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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.
Qn;-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.
Qn:-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 ho
w 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:-
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.
  1. 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.
  1. 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.
(c) The chemical bonds i.e hydrogen and phosphodiester bonds, increase mechanic 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:-
    1. A representative portion of DNA, which is about to undergo replication is shown.
    2. DNA polymerase causes the two strands of the DNA to separate.
    3. The DNA polymerase completes the splitting of the strand. Meanwhile free nucleotides are attracted to their complementary bases.
    4. Once the nucleotides are lined up joined together. The remaining unwinded bases continue to attract these complementary nucleotides.
    5. 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

(c) Summarize the structural differences between DNA and RNA.
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 t
he 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 Xy
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).
(c) Combination of those amino groups with a carbohydrate skeleton (eg. α – ketoglutarate from Krebs cycle).
(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. Th
is 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;-
    1. Binding of mRNA to ribosome.
    2. Amino acid activation and attachment to tRNA.
    3. Polypeptide chain initiation.
    4. Chain elongation.
    5. Chain termination.
    6. 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
  1. 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 1st 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.
(c) Mendelian crosses/genetics diagrams
(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.
(c) If she is homozygous tall, then half the offspring will be phenotype tall and the half short(dwarf) above reveals.
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:-

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

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


gametes
PS
Ps
ps
ps
PpSs
Ppss
ppss

gametes
IA
IB
gametes
IO
IA IO
IB IO

Individual
Genotype
1
Asay
2
Asat
3
ayay
4
atat
5
ayat
Gamete
WB
Wb
wB
wb
WB
WWBB
WWBb
WwBB
WwBb
Wb
WWBb
WWbb
WwBb
Wwbb

ecolebooks.com
Wb
WWBb
WWbb
WwBb
Wwbb
wb
WwBb
Wwbb
wwBb
wwbb
Phenotype
Possible genotype
Pea
PPrr or Pprr
Rose
ppRR or ppRr
Single
pprr

walnut
PPRR, PPRr, PpRR or PpRr




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3 Comments

  • Forbi beltus fuhnwi matute, August 1, 2024 @ 12:41 pm Reply

    Very. Thank you good

  • Phibean Ogunlade, June 3, 2024 @ 11:50 am Reply

    Very comprehensive and educative.

  • Jovan Pro, February 8, 2024 @ 4:36 am Reply

    I have liked your approach to biology

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