible with the blood of the recipient, agglutination occurs.

The ABO blood group system and the Rhesus antigens are used to settle parentage disputes.

3. Genetic counseling

Genetic information is used to advise couple who have hereditary disorders about the chances of children inheriting the disorders.

Genetic information could also be used in choosing marriage partners.

Symptoms of Down’s syndrome:

Mental retardation.
Straight hair.
Increased risk of infections particularly respiratory and ear infections.
Short stature.
Heart defects.

PLEIOTROPY

Pleiotropic genes are genes that code for a specific mutabollic process and at the same time affecting other metabolic processes.

eg. In cystic fibrosis, A gene that codes for secretion of Cl^– also induce secretion of viscous (thick) mucus in the lungs, pancreases and gut.

Question

(a) Define the term sex limited traits giving one example of a disorder that expresses it.
Define the term sex influenced characters giving one example of a disorder that expresses it.
Describe the following disorders showing clearly its mode of inheritance;
Tuner’s syndrome.
Klinifelter’s syndrome.
(a) Explain the following ratios using genetic crosses;

9:3:4 cross (2) gives answer.
9:7
9:3:3:1 under non Mendelian trend of inheritance.
3:3:2
2:1
1:2:1 under non Mendelian trend of inheritance cross (3).

(b) Outline;

(i) How a peptide bond is formed.

(ii) The characteristics of genetic material.

7. Why are most lethal genes recessive?

Perform below crosses and see if ratios can provide solutions to question (2) (a)

Agouti x Albino. Where; A-Agout.

AaCc Aacc C-Black.

Agouti x Agouti.

AaCc AaCc

Agouti x Albino.

AaCa aacc

Heterogemetic sex (XY). In some insects, while the female is XX, the Y chromosome is absent in the male, which is therefore XO. In the fruit fly Droasphila, the female is XX and the male is XY.

SEX-LINKAGE

Sex linkage refers to the carrying of genes on the sex chromosomes.
The X chromosome carries many such genes, the Y chromosome has very few.
The non – homologous part of the X – chromosome with Y – chromosome is the one which bears extra genes.

Those genes that are inherited together with sex chromosomes are called sex – linked characters (traits).
Two well known sex-linked genes in humans are those causing haemophilia and pale green colour blindness. Both are linked to the X-chromosome and both occur almost exclusively in males.

For the condition to arise in females, requires the double recessive state and as the recessive allele is relatively rare in the population, this is unlikely to occur. In females the recessive allele is normally masked by the appropriate dominant allele which occurs on the other X – chromosome. These heterozygous females are not themselves affected but are capable of passing the recessive allele to their offspring.

For this reason such female are termed Carriers.

When the recessive allele occurs in males it expresses itself because the Y – chromosome cannot carry any corresponding dominant allele.

HAEMOPHILIA

Hemophilia is the inability of blood to clot, leading to slow and persistent bleeding especially in the joints. Unlike colour blindness it is potentially lethal.

Hemophilia is a sex linked character caused by a recessive allele which is carried by the X – chromosome.

Consider the following genetic attributes:-

X^HX^H – Normal female.

X^HX^h – Normal but carrier female.

X^hX^h – Hemophilia female.

X^HY – Normal male.

X^hY – hemophilia male.

Hemophilia females are rare in nature because:-

Mostly, they do not grow beyond the first menstrual flow.
For a female to have haemophilia, both X – chromosomes must bear the allele h, in the heterozygous state, one is normal.

Example

Consider a cross between a carrier female and a normal male.

From the cross,

All the females are phenotypically normal.
50% of the males, are hemophiliac.

Considering a reciprocal cross we have

Topic 4: Genetics - Biology Notes Form 6

Topic 4: Genetics - Biology Notes Form 6

GENETICS

Characteristics of hereditary materials:-

SPECIES CONCEPT:

EXTRA OF HEREDITARY MATERIALS

CHROMOSOMES AND THEIR STRUCTURE

Structure of chromosome:-

THE NUCLEIC AIDS, TYPES OF HEREDITARY MATERIALS

Chemical nature of nucleic acids:-

1. Pentose sugar – This is a five carbon sugar.

2.Nitrogenous (organic) base

3. Phosphate group:-

Structure of a typical nucleotide:-

(A) RIBONUCLEIC ACID (RNA)

Diagram to show structure of RNA:Role of RNA.

Types of RNA

(a) Messenger RNA (mRNA).

(b) Ribosomal RNA (rRNA).

(c) Transfer RNA (tRNA)

Role of tRNA

(B) DEOXYRIBONUCLEIC ACID (DNA)

Role of DNA in protein synthesis.

Evidence for the role of DNA in inheritance

Evidence from bacteria:

DNA REPLICATION

Significance of DNA replication:-

Mechanism of DNA replication:-

Semi – conservative replication

Study Questions:-

The nature of genes:-

Below are ways of overcoming this objection.

Problems.

Features (Characteristics) of the genetic code:-

PROTEIN BIOSYNTHESIS. ‘DNA makes RNA and RNAmakes Protein’

Adaptations of the ribosome to protein synthesis

Mechanism of protein synthesis:-

Introns and exons.

Summary:

MENDELIAN GENETICS

Mono hybrid inheritance and the principle of segregation:-

Glossary of common genetic terms:-

Basic Monohybrid ratio

Mendel’s experiment and the Monohybrid ratio

Assumptions:–

Mendel’s 1st law of inheritance (Law of segregation)

Meiotic explanation of Mendel’s first law.

CROSS I

CROSS II

MERITS AND DEMERITIS OF MENDEL

NON – MENDELIAN GENETICS

INCOMPLETE DOMINANCE (under gene interactions)

LINKAGE

Gene mapping

Sex determination

MULTIPLE ALLELISM

Dominance series

Worked examples

Sex Limited and sex influenced characters:

PEDIGREE ANALYSIS

Worked examples

More examples:

GENE INTERACTION

(a) Lethal genes.

(b) Epistasis.

(c) Collaboration (complex gene/simple Gene Intervention).

(d) Complementary genes.

(e) Polygenic inheritance (Multiple gene interaction).

VARIATION

Discontinuous Variation

HISTOGRAMS

Mutations and deleterious genes

Significance of mutations

Causes of mutations:

Types of mutations:

1. Albinism

2. Sickle cell anaemia

(b) Chromosome Mutations

Significance of polyploidy:

Auto polyploidy

Diagram to show structure of RNA:

Role of RNA.

PROTEIN BIOSYNTHESIS. ‘DNA makes RNA and RNA
makes Protein’

Mendel’s 1^st law of inheritance (Law of segregation)