Enriched genetics notes 2021 @kingdom solutions

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GENETICS, INHERITANCEAND
VARIATION
Inheritance
Inheritance is the passing of parental
characteristics to their offsprings /next generation.
Genetics is the scientific study of inheritance or
heredity.
The importance of genetics
 It is applied in genetic engineering to produce
better breeds and varieties of plants and
animals by altering their genetic constitution.
 It is important in courts of law to determine the
paternity of the child.
 Genetics forms the basis of blood transfusion to
determine compatible blood groups.
 Genetic counseling is important in preventing
the transmission of genetically determined
diseases among married couples. This will help
to relieve the families and community of the
costs of treatment as well as the suffering of the
sick and their families.
 It can be used in the identification of criminals
by use of fingerprints and DNA profiling.
 It is used in molecular biology to manufacture
artificial enzymes, hormones, and vaccines by
manipulating responsive genes from
organisms.
 Forms the basis of cloning to increase the
number of genetically important plants and
animals.
Terms used in genetics.
1. Chromosome. These are thread-like
structures bearing genes and located in the
nucleus.
2. Chromatid. This is half of a chromosome split
longitudinally.
3. Bivalent. This is a pair of homologous
chromosomes.
4. Gene. This is a unit of the hereditable material
found on the chromosome and responsible for
controlling a particular trait/character.
5. Allele. This is the alternative form of the same
gene. Most genes are made up of two alleles.
Alleles of the same gene are represented by the
same letter but the dominant allele is
represented by a capital letter and the recessive
allele by a small letter in the case of dominant
recessive characters
6. Diploid. This is a description of a cell, which
has a whole set of chromosomes.
7. Haploid. This refers to a cell with half the set
of chromosomes.
8. Genotype. This refers to the genetic
composition of an organism.
9. Phenotype. This is the physical appearance or
the outward expression of an individual.
10.Dominance. Asituation in whichonemember
of a pair of allelic genes expresses itself as a
whole (complete dominance) or in part
(incomplete dominance).
Dominant gene/dominant allele. This is a
description of a gene /allele whose effect is seen in
the phenotype of the heterozygous individual. The
effect of the dominant gene/allele is seen in the
phenotype even in the presence of another
gene/allele.
Incomplete dominance. This is a condition
where neither of the genes is dominant over the
other.
Recessive. This is a description of a gene whose
effect is not phenotypically expressed in the
heterozygous state. The effect of a recessive
gene/allele is not seen in the presence of another
(dominant) gene/allele.
Homozygous. This refers to a gene with two
identical alleles for exampleif T represents thegene
for height where tallness is dominant to shortness
thentheallele for tallness is T andthatforshortness
is t. an individual with TT is said to be homozygous
tall and tt is said to be homozygous short.
Homozygous dominant. This is where both
alleles of a gene determine a dominant character.
Homozygous recessive. This is where both
alleles of a gene determine a recessive character.
Heterozygous. This refers to a gene with two
different alleles for example if T represents the
allele for tallness and t for shortness then Tt is the
heterozygous state of this gene.

Hybrid. This is an offspring produced by parents
of two different pure lines.
Gametes. These are reproductive cells.
Fertilization. This is the fusion of the male and
female gametes to form a zygote.
Monohybrid inheritance. This is a type of
inheritance,whichinvolvesstudying a singlepair of
contrasting characteristics.
11. Dihybrid inheritance. This is a type of
inheritance, which involves studying two pairs
of contrasting characteristics at ago
12.Test cross
This is a type of back cross which involves crossing
an offspring having a dominant character with its
recessive parent to determine the test of that
offspring.
13.Back cross
This is the mating of an offspring with one of its
parents.
MENDEL’S EXPERIMENT IN GENETICS
Johann Gregor Mendel (1822-1882) was an
Austrian monk, who studied the process of heredity
in selected features of garden pea (Pisum sativum)
For his experiment, he collected one of the varieties
of garden peas (Pisum sativum) with contrasting
features such as one variety was producing tall
plants when stems are about 200cm and another
short plant with stems of 25cm. He crossed these
plants for his experiments.
He crossed pure tall pea plants with pure short pea
plants and all the offsprings were tall (F1
generation)
Tallness was the dominant character and shortness
the recessive character.
The dominant character is represented using a
capital letter while the recessive character is
represented using a small letter.
Offspring phenotype: All tall
Mendel then selfed the plants of the F1 generation
and obtained an F2 generation with tall and short
plants in a ratio of 3:1
Mendel’s conclusions
Mendel suggested the following to explain
his results.
1. Gametes like pollen grains and ovules of the
garden peas carry character determining factors
through which resemblance is passed on from one
generation to the next.
2. A character like the height of the garden pea is
controlled by a pair of genes. These separate during
the formation of gametes and only one goes into
each gamete. This means that only half of the usual
number of genes is present in the gametes.
However, the normal number is restored at
fertilization by the fusion of the two gametes

3. He named a gene determining a dominant
character as a dominant gene and one determining
a recessive character as a recessive gene. In his
representation, dominant genes were given capital
letters and recessive genes were given small letters.
Mendel’s laws of inheritance
From his observations, Mendel put up two laws of
inheritance.
First law: The law of segregation.
This law states that the character of an organism is
determined by a pair of alleles. Only one allele of
such a pair is carried in a gamete.
Second law: The law of independent
assortment.
This states that each of the alleles in a pair may
combine with another allele from another pair
randomly.
Conclusions from Mendel's crosses.
1. A character can be transmitted from parent to
offspring independent of other characters.
2. Genes occur as a pair of alleles.
Only one allele of the same gene is carried in a
single gamete.
3. If an organism has two unlike alleles for a given
character, one may be expressed (dominant) to
exclusion of the other (recessive allele).
4. During meiosis, each pair of the allele separates
and each gamete receives one of each pair of alleles.
5. Each allele is transmitted from one generation to
another as a discrete (separately) unchanging unit
SOME PRINCIPLES ESTABLISHED BY
MENDEL
1. Principle of dominance
Varioushereditary traits arecontrolled by factorsin
pairs. Only one of the pairs of the factors expresses
itself in the F1 generation of pure breeding
individuals with contrasting characteristics. This is
the dominant factor. The recessive factor will
appear in the F2 generation.
2. Principle of segregation
A pair of factors segregate during gamete formation
and the paired condition is restored by random
fusion of gametes.
3. Principle of independent assortment
For dihybrid crosses, he found out that inheritance
of two or more contrasting factors at a time and
their distribution in the gametes and subsequent
generation is independent of each other.
4. Principle of chance transmission of
characters.
Mendel also observed that all offsprings had a
chance of inheriting a given trait from the parent.
Mendel`s consideration about the choice of
material to study
Several reasons were given for the choice of
breeding material or organism for genetic
experiments by Mendel and include the following
reasons:
1. Variation. The organisms which are to be
chosen for the genetic experiments should have
many detectable differences and at a time only a
single detectable character should be considered.
2. Reproduction. The chosen organisms should
be sexually reproducing because only then the
offsprings will be able to receive different
characters from both the male and female parents.
3. Controlled mating. The chosen organisms
should beable to mateincontrolledor well-planned
conditions. Because in genetic experiments
sometimes there is the rearing of genetically pure
parents by methods of controlled mating.
4. Short life cycle. The chosen organisms should
have very short life cycles.
5. Number of offsprings. The organisms which
have been chosen for the genetic experiments
should produce a large number of offsprings after
each successive mating because it will help in
deducing the correct conclusions.
6. Convenience in handling. The experimental
species should be of a type that can be raised

and maintained conveniently and inexpensively in
the laboratory
MONOHYBRID INHERITANCE.
Is the passing of one pair sharply contrasting
characteristics of the parents to their offsprings.
Monohybrid inheritance involves the study of how
one character is inherited from the parents to the
offsprings. Mendel carried out several experiments
on peas to study monohybrid inheritance.
Reasons why Mendel chose Garden peas
1. The Pea plant has several sharply contrasting
characteristics e.g
(i). position of flowers (axial or terminal)
(ii). Length of stems (tall or dwarf)
(iii). The shape of seeds (round or wrinkled)
(iv). Colour of pods (green or yellow)
(v). color of flower (red or white)
2. It naturally pollinates itself and can also be
cross-pollinated
3. It matures at a faster rate /has a short life cycle
4. It produces many seeds and hence many
offsprings could be obtained in a single season
5. Garden pea plants were small in size and
easy to cultivate in small experimental
plots.
Mendel’ experiment on Monohybrid
Inheritance
 Mendel identified two pure /true-breeding pea
plants, one homozygous dominant tall and
another homozygous recessive short and cross-
pollinated;
 The F1 generation were all tall pea plants.
 When he selfed/self-pollinated the F1
generation offspring, he obtained both tall and
short pea plants.
 Relative numbers of these plants were in the
ratio of 3 tall pea plants to 1 short pea plants,
 Described the above ratio as the monohybrid
inheritance ratio.
Using the genetic diagram, Mendel explained the
occurrence as follows:
Let T represent the allele for Tallness in a pea
plant be
Let t, represent an allele for shortness in a pea
plant
2n –diploid state (with a pair of chromosomes)
n-haploid state (with single chromosomes)
EXPLANATION
All theoffsprings in the F1 generationaretall pea
plants, there is no dwarf/short plant pea because
the allele for tallness is dominant over that for
shortness, therefore suppressed the recessive allele
for dwarfism/shortness.
Table showing a summary of results
obtained by Mendel after investigating
inheritance (monohybrid using various
characteristics in
Mendel’s Conclusion from the above
investigation
1. There are alternative forms for genes i.e. a
single characteristic exists in two forms e.g.
Gene that determines pod color can either

be (G) for green pod color and (g) for yellow
pod color.
2. For each characteristic /trait, the organism
inherits two alternative forms of that gene,
one from each parent
3. During Gamete formation /meiosis, allele
pairs separate/segregate leaving each
gamete with one of the alleles.
4. During Random fertilization/Random
fusion, the resultant offspring will contain
two sets of alleles, one from each parent
5. When two alleles of a pair are different i.e.
One is dominant and the other is Recessive
one, recessive one is NOT lost in the F1
generation,butis suppressed andreappears
once F1 generation is selfed.
EXAMPLES OF MONOHYBRID
INHERITANCE IN MAN
There are many genetically determined
abnormalities and diseases that affect man (and
other animals). Since these are genetic diseases,
they can only be inherited from parents and their
occurrence is determined by those genes inherited
from parents during fertilization
Examples of such diseases include:
 Sickle-cell anemia
 Albinism
 Achondroplasia
 Cystic fibrosis and many more
NB: Research has shown that most of, though not
all thegeneticabnormalitiesarecausedby recessive
genes (alleles) andthe genesresponsible fornormal
conditions are dominant. This implies that for an
individual to suffer from such diseases, they must
have two copies of the responsive genes
(homozygous recessive). The heterozygotes and the
homozygous dominant individuals are normal.
Though the former are phenotypically normal but
their cells contain a copy of the recessive allele and
are described as carriers.
INHERITANCEOFSICKLE-CELLANAEMIA
Sickle-cell anemia isa recessive charactercausedby
a point substitution mutation in which the amino
acid called glutamic acid in normal hemoglobin
is replaced by valine. Normal hemoglobin (HbA)
contains an amino acid glutamic acid at position 6
of the β-chain. The amino acid is polar and
hydrophilic which makes normal hemoglobin
soluble in water. It is coded for by the DNA
triplet CTT which changes to CAT by substitution
of the base T with base A. This triplet codes
for valine which is non-polar and hydrophobic
hence reduces the solubility of haemoglobin,
especially at low oxygen tensions.
This abnormal haemoglobin crystallizes into rigid
rod-like fibres which distort the normal biconcave
shape of RBCs into a crescent/sickle shape. Such
abnormal haemoglobin is called HbS, It has a very
low oxygen-carrying capacity leading to symptoms
of anemia and the disease is known as sickle-cell
anemia.
Being a recessive character, for a person to be a
sufferer they must possess two copies of the faulty
gene (homozygous recessive, i.e., HbSHbS or
ss). Heterozygotes (carriers, i.e., HbAHbS or
Ss) have one copy of the responsive gene whose
effects are masked by the other dominant gene.
They don’t sufferfrom thedisease symptomsexcept
at exceptionally low oxygen tensions; this is known
as the sickle-cell trait.
It is therefore advisable to avoid exposure of such
people to low oxygen environments like crowded
places, high altitudes, and flying in unpressurized
aircraft.
Question; if two people suffering from sickle cell
trait are married, what is the probability that they
will produce an anaemic child?

Genotypic ratios:
1HbAHbA:2HbAHbS:1HbSHbS
Phenotypic ratios: 1normal:2 carriers: 1sickler
Probability of a sickler is 1⁄4 = 0.25
Complications due to sickle cell anemia
1. Anaemia this is occurring because the sickle cells
aredestroyed whichlowerstheamountofoxygento
be carried leading to acute anemia. This leads to;
 Fatigue (weakness)
 Poor physical development
 Dilation oftheheartwhichmayleadtoheart
failure
2. Sickled RBCs block small capillaries which
impede blood flow. This leads to;
 Heart damage which leads to heart failure
 Lung damage which leads to pneumonia
 Kidney damage which leads to kidney
failure
 Liver damage
3. Enlargementofthespleenbecausethe sickle cells
collect in the spleen for destruction
The effects above make the homozygous sufferers
often die before reproductive age.
Physiological effects;
 Stress to the heart because it has to pump
faster to supply blood, leading to failure of
heart function.
 The spleen and liver are
overtasked/overworked to break down
defective RBCs leading to spleen and liver
failure
 Impairment of the brain causes strokes and
paralysis.
 Decreased blood flow to muscles causes
rheumatism.
NB: Despite the above complications suffered by
sufferers of sickle cell anemia, the heterozygotes
tend to have an advantage of showing increased
resistance to the plasmodium parasite that causes
malaria much more thanboth the sufferers and the
normal.
Carriers (heterozygotes) of sickle cell anemia show
the sickle cell trait, a co-dominant condition, in
which most of the red blood cells have normal
hemoglobin and only about 40% of the red blood
cells have abnormal hemoglobin S.
This produces mild anemia and prevents carriers of
the sickle cell trait from contracting malaria. This is
because
when the plasmodium that causes malaria enters a
red blood cell with hemoglobin S, it causes
extremely low oxygentensioninthecellwhichleads
to the cell sickling in heterozygotes. These sickled
cells are quickly filtered out of the bloodstream by
the spleen, thus eliminating the
parasites.
This resistance is due to the consistent change in
oxygen levels between normal and sickle cells
makes it difficult for the parasite to adapt. In such
cases, the immune system of the body eliminates
the parasites before the disease is established
rendering resistance to the heterozygotes
This is referred to as the heterozygous
advantage which increases chances of survival for
heterozygotes especially in the tropics where
malaria is one of the leading causes of death
INHERITANCE OF ALBINISM
Albinism is a recessive character that fails in the
formation of body pigments. Albinos have the
following characteristics as a result;
 Light-colored skin
 White hair
 Pink eyes
Albinism results from one of several different
mutations in a gene coding for melanin production.
The mutant allele is recessive and does not allow
melanin to be produced. Mammals homozygous for
such an allele have no melanin in their coats or
irises andare whitewith pinkeyes. Severaldifferent
mutant alleles can have this effect. One of them

prevents the formation of the enzyme tyrosinase.
Tyrosinase enzyme is the one whois responsible for
the production of melanin if tyrosinase is not there,
there will be no melanin inside of the body.
Tyrosinase enzyme catalyzes the conversion of the
amino acid, tyrosine to melanin.
Sample question
A manwithnormalskinmarriesa carrierfor albino
skin. What is the probability that some of their
children will be albinos?
INHERITANCE OF CYSTIC FIBROSIS
This is a recessive character caused by a mutation
resulting in the accumulation of abnormally thick
andsticky mucus thatblocksthe pancreaticduct,
bile duct, and air passages.
The mutation occurs on an autosomal
chromosome 7 affecting the gene that codes for a
chloride channel protein in epithelial cells.
This results in the total absence or malfunctioning
of this channel protein hence interfering with
chloride ion flow.
Chloride ions accumulate in the cells and attract
sodium ions towards the opposite charge; this
increases the ion concentration, hence the osmotic
potential of the cells which prevents osmotic
outflow of water.
As a result, the mucus secreted is dry, thick, and
sticky; blocking small tracts of some body organs.
This is known as cystic fibrosis.
In the pancreas, fibrous patches
called cysts develop (hence the name) and
complications include digestive problems due to
poor release of pancreatic enzymes, poor
absorption of digestive products, chronic lung
diseases, reduced fertility, etc.
ACHONDROPLASIA (DWARFISM)
Achondroplasia is caused by a mutation in the
fibroblast growth factor gene for the protein
required for the production of collagen and other
structural components in tissues and bones in
which the amino acid glycine is replaced with
arginine. This produces a faulty protein.Whenthe
gene is mutated, it interferes with how this protein
interacts with growth factors, leading to
complications with bone production. Cartilage is
not able to fully develop into bone, causing the
individual to be disproportionately shorter in
height. Other phenotypic effects are: enlarged
head, prominent forehead,
The effect is genetically dominant, with one mutant
copy of the gene being sufficient to cause
achondroplasia, while homozygous recessive
condition of the mutant gene is fatal (recessive
lethal) before or shortly after birth. In couples
where one partner has achondroplasia there is a
50% chance of passing the disorder onto their child
every pregnancy. In situations where both parents
haveachondroplasia thereisa 50%chancethechild
will have achondroplasia, 25% chance the child will
not, and a 25% chance that the child will inherit the
gene from both parents resulting in homozygous
condition and leading to lethal condition.
One example is achondroplasia, a form of dwarfism
that occurs in one of every 25,000 people in the
world. Heterozygous individuals therefore have the
dwarf phenotype as shown below.

Consider the illustration below:
Since this character is dominant (caused by a
dominant allele), all people who are not
achondroplastic -99.99% of the population-are
homozygous for the recessive allele. Like the
presence of extra fingers or toes mentioned earlier,
achondroplasia is a trait for which the recessive
allele is much more prevalent than the
corresponding dominant allele.
Polydactyly
Is concerned with the presence of the sixth fingers
on each arm and is caused by a dominant gene. It is
a single gene disorder, manifesting by itself, then it
is due to an error in the finger gene, and it shows an
autosomaldominantinheritability.Hence,only one
copy of the mutated gene is enough to induce this
condition. Also, if at least one parent has this
presentation, then each progeny has a 50% chance
of inheriting the trait
PHENYLKETONURIA (PKU)
Is a condition of severe mental retardation light-
colored hair, a musty odor in the breath, skin or
urine . It is caused by a recessive mutation defect in
amino acid metabolism of the amino acid
phenylalanine not converted to tyrosine because
the necessary enzyme, phenylalanine
hydroxylase is lacking or ineffective. High
amount of phenylalanine or derivatives in mother’s
blood affects fetus. Phenylalanine or derivatives
excreted at a high level in urine.
Treatment: phenylalanine in diet-restricted
during childhood
Question:
1. The fruit fly (Drosophila melanogaster)
usually has wings twice as long as its
abdomen but some drosophila has very
short or vestigial wings. Long winged
drosophila (male) was crossed with a
vestigial winged female drosophila and all
theF1 offsprings werelong winged.Thelong
winged F1 generation were then mated.
I. How can the cross be represented
diagrammatically
II. State thephenotypesofthe offsprings in the
F2 generation and state their genotypic
ratio.
III. What is the percentage of the vestigial
winged drosophila flies in the the F2
generation.
IV. A drosophila is normally used in
experiments on heredity, why do you think
it is suitable for such experiments.
2. In cattle, the gene for hornless condition is
dominant over one for horns. A hornless
cow was mated with a horned bull. Using

genetic symbols, show the possible
phenotype and genotype of the F1 offspring.
Let h represent the allele for horned bull.
Let H represent the allele for hornless bull
All were horned cows.
A bull whose horns were removed was mated to a
horned cow. Show the possible genotypes and
phenotypes of the F1 offsprings. Give a reason for
your answer.
Let h represent the allele for horned bull.
Let H represent the allele for hornless bull
All are horned
Because the bull with cut off horns still has the
genes for horned and cutting off the horns doesn’t
change the genes.
TEST CROSS.
- Is a test carried out to establish whether an
organism ofa dominantphenotypeheterozygousor
homozygous.
- Involves crossing an organism showing a
dominant characteristics with another organism
that is homozygous recessive
- Two possible results can be obtained e.g , in
establishing the genotype of pea plant showing the
dominant phenotype of round seed coat. i.e .
(i). if the off springs all show dominant trait, then
genotype of the organism of a dominant phenotype
is homozygous.
Let the dominantallele forroundseed coat
be R
Recessive allele for wrinkled seed coat be r
Test cross phenotype; homozygous round seed
coat × homozygous wrinkled seed coat.
(ii). If the off springs are a mixture of phenotypes
in ratio of 1:1, then the genotype of the organism of
a dominant phenotype is heterozygous.
EXAMPLE OF A TEST CROSS
Ina species ofmammal,thegeneforhaircolourhas
two alleles,
B and b. Allele B gives brown fur and allele b gives
white fur.
If an animal has brown fur, we do not know if its
genotype is BB or Bb. We can find out by breeding

it with an animal with white fur, whose genotype
must be bb.
If there are any white offspring, then the unknown
animal must have the genotype Bb, as it must have
given a b allele to these offspring.
If there are no white offspring, then the unknown
animal probably has the genotype BB. However, it
is still possible that it is Bb and, just by chance,
none of its offspring inherited the b allele from it.
BACK CROSS OR TEST CROSS
A test cross is used to distinguish between
homozygous and heterozygous dominant forms.
This is when an F1 individual with the phenotype of
the dominant parent is crossed with the recessive
parent to determine the phenotype of the parent.
 If the F1 is homozygous dominant, all the
offsprings will show the dominant
character.
 If the F1 individuals are heterozygous, a 1:1
ratio of dominant or recessive characters is
obtained e.g:
Let T represent the allele for tallness
Let t represent the allele for shortness
Two offsprings will be heterozygous tall and 2
will be homozygous short.
Heterozygous tall
Question.
In pea plants, the allele for purple flowers is
dominant over the allele for white flowers. How
would you find out if a purple-flowered plant is
homozygous or heterozygous?
Solution
Carrying out a test cross, by crossing
with a white-flowered plant
If some offsprings are white-flowered,
then the plant was heterozygous
If all purple-flowered offsprings are
produced, then the plant was homozygous
FACTORS THAT CAN MODIFY THE
MENDELIAN MONOHYBRID
RATIO/EXCEPTIONS
1. LETHAL GENES: These are genes that when
expressed in homozygous dominant form, lead to
the death of the bearer. Lethal genes are divided
into 3 major categories;
a. Gametic lethal genes. These are genes that
kill the gametes and therefore prevent fertilization.
b. Zygotic lethal genes. These are genes that kill
the zygotes and embryos before birth e.g. the gene
that determines coat color in mice.
c. Infanticlethalgenes. These aregenes thatkill
individuals between birth and reproductive stages
e.g. the gene that determines chlorophyll formation
in maize, sickle cell anemia in man e.t.c.
in humans, the following conditions are controlled
by lethal genes:
i. Congenital ichthyosis
ii. Infantile amaurotic idiocy
iii. Thalassemia
Lethal genes in mice
The gene that determines coat color in mice is a
zygotic lethal gene. In mice, there are two colors
determined by these genes i.e. yellow and grey
(agouti).
Example 1
If two yellow mice are crossed they produce both
yellow and grey offspring however these offspring
appear in a phenotypic ratio of two yellow; 1 grey
instead of 3:1.
This is because the homozygous dominant yellow
micedie in the uteruswhich reducesthephenotypic
ratio. The yellow mice produced are always

heterozygous and this changes the monohybrid
genotypic ratio from 1:2:1 to 2:1.
Example 2:
In chicken, a normal condition of legs and an
abnormalleg growth,calledcreepers,arecontrolled
by two genes, which express incomplete dominance
and lethalgenes. Thenormalconditionis dominant
over the creeping condition. However,
heterozygotes of this condition all appear to have
thecreeping condition.Whentwoheterozygotesare
crossed, the 3:1 ratio is never obtained, since the
recessive homozygous condition is lethal during
incubation as shown below:
The ratio of the F1 phenotypes is 2: 1 since the
homozygous recessive chick is never hatched due to
the gene for the creeping condition being lethal
gene:
Numerical problem:
1. In maize, the amount of chlorophyll on the
plant is controlled by a gene, such that the allele
for green leaves is dominant over the allele for
white leaves. When the allele for white leaves is
expressed as a homozygous state, the plants fail to
develop and die due to the absence of chlorophyll
for the synthesis of food. Using genetic symbols,
well defined, carry out a cross between a pure
breeding normal plant and a white plant to
determine the phenotypes of the F1 and F2 plants
2. In snapdragon plants, three phenotypes of
plants are green, golden/Auria plants, and white
plants, and the gene for white plants is lethal in
the homozygous state. The heterozygotes are
normal and are called golden plants, though they
have patches of white and green mixed. When two
heterozygotes were bred, the expected
monohybrid ratio was not obtained. Using
suitable genetic symbols, carry out the cross to
determine the phenotypes and genotypes of the
plants from the breeding. Explain your results.
Note:
a) Dominant lethal genes are very rare in a
population because they are usually manifested
easily in the growth and development of the
offspring at an early age and hence easily
eliminated.
b) A pleiotropic gene is the one that controls
more than one aspect or characteristic in the
metabolism of an organism
or it is a gene that has several phenotypic effects on
an organism e.g. the Y gene in mice is controlling
both viability and coat color, for viability the Y gene
acts as a recessive gene since homozygous YY mice
die in the uterus and since Yy mice are yellow this
phenomenon is called pleiotropy.
Other examples of pleiotropic genes include sickle
cell anemia and the gene for the production of
earwax in man.
2. CO-DOMINANCE
This is a phenomenon whereby the alleles
controlling a particular characteristic have equal
powers of expressing themselves in the phenotype
in the heterozygote. Therefore the offsprings
produced will have a mixture of the two parental
characteristics in the phenotype.

Co-dominance is taken to be a form of incomplete
dominance since no allele suppresses the
phenotypicexpression ofanother.Inco-dominance
which uses a capital letter to represent all the two
alleles each letter corresponding to each of the two
characteristics.
Examples of co-dominance include the
following;
a. The gene that determines coat color in
cattle
b. Inheritance of blood group AB in man
c. Inheritance of sickle cell trait
Given that both alleles of the same gene are
dominant, we let a single letter for the gene and
alleles be attached as superscripts. I.e. CR and CW
or simply R and W represent alleles for red and
white petals respectively. The third phenotype
results from flowers of the heterozygotes (CR CW or
simply RW) having less red pigment than the red
homozygotes.
Inheritance of coat color in cattle.
Consider a cross between a red bull and a white cow
whose F1 offsprings are selfed. Workout the
genotypes and phenotypes in F1 and F2 generation
stating in each case the ratios.
Let R represent the allele for red coat
color
Let W represent the allele for white coat
color.
F2 phenotype. 1 red, 2 roan, and 1 white.
Inheritance of sickle cell anemia.
This is an abnormal condition in which the red
blood cells collapse into a sickle shape under low
oxygen concentration due to the presence of
abnormal hemoglobin (Hbs) in the red blood cells.
Normal hemoglobin is found in red blood cells with
a biconcave disc shape.
3. INCOMPLETE DOMINANCE
This is a condition in the heterozygous where
neither of the alleles is dominant over the other and
the phenotype of the offspring is an intermediate
between that of the parents. It mainly occursin
plants. E.g. in plants, when a red-flowered plant is
crossed with a white-flowered plant, the offspring
produced pink and white flowers in a ratio of 1:2:1
respectively.
Example 1.
Consider petal color in flowers.
Let the gene for red petal flowers be R.
let the gene for white flowers be W
F1 phenotype: all pink petals.
Selfing F1. (Cross between offspring in F1)
RW

F2 phenotype. 1 red, 2 pink, and 1 white.
Phenotypic ratio; 1 red: 2 pink: 1 white. (1:2:1)
Other examples of incomplete dominance
include:
Characterist
ic
Allelomorph
ic
characteristi
cs
Heterozygo
us
phenotype
Mirabilis
Japalla (4-
oclock plant)
Red and White Pink
Angora rabbit
hair length
Long and short Intermediate
Plumage color
in Andalusian
fowls
Black and
splashed white
Blue
MULTIPLE ALLELES
These are three or more forms of the same gene
occurring at the same locus.
Most genes are known to occur in two alternative
forms (allelic forms) located on the same locus of
homologous chromosomes. Some genes are known
to occur in more than two allelic forms called
multiple alleles of which any two can occupy the
gene locus in a diploid organism. This is easily
noticed for the gene responsible for blood groups in
man.
Examples of multiple genes include;
 The ABO blood group system.
 Eye color in mice
 Coat color in mice
Inheritance of Blood Groups
ABO Blood group
 The ABO is determined by an autosomal gene
whose locus is represented by a symbol I
standing for isohaemagglutinogen. The
gene has more than three alleles or it has
multiple alleles; A, B, and O, whereby alleles A
and B are codominant, while O is recessive to
both A and B.
 When an individual inherits any one of the
following combinations of alleles via
fertilization, he or she tends to possess any of
the four blood groups A, B, AB, and O;
 Genotype IAIA or IAI0 determine the
formation of agglutinins/antibody b in the
plasma and antigen/agglutinogen A on the
surfaceofthe erythrocyteandtheindividual
has blood group A
 Inheritance of BB or BO leads to the
production of agglutinins/antibodies `a` in
the plasma and antigen `B` on the
erythrocyte hence the possession of blood
group B
 Genotype IAIB leads to the production of
both antigens A and B on the erythrocytes
without agglutinins or antibodies in the
plasma, hence the individual has blood
group AB.
 Genotype IOIO codes for the production of
agglutinins `a` and `b` without
antigens/agglutinogens on the red blood
cells, enabling the individual to have blood
group O.
Alleles A and B show co-dominance to each other,
yet both are dominant to allele O. The three alleles
give 6 possible genotypes and four phenotypes.
Genetic diagrams involving multiple alleles are
constructedin thesameway asbefore.Forexample,
you couldbe asked touse a geneticdiagram toshow
how parents with blood groups A and B could have
a child with blood group O.

Example:
1. Work out the possible blood groups of the
offsprings produced if a man of blood group
A marries a woman of blood group AB. The
man can have two possible genotypes, i.e.
2. A man having blood A marries a woman
having blood group AB. What are the
possible genotypes and phenotypes of their
offsprings if the man is heterozygous for
blood group A?
Let IA represent the allele for the formation of
antigen A
Let IB represent the allele for the formation of
antigen B
Let IO represent the allele for no formation of
antigen A or B.
Results from other crosses
Parents
blood
group
Parents
phenotypes
Offspring
genotype
Offspring
blood
group
A X O IAIA X I0I0
IAIO X IOIO
IAIO
I0I0 and
IOIO
A
A and O
B X O IBIB X I0I0
IBI0 X I0I0
IBIO
IBI0 and
IOIO
B
B and O
A X B IAIA X IBIB
IAIO X IBIB
IAIA X IBIO
IAIB
IAIB, ; IBIO
IAIB; IAIO
AB
AB and B
AB and A
AB X O IAIA X I0I0 IAIO, IBIO A and B
Physiology of the blood groups in humans
Human blood contains blood group antigens and
blood group antibodies. Some of these specifically
determine blood groups e.g. allele A determines the
production of antigen A, allele B determines the
production of antigen B and allele O does not code
for the production of any antigens. Antigens A and
B occur on the plasma membranes of red blood
cells. These antigens have corresponding protein
molecules known as blood group antibodies
(agglutinins) in blood plasma. These antibodies can
react with the antigens under the lock and key
hypothesis should they be similar to the antigens
brought into the recipient’s blood, leading to the
formation of a precipitate or an agglutinate in
blood.

Therefore an individual should not have
blood group antibodies corresponding or
similar to his bloodgroup antigensto avoid
agglutination. Consequently, individuals should
have the following antibodies not corresponding to
their antigen to avoid blood clotting.
The importance of blood groups
a) They are important during blood
transfusion where they are used to
prevent agglutination (precipitation) of the
blood of the recipient. To avoid
agglutination, the donor's blood group
should be compatible (matching with) that
of the recipient by having the donor's blood
group antigen that is different from the
blood group antibody of the recipient.
When the recipient gets antibodies from the donor,
such antibodies become diluted in the recipient’s
blood and so cause either minor clotting of blood or
no blood clotting at all and so cannot lead to the
death of the recipient. However, in case the donor
introduces anantigenthatis similar to theantibody
of the recipient, it stimulates the recipient’s blood
to produce more antibodies which attack and react
with the donor’s antigen to cause severe blood
clotting. Therefore an individual with a specific
antigen on the red blood cell membrane does not
possess its corresponding antibody in the blood
plasma to avoid agglutination.
Blood plasma permanently contains two blood
group antibodies a and b which do not correspond
with a specific antigen in blood to avoid
agglutination e.g. a person with blood group A has
antigen A and antibody to avoid agglutination. A
person with blood group B cannot donate blood to
a person ofblood O becauseantigenBinthedonor’s
blood will be attacked by antibody b in the
recipient’s blood leading to agglutination.Thesame
applies to blood group A and blood group AB
donors to blood group O recipients.
b) They are used in settling court cases
about who the father of the child is (i.e.paternity
suits). Although blood groups cannot prove
below reasonable doubt who the father of the
child is it is possible to use their inheritance to
show thatanindividualcould be thefatherof the
child. Consider a mother who is of blood group
O having a child of blood group O and the child
produced also with blood group O. she claims
thatthe fatheris a manwhosebloodgroup is AB.
Since the child is blood group O its only possible
genotype is IOIO and it must therefore have
inherited one IO allele from each parent. Since
the man is of blood group AB he cannot donate
the IO to the child and therefore he cannot be the
father of the child.
Even if the father was found to be of another blood
group such as blood group A still the evidence will
be insufficient because any other man can possess
such a blood group and donate the IO allele to the
child. Therefore a DNA test should be carried out to
confirm who the father of the child is.
c) Blood groups can also be used as evidence
of evolution. This is because organisms of
different species having similar blood group
systems such as the ABO system are
believed to have originated from the same
ancestor in the course of evolution for
example humans, chimpanzees, gorillas,
Baboons e.t.c.
Questions:
1. A man of blood group A married a woman
homozygous for blood group B, and they produced
a son of blood group B.
(a) Work out the genotypes of the father and the
son

(b) The son married a wife of blood group O,
showing your working, show the percentages of the
possible phenotypes of their offsprings.
2. What are the genotypes of blood groups of the
children borne to a man of blood group A and a
woman of blood group B both of whom are
heterozygous.
3. A man who is homozygous for blood group A
married a woman who is homozygous of blood
groupB.Whatarethegenotypesoftheir offsprings?
DIHYBRID INHERITANCE AND MENDEL’S
SECOND LAW OF INHERITANCE
Dihybrid inheritance refers to the inheritance of
two pairs of contrasting characteristics
simultaneously.
For instance, in one of his experiments; Mendel
crossed pure breeding tall pea plants with red
flowers with pure breeding dwarf plants having
white flowers. All in the F1 progeny were tall with
red flowers. This showed just like Mendel had
discovered before that the alleles for tallness and
red flowers were dominant to those for dwarfs and
white flowers respectively.
Mendel went ahead to self-pollinate the F1 plants
and obtained an F2 progeny, this comprised of a
variety of phenotypes as summarised in the table
below.
 315 Tall with red flowers
 101 Tall with white flowers
 108 Dwarf with red flowers
 32 Dwarf with white flowers
These give the respective phenotypic ratios as
9:3:3:1. This is known as Mendel’s Dihybrid
ratio; the ratio of phenotypes in the F2 generation
for a Dihybrid cross.
From this and many other similar crosses, Mendel
was able to make the following observations:
 Both phenotypes/characters (height and
flower color) combined in the F1 but
separated andbehavedindependently in the
F2.
 Two of the F2 phenotypes resembled one or
the other of the parental phenotypes
WHILE two new combinations of
phenotypes appeared in the F2; (Tall/white
and Dwarf/red). These are known as
recombinants.
 The allelomorphic pairs of characteristics
(controlled b different alleles of the same
gene) occurred in a phenotypic ratio of 3
dominant: 1 recessive.E.g.3tall: 1 dwarfand
3red: 1 white.
Basing on these observations, Mendel formulated
his second law known as the law of independent
assortment. The law states that; “Any one of a
single pair of characters may combine
randomly with either one from another
pair”
Below is a full genetic explanation of the 9:3:3:1
ratio of phenotypes in the F2generatioon of a
dihybrid cross.
Phenotypic ratios: All Tall with red flowers.
By selfing F1 plants;

(c) By carrying out a test cross between F1 tall red-
flowered pea plant with short flowered pea plants.
Parental phenotype: tall red-flowered pea plant x
short whit flowered pea plant
Parental genotype: TtRr x ttrr
Test cross genotypes (2n): punnet square to show
the fusion of gametes for genotypes and
phenotypes.
Punnett square to show the fusion of gametes to
form F2 genotypes and phenotypes.
Test cross phenotypic ratio: 1 tall red-flowered: 1
tall white-flowered: 1 short red-flowered: 1 short
white-flowered.
This ratio proves that the F1 plants were
heterozygous (TtRr).
Note: In dihybrid inheritance, some of the
offsprings formed in the F2 have a mixture of the
two parental phenotypes that gave rise to F1, and
such offsprings areknownasrecombinants while
other offsprings in F2 resemble one of the two
parental phenotypes that gave rise to F1 and such
offsprings are known as parentals.
Recombinants arise when crossing over takes place
during the formation of gametes in meiosis which
leads to the mixing of the two parental
characteristics. The number of recombinants in F2
is usually smaller than that of the offsprings which
resemble the parental phenotypes (parental
offsprings). This is because crossing over occurs by
chance which reduces the number of recombinants
formed.
NB: When performing a dihybrid cross;
 Alleles of the same gene cannot pass into the
same gamete (they segregate during meiosis).
I.e. T can only be present with Y or y but not t
while t can only be present with or y but not T
as in the above case
 The possible combination of gametes during
fertilization is shown in a Punnett square
(after the Cambridge geneticist R. C.
Punnett). This minimizes errors when listing
the combinations.
In summary; the following can be noted from
Mendel’s hypotheses:
 Each characteristic of an organism is
controlled by a pair of alleles.
 During meiosis, each pair of alleles
segregate (separates) and each gamete
receives one of each pair. This is known as
the law of segregation.
 During gamete formation, either one of a
pair of alleles can pass into the same
gamete with either one from another pair.
This is known as the law of independent
assortment.

 Each allele is transmitted one generation
to the next as a discrete unit
 Each diploid organism inherits one allele
for each character from each of the two
parents.
 If an organism has twounlike alleles for a
given gene, one may be expressed
(dominant) at the total exclusion of the
other (recessive).
EXPLANATION OF THE LAW OF
INDEPENDENT ASSORTMENT:
Mendel’s second law can be explained/accounted
for on the chromosomal basis by meiosis.
During the formation of gametes by meiosis, the
distribution of each allele from a single pair is
entirely independent of alleles from other pairs.
This in turn depends on the random orientation of
homologous chromosomes onto the equatorial
spindle in metaphase I. Subsequent separation
during anaphase I lead to a variety of allele
combinations in gametes. In this process; any one
of a single pair of alleles can combine randomly
with either one from another pair.
Illustration of the law of independent
assortment.
Example 2
In Drosophila melanogaster flies, the gene
determining the size of the abdomen occurs on the
samechromosomewith thatdeterminingthelength
of the wings. When pure breeding broad and long-
winged female fly was crossed with a narrow and
vestigial winged male fly all the F1 offsprings
obtained head broad abdomen and long wings. If
the F1 offsprings were selfed to obtain F2.
a. Using suitable genetic symbols to work out the
phenotypes and genotypes that were obtained in
the F2 generation.

b. Suppose 480 flies were obtained in F2 to work
out the numbers of the flies for each phenotype
class.
c. How many of these flies were recombinants.
Approach
Let B represent allele for broadness
Let b represent allele for narrow
Let N represent allele for long
Let n represent allele for vestigial winged.
F2 genotypic ratio: 9B_N_:3B_nn: 3bbN_:1bbnn
F2 phenotypic ratio: 9 broad long-winged: 3 broad
vestigial winged: 3 narrow vestigial winged: 1
narrow short-winged
Example 3
Whenpure breeding broad andlong-wingedfemale
fly was crossed with a narrow and vestigial winged
male fly all the F1 offsprings obtained head broad
abdomen and long wings.
a) Using suitable genetic symbols work out the
phenotypes and genotypes that were obtained
in the F2 generation.
b) Suppose 480 flies were obtained in F2 to work
out the numbers of the flies for each phenotype
class.
c) How many of these flies were recombinants.
Solutions:
Obtaining the F2 generation:

(b) Phenotypic ratios = 9:3:3:1, Total ratio =
(9+3+3+1) = 16
Number of flies = (
𝑅𝑎𝑡𝑖𝑜
𝑇𝑜𝑡𝑎𝑙
)𝑥480 Flies
i. Broadabdomen,long winged=
9
16
X
480 = 270 flies
ii. Broad abdomen, vestigial winged =
3
16
X 480 = 90 flies
iii. Narrow abdomen, long winged =
3
16
X 480 = 90 flies
iv. Narrow abdomen, vestigial winged
=
1
16
X 90 flies
(c)Number of recombinants = (90 + 90) flies = 180
flies
Example 2
A species of mammal has either white or grey fur,
or either a long or short tail. Several crosses
between a heterozygous animal with white fur and
long tail, and an animal with grey fur and short tail,
produced these offspring:
White fur, long tail 12
White fur, short tail 2
Grey fur, long tail 2
Grey fur, short tail 12
Carry out a genetic cross to determine whether
these results show that the genes for fur color and
eye color are on the same chromosome.
EXCEPTIONS TO MENDEL’S LAWS
It should however be noted with concern that
Mendel’s laws of inheritance are not of universal
application to all processes of inheritance in
organisms, For the work that led to his two laws of
inheritance,Mendel chosepea plantcharactersthat
turn out to have a relatively simple genetic basis:
Each character is determined by one gene, for
which there are only two alleles, one completely
dominant and the other completely recessive. But
these conditions are not met by all heritable
characters, and the relationship between genotype
and phenotype is rarely so simple. In this section,
we will extend Mendelian genetics to hereditary
patterns that were not reported by Mendel.
These arereferred to asexceptions to Mendel’s laws
ofinheritancebecausethey never producethe 3:1 or
the 9:3:3:1 ratios of phenotypes in monohybrid and
dihybrid crosses respectively.
1. LINKAGE
Linked genes are more than twogenes located on
the same chromosome but controlling different
characteristicsandareinherited togetherasa single
black.
Linkage is the occurrence of more than one gene
on the same chromosome which are inherited
together along with the chromosome as a single
block.
Linked characteristics are the ones controlled
by genes located on the same chromosome and so
are transmitted together with the chromosome
from generation to generation.
The process of linkage was explained by an
American scientist Thomas H. Morgan. In a cross
between a grey, long-winged drosophila
heterozygous for both traits with a black, vestigial
winged drosophila (This is a test cross); Morgan
predicted that in the normal Mendelian
inheritance; parental phenotypes and
recombinantswouldbeobtained ina ratioof 1:1:1:1.
In an experiment with Drosophilla, the genes for
wing length and body colour were linked and were
expected to produce parental phenotypes in a ratio
of 1:1.However,evenafterperformingthetestcross
several times, Morgan never obtained the predicted
outcomes. He instead obtained approximately
equal numbers of the parental phenotypes with
significantly few recombinant phenotypes also in
approximately equal numbers as summarised
below.
41.5% grey, long winged
41.5% black, vestigial winged
8.5% grey, vestigial winged
8.5% black, long winged
Morgan made the following deductions from his
findings:
i. Genes located in the same chromosome
tend to stay together during inheritance,
and are said to be linked.
ii. Genesarearrangedin a linear fashionin the
chromosomes;
iii. The linkage is broken down by the process
of crossing over occurring during meiosis;
iv. The intensity of linkage between two genes
is inversely related to the distance between
them in the chromosome;
v. Coupling and repulsion phases are two
aspects of linkage

Forms of linkage
a) Complete/total linkage; is a condition
where the genes responsible for different
characters fail to assort independently and
segregate during meiosis I of cell division.
Total/complete linkage is when the distance
between linked genes is not sufficient to allow
for successful crossing over. Such genes form a
linkage group and pass into the same gamete
during meiosis and are therefore inherited
together. As a result, these genes do not show
independent assortment (applies to genes on
non-homologous chromosomes) and fail to
produce the 9:3:3:1
Example
In drosophila flies, the genes controlling body
color and the length of wings occur on the same
chromosomes and are linked together. Consider a
cross between a pure breeding grey-bodied long-
winged fly with a black-bodied vestigial
winged fly whereby the grey-bodied is female
while the black-bodied is female. If all the F1 flies
obtained here grey bodied and long-winged what
are the phenotypic and genotypic ratios of the F2
flies.
Let G represent the allele for grey body
Let g represent the allele for black body
Let L represent the allele for long wings
Let l represent the allele for vestigial wings.
In the F2, a 3:1 ratio of grey long-winged and
black vestigial winged (the original
parental) phenotypes are obtained.
b) Incomplete linkage; is a situation where the
two linked genes assort independently and
segregate during meiosis I of cell division,
resulting in the formation of new recombinants
in the gametes. In this case, crossing over
between homologous chromosomes takes place
at the chaisma and recombinant gametes are
formed. E.g in the above example, the
phenotypic ratio is 3 greys long-winged: 1 black
vestigial winged. However, in the case of
incomplete linkage, the 3:1 ratio of parental
phenotypes is neverobtained in practice. This is
because the total linkage is rare. Instead,
approximately equal numbers of parental
phenotypes are obtained with significantly few
recombinant phenotypes also in approximately
equalnumbers.Twoormoregenesaresaidto be
linked if recombinant phenotypes occur much
less frequently than parental phenotypes.
Example 1:
A plant has genes for flower color (allele Y for
yellow and y for white flowers) and petal size (P for
largepetals andp forsmallpetals). A cross between
two plants that are heterozygous for both alleles.
Construct two genetic diagrams to show the
expected offspring
ratios from this cross

(a) If the genes are linked with YP on one
chromosome and yp on the other, and;
(b) If the genes are not linked.
Approach
This would give a ratio of 3 yellow, large: 1
white, smallifthe genesare linked. Ifthereis some
crossing over, we would expect to get small
numbers of other phenotypes.
b) If there is no linkage, these alleles show
independent assortment, and the 9: 3: 3: 1 dihybrid
ration will be obtained as shown below.
Example 2
In sweet peas, the genes for flower colour and
shape of pollen grains are linked. When a cross is
made between blue flower and long pollen (BBLL)
with red flower and round pollen (bbll) all the F1
offsprings had a blue flower and long pollen (BbLl)
in heterozygous condition.
a) Carry out a genetic cross to obtain the F1
and and F2 offsprings, assuming that
linkage was not complete
Solution
b) When a test cross between blue and long
(BbLl) anddouble recessive (bbll) wasdone,
the following results were obtained:
 Blue long (43.7%),
 Red round (43.7%),
 Blue round (6.3%) and
 Red long (6.3%).
The parent combinations are 87.4% are due to
linkage in genes on two homologous chromosomes,
while in case of new combinations (12.6%) the
genes get separated due to breaking of
chromosomes at the time of crossing over in
prophase-I of meiosis. New combinations in the
progeny appeared due to incomplete linkage
Question:
A homozygous purple-flowered short-stemmed
plant was crossed with a homozygous red-flowered
long-stemmed plant and all the F1plants had purple
flowers and short stems. When the F1generation
was taken through a test cross, the following
progeny was produced
53 purple-flowered short-stemmed
47 purple-flowered long-stemmed
49Red flowered short-stemmed
45 red-flowered long stems.
Explain the results fully.
CROSSING OVER AND CROSS OVER
VALUES
During crossing over, the frequency of crossovers
that take place was found to be dependent on the
distribution and arrangement of chromosomes.
This is given by the cross-over value/frequency aka
recombination frequency. This is calculated as a
percentage ratio of recombinants to the total
number of offsprings.
𝐶𝑜𝑉 =
𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑟𝑒𝑐𝑜𝑚𝑏𝑖𝑛𝑎𝑛𝑡𝑠
𝑇𝑜𝑡𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑜𝑓𝑓𝑠𝑝𝑟𝑖𝑛𝑔𝑠
𝑥100
Example
In a test cross carried out on a grey long-winged
drosophila, the following results were obtained
Phenotype Number of
offsprings

Grey, long-winged 965
Black, vestigial 944
Black, long-winged 206
Grey, vestigial
winged
185
Calculate the cross over value
𝐶𝑜𝑉 =
𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑟𝑒𝑐𝑜𝑚𝑏𝑖𝑛𝑎𝑛𝑡𝑠
𝑇𝑜𝑡𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑜𝑓𝑓𝑠𝑝𝑟𝑖𝑛𝑔𝑠
𝑥100
𝐶𝑜𝑉 =
206 + 185
(965 + 944) + (206 + 185)
𝑥100
= 17%
The COV also indicates the relative distance
between linked genes and the possibility of
successful crossing over during meiosis, in the
above case the distance between adjacent genes is
17 units. These values can also be used to position
genes along the chromosome a process called gene
mapping.
NB:
The larger the C.O.V, the more separated the two
genes are on the chromosomes and the higher the
chances of crossing over taking place. The
illustration of the distance between the genes on
the chromosome gives the chromosome map
How to construct a gene map
Consider the cross-over values involving different
genes P, Q, R, and S.
The distance separating these four genes is shown
below;
P-Q = 24%
R-P = 14%
R-S = 8%
S-P = 6%
Draw the chromosome map to show the position of
these chromosomes.
Answer.Draw thechromosomemapforthesegenes
a. Insert the positions of the genes with the
smallest crossover value first in the middle of the
chromosome map
b. Examine the next largest cross-over value
andinsert both possible positions of its geneson the
chromosomes relative to either S or P.
c. Repeat the procedure for the entire
remaining cross-over values until you reach the
largest cross-over values.
Example 1
In drosophila flies, the genes controlling body color
and eye color occur on the same chromosome and
are linked together. In an experiment, a
heterozygous female fly for grey body and normal
eyes was crossed with a black body
and purple-eyed fly. In these flies, the grey body is
dominant over black while normal eyes are
dominant over purple flies. If 1000 offsprings were
obtained from this cross as shown in the table
below;
a) Carry out a genetic cross using suitable
symbols to determine the test cross ratio
b) Calculate the cross-over value.
c) Draw a crossing over map for the above
results
Solution
a)
Parental phenotype: grey body normal eyed fly x
black body purple-eyed fly.
Test cross genotypes: Punnet square to show the
fusion of gametes
Test cross phenotypic ratio
1 grey normal eyed fly: 1 grey purple-eyed: 1
black normal eyed: 1 black purple-eyed.
The obtained results in the test cross differ from the
expected ones becausethegenes arelinked together
on the chromosomes and were separated by
crossing over which occurs by chance hence
resulting in the formation of fewer recombinants
compared to the parents.
b)

c)
Example 2
A further experiment on these flies indicated that
the genes for the body color, length of wings, and
eye color are on the same chromosomes. Using the
information in the table below calculates the cross-
over value and illustrates the distance between the
genes.
Solution
FACTORS THAT AFFECT CROSSINGOVER
1) The relative distance between the genes
on the chromosome. When the genes are far
apart from each other on the chromosome, they
have high chances of forming chiasmata in between
thereby leading to genetic exchange on the other
hand when genes are very close to each other on the
chromosome, their chances of forming chiasmata is
limited.
2) The position of the centromere on the
chromosome. If the genes are very close to the
centromere their chances of undergoing genetic
exchange are limited. However, if the genes are far
away from the centromere, there are high chances
that they can be exchanged by crossing over.
3) Temperature. Crossing overdecreases with an
increase in temperature because the process of
meiosis requires a suitable temperature that can
promote efficient crossing over.
4) Age of the organism. An increase in age
lowers the chances of crossing over.
Meiosis is more efficient in grown-up adults before
the menopause stage in females and before
senescence in males.
5) Mutagens. These can decrease or increase the
rate of crossing over. The chances of crossing over
are greatly reduced by the presence of chemical
substances that inhibit chiasmata formation
thereby preventing cross over e.g. in drosophila
flies.
Exercise
In maize, the genes for colored seed and full seed
are dominant to the genes for colorless and
shrunken seed. Pure breeding strains of double
dominant variety were crossed with a double
recessive variety anda test cross ofthe f1 generation
produced the following results
Colored full 380
Colorless shrunken 396
Colored shrunken 14
Colorless full 10
Calculate the distance between the genes for
colored seed and seed shape.
2. INHERITANCE OF
COMPLEMENTARY GENES
These are two or more genes that interact together
to control a single character in an organism.
Inheritance of these genes, therefore, does not
agree with Mendel’s laws of inheritance. Although
these genes control a single characteristic, they
show independent assortment. Therefore these
genes are passed on from the parents to the
offspring in a normal Mendelian fashion. The best
example of complementary genes are genes that
control the shape of combs in chicken.
In chicken there are four types of combs namely;
i. walnut comb
ii. single comb
iii. pea comb
iv. rose comb

These four types of combs are controlled by the two
genes located at two loci situated on different
chromosomes and which interact together to give
rise to the four comb types. The shape of the combs
is controlled by two genes which are represented by
two alleles shown below;
Let P represent the allele for pea comb
Let R represent the allele for rose comb
The pea comb develops in the presence of the P-
allele and the absence of the R-allele while the rose
comb develops phenotypically in presence of the R-
allele and the absence of the P-allele. When both
alleles, P and R,are present together a walnutcomb
develops. A single comb appears only in the
homozygous double recessive condition
Consider a cross between a pea comb-shaped crock
with a rose combed hen whose F1
offspring is then selfed.
What is the phenotypic ratio obtained in F2?
Punnett square to show the fusion of gametes to
form F2
genotypes and phenotypes
The walnut and single combs are produced by the
interaction of the genes at both loci as summarised
below:
Name
of
comb
Production Possible
genotypes
Pea
comb
Dominant allele P
but without
dominant allele R
PPrr, Pprr
Rose
comb
Dominant allele R
but without
dominant allele P
ppRR, ppRr
Walnut
comb
Dominant alleles
for both P and R
PPRR,
PpRR, PPRr,
PpRr
Single
comb
Only by
homozygous double
recessive condition
pprr
F2 genotypic ratio: 9PPRR :3PPrr: 3ppRR
:1pprr
F2 phenotypic ratio: 9 walnuts combed: 3
peas combed: 3 rose combed: 1 single
combed.
In this inheritance, the genes are usually situated at
different loci at different chromosomes from where
they interact together and give rise to four distinct
phenotypes for a single characteristic.
The walnut comb results from a modified form of
co-dominance in which at least one dominant allele
of either pea comb or rose comb is present.
This is an incidence whereby a 9:3:3:1 phenotypic
ratio is obtained for a single characteristic.
Although this ratio is this pattern of inheritance
differs from the hybrid inheritance because;
1. The F1 progeny (offsprings) resembles
neither parent i.e. they are all walnut comb-
shaped, unlike their parents.
2. The F2 progeny also contains two new
phenotypes which do not exist in the F1
parents namely walnut and a single comb-
shaped and these appear in a higher ratio as
compared to the rose and the pea comb.

Example 2
the gene for flower color in sweet pea is controlled
by two dominant alleles P and C, which interact
together
to produce a purple color. The absence of both
dominantalleles togetherwillproduce a whitecolor
in the flower petals. i.e;
PPCC
PPCc
PpCc
PpCC
(P and C are both present for purple to be formed)
ppCc
PPcc
Ppcc
ppCC
(P and C are not present)
An explanation for the inheritance of color
of petals
The production of colored pigment is regulated by
an enzyme, which is responsible for the production
andpresence ofcolored pigmentinpetals hencethe
purple or red color of flowers. The dominant allele
of the gene is responsible for the presence or
deposition of pigment, while the homozygous
recessive alleles of this gene are responsible for the
absence of pigment. Likewise, the dominant alleles
of a second gene in homozygous or heterozygous
conditions causes the production of an enzyme that
is necessary for color production from pigment,
while homozygous condition does not produce any
such enzyme. At least one of each allele must be
present if a colored pigment is to be formed and
deposited on the petals.
Question
A cross between pure breeding varieties for purple
andwhiteflowersof sweet pea wasdone and selfing
the F1 generation, offsprings did not give F2
generation phenotypes in the ratio 9:3:3:1, instead
gave 9:7 ratio. Explain your answers with suitable
genetic symbols.
Solution
Let PPCC represent genotype for purple flowers
Ppcc represents the genotype for white flowers.
SUPPLEMENTARY GENES
These aregenes with noeffecton their ownbutwith
another gene to produce a new phenotype e.g the
common mice have three coat colors ie agouti and
albino. In mice, two genes affect fur color:
Gene for distribution of melanin in hairs: A
produces banding (agouti), a produces a uniform
distribution.
Gene for the presence of melanin in hairs: B
produces melanin, b does not produce melanin.
Clearly, gene A/a can only have an effect if the
pigment melanin is present. If there is no melanin,
the fur color is white. The genotypes and
phenotypes from this interaction are as shown
below:
Question:
a) Construct a genetic diagram to predict the
ratios of phenotypes resulting from a cross
betweentwomicewith thegenotypesAaBB
and AaBb.
b) How could you use a test cross to find out
the genotype of a white mouse?
Approach
a)
b) A white mouse could have the genotype
AAbb, Aabb, or aabb. We could cross this
Purple
colour
genotype
White
Flowers

mouse with a pure-breeding black mouse,
with the genotype aaBB.
If the unknown mouse has the genotype
AAbb, then all the offspring will be agouti
(AaBb).
Aabb, then half the offspring will be agouti
(AaBb) and half will be black (aaBb).
aabb, then all the offspring will be black
(aaBb).
3. EPISTASIS
This refers to a condition in which non-allelic genes
interact during which the epistatic allele of the
epistatic gene on one locus suppress the phenotypic
expression of the hypostatic allele of the hypostatic
gene on another locus.
An epistatic allele is the one which suppresses
another allele in the phenotype though they are not
located on the same locus and the suppressed gene
or allele at another locus is the hypostatic allele.
They are 3 types of epistasis which include the
following;
i. Dominant epistasis
ii. Recessive epistasis
iii. Isoepistasis
Dominant epistasis. This is the type of epistasis
where the epistatic allele is dominant such that is
presence suppresses the phenotypic expression of
the recessive allele on another locus. This type of
epistasis changes the phenotypic ratio from 9:3:3:1
to 12:3:1.
Examples
One of the best known examples of a 12:3:1
segregation ratio is fruit color in some types of
squash. Alleles of a locus that we will call B produce
either yellow (BB) or green (bb) fruit. However, in
the presence of a dominant allele at a second locus
that we call A, no pigment is produced at all, and
fruit are white. The dominant A allele is therefore
epistatic to both BB and bb combinations. One
possible biological interpretation of this
segregation pattern is that the function of
the A allele somehow blocks an early stage of
pigment synthesis, before neither yellow nor green
pigments are produced.
Illustration.
Parental phenotypes: Black Male X Black
Female
Parental genotypes: AaBb X AaBb
Gametes
Using Punnet squares
Recessive epistasis.
This is thetype of epistasis wheretheepistatic allele
is recessive, such that its presence in homozygous
condition, suppresses the phenotypic expression of
the dominant allele located on another locus. This
type of epistasis changes the dihybrid phenotypic
ratio from 9:3:3:1 to 9:3:4.
Example:
In the Labrador Retriever breed of dogs. the B locus
encodes a gene for an important step in the
production of melanin. The dominant allele, B is
more efficient at pigment production than the
recessive b allele, thus B_ hair appears black,
and bb hair appears brown. A second locus, which
we will call E, controls the deposition of melanin in
the hairs. At least one functional E allele is required
to deposit any pigment, whether it is black or
brown.Thus,allretrievers thatare ee fail to deposit
any melanin(andsoappearpale yellow),regardless
of the genotype at the B locus.
In a cross between two heterozygotes for the genes
was done as follows:
Parental phenotypes: Black Male X Black
Female
Parental genotypes: EeBb X EeBb
Gametes

Using Punnet squares
Phenotypicratio: 9black: 3 brown: 4 brown
Isoepistasis. This is the type of epistasis in which
both alleles and the non-allelic genes have equal
powers of suppressing each other in the phenotype.
This modifies the d ihybrid phenotypic ratio to 15:1.
Summary of F2 genotypes due to epistasis
 Complementary genes (9:7)
 Supplementary genes (9:3:4)
 Duplicate genes (9: 6: 1)
Cumulative effect
 duplicate dominant gene (16: 1)
 dominant gene (12 : 3 : 1)
 dominant and recessive interaction (
13:3)
BIOCHEMICAL EXPLANATION OF
EPISTASIS
The expression of a character is as a result of series
of enzymes-mediated biochemical reactions.
Production of each enzyme depends on specific
gene for example; gene pathways for synthesis of
sweet pea pigments.
P A C
E1, E2 and E3 are specific enzymes whose synthesis
is controlled by genes G1, G2 and G3
A and B are intermediate white compounds.
C is the coloured compound and the product
pigment.IfG2becomesa mutatedgene,then Bwill
not be formed and C would also not be formed.
The white compound A will therefore accumulate.
G3 (gene 3) would therefore not have any effect on
the phenotype. Therefore, gene G2 is the epistatic
gene while G3 is hypostatic gene.
Examples
In oats the inheritance of color is controlled by the
epistatic gene which hastwoalleles,oneallele being
dominant for color appearance while the other
allele is for no color formation (white or albino) i.e.
the hypostastic gene is responsible for color
deposition or type of color,Where by black is
dominant over white.
Consider a cross between homozygous black oat
plant with a homozygous white oat plant and then
the F1 plants are selfed to get F2.
a. Were out the phenotypic ratio of the F2
generation
b. How many individuals are found in each of the
phenotypic classes obtained in F2 if 130
individuals were found in F2?
Solution
Let G represent the allele for color formation
(epistatic)
Let g represent the allele for no color formation
(albino white)
Let B represent the allele for black
Let b represent the allele for grey.
G1
E1
G2
E2
G3
E3
G1
E1
G1
E1
Precursor
Intermeediate
cpd (white)
Colour
pigment (e.g
purple
B

INHERITANCE OF SEX AND SEX
DETERMINATION
Some ancient Greeks thought that sex depends on
the testicle from which the sperm comes, some
European kings tied off or removed their left testes
to ensure a male heir to the throne. Others believed
that the sex depends on the phase of the moon
during conception, wind direction or speaking
certain words. Currently we know that sex is
determined by sex chromosomes.
 Sex in man is determined by sex chromosomes
X and Y carried in the gametes. During
fertilization, an embryo may either inherit XX
genotype if a sperm containing the X sex
chromosomeorXY genotypeifthereis fusion of
sperm having the Y-sex chromosome.
 The XX genotype determines the female
homogametic sex, while the XY genotype
determines the male heterogametic sex.
 Since all the embryo possess an X chromosome
which carries the testicular feminization
gene for production of a protein in form of
testosterone receptor, all of them have this
receptor.
 Because the would be embryos have the Y
chromosome with testes determining gene for
production of a protein called H-Y antigen/
testes determining factor (TDF) on the body
cells of a developing male. The non-
differentiated gonads/genital ridges
differentiate into testes/seminiferous tubules
and interstitial cells.
 Without the H-Y antigen/TDF, the non-
differentiated gonads will differentiated into
the ovary of the female embryo.
 The developing testes of the male embryo
secrete testosterone hormone into the blood
stream which binds on the testosterone
receptor to form the testosterone-receptor
complex which moves from the cytoplasm into
the nucleus and activates the genes which code
for the development of the tissues which give
rise to male reproductive system.
 The tissues of the developing female are not
activated due to lack of testosterone, hence a
female reproductive system develops.
Inman,thereare23pairs of chromosomes; ofthese
only one pair carries genes for sex determination.
These are called sex chromosomes (heterosomes)
designated X and Y, and the other 22 pairs are
called autosomes.
In some animals like birds (including poultry),
moths and butter flies; the sex genotypes are
reversed. The homogametic genotypes (XX) are
male while the heterogametic genotype (XY) is
female.
In some cases, the Y chromosome is completely
absent andtheheterogameticsex (XO) ismale.This
is the X-O system as in grass hoppers, cockroaches
and some insects. The sex of the off springs is
determined by whether the sperm cell contains an
X chromosomeornosex chromosome.Thisimplies
thattheY chromosomedoesnotcarry genesneeded
for survival of the organisms.
In some species of bees and ants, there are no sex
chromosomes.Femalesdevelopfrom fertilizedeggs
and are thus diploid while males develop from un
fertilized eggs and are haploid, without feathers.
Example:

Genotypic ratios: 1XX: 1XY
Phenotypic ratios: 1female: 1male
This shows that there is a 50% chance of any child
being a male or female.
Environmental determination of sex
Sex is primarily genetically determined as
described above but in some lower animals, sex can
be determined by environmental factors such as
temperature, salinity, type of food etc. for example
in some turtles the eggs laid warm sand develop
into females while those laid in cool sand develop
into females.
SEX CHROMOSOMES
The sex chromosomes are called heterosomes
because they are non-identical and are designated
X and Y. The X chromosome is rod shaped and
much bigger than the Y chromosome which is hook
shaped.
The Y chromosome carries genes responsible for
secondary male sex characteristics, differentiation
of testes and development of genital organs in
humans. Actually in some organisms, the Y
chromosome is absent and is believed not to carry
genes necessary for survival of the organism and is
described as genetically inert.
SEX LINKAGE:
In humans, there are several thousands of
characteristics each genetically controlled. With
only 23 pairs of chromosomes, each chromosome
must therefore carry many genes; a phenomenon
that does not exclude sex chromosomes. These in
addition to genes responsible for sex differences
may carry genesdetermining someotherfeaturesin
the body.
Sex-linked genes are genes carried on sex
chromosomes and inherited together with those
determining sex. Sex linked traits (characters) are
traits determined by genes carried on sex
chromosomes and inherited together with those
determining sex.
Note: The Y chromosomes don’t carry genes, sex
linked genes are specifically carried on the X sex
chromosomes but not on the Y chromosome.
Many experiments were carried out by Thomas
Morgan about sex-linked genes in drosophila.
Inoneofhisexperiment,Morganmateda wildtype
(pure breeding) red-eyed female with a mutant
(white eyed) male. All the F1 hybrids were red eyed.
He went on to interbreed the F1 males and females
to obtain an F2 generation which consisted of red
eyed and white eyed offsprings in a ratio of 3:1
respectively. However, all female were red eyed and
all the white eyed flies were males though some
males were red eyed.
In conclusion, all the F1 were red eyed; implying
thatthis allele is dominantoverthatforwhite.Since
in the F2 all the white eyed were males, this
indicates that the gene for eye colour is located on
the X chromosome and there is no corresponding
locus on the Y chromosome; otherwise some
females would also be white eyed.
Phenotypic ratios: 3 red eyed: 1 white eyed
Note that all the white eyed are males yet some red
eyed are males
Examples of sex linked characters in man
include the following

 Haemophilia
 Colour blindness
 Pre-mature balding
 Eye colour in drosophila
Most of these characters are caused by recessive
alleles and in a genetic cross, these must be
represented as superscripts on the x sex
chromosome.
HAEMOPHILIA (BLEEDERS’ DISEASE)
Haemophilia is a recessive sex-linked blood
disorder that leads to absence of one or more blood
clotting factors, leading to prolonged bleeding even
from minor cuts.
Just like other sex-linked traits, haemophilia is
carried on the X chromosome and the responsive
allele is recessive to thenormalallele.The condition
interferes with formation of blood clotting factors;
commonly factor VIII (Anti-Haemophiliac
Globulin) whose absence greatly delays the blood
clotting process. This results into prolonged
bleeding and excess blood loss even from minor
cuts which may lead to death.
The allele being recessive, haemophiliac females
must inherit two copies of the defective allele while
males inherit one copy. The heterozygous females
show normal blood clotting and are described as
carriers. This is because the other X chromosome
carries a dominant allele needed for normal blood
clotting which suppresses the recessive allele for
haemophilia. The males lack the alternative allele
and the recessive allele is automatically expressed
phenotypically.
Example: When a carrier woman is married to a
normal man
It can be noted that there is a 50% chance of a
daughter being a carrier and a 50% chance of a son
being haemophiliac. Sons can only inherit
haemophilia (andothersex linked traits) from their
mothers but not fathers as they only inherit the
father’s Y chromosome and not the X chromosome
that carries sex linked genes. Girls can inherit from
both parents.
Today, people with hemophilia are treated as
needed with intravenous injections of the missing
protein.
COLOUR BLINDNESS
It is a recessive sex linked character that leads to
inability of the individual to distinguish between
colours.
It is caused by a recessive allele, carried on the X
chromosome and inherited in the same way as
haemophilia.Colourvisionis due to presence in the
retina of red, blue and green cones needed for
seeing the respective colours. The recessive alleles
result into absence of some of these cones which
renders inability to identify such coloursfrom other
related colours. This is called colour blindness; the
commonest being red-green colour blindness
where individuals lack red and green cones in their
eyes.
Example
Green colour blindness is sex linked in man. A
normal man married a colour blind woman. Using
suitable genetic symbols workout the genotypes
and phenotypes of their children

Colour blind individuals are more common in
the population than haemophiliacs despite the two
being inherited in the same way. This is because
haemophilia is associated with many lethal effects
due to excessive internal and external bleeding
which increases chance of dying before
reproductive maturity to pass on their genes to the
nextgenerations.Colourblindness exerts less lethal
effects as colour vision is not much necessary for
survival. Colour blind individuals usually survive to
reproductive age and pass the allele to subsequent
generations hence increasing the number of colour
blind individuals in the population. Also
haemophiliacs are advised to desist from
reproducing as they may end up bleeding to death
which further reduces the numbers of
haemophiliacs.
NB: Sex linked charactershavebeenfoundtooccur
morecommonlyinmalesascomparedtofemalesin
the human population. Being caused by recessive
alleles, the other X chromosome in females may
carry a dominant allele to mask the defective allele
hence preventing its phenotypic expression in the
population. In males however, these genes are
carried on the non-homologous portion of the X
chromosome for which there is no alternative gene
onthe Y chromosome.Suchgenesareautomatically
expressed in males leading to higher frequencies in
males as compared to females.
SEX LIMITED CHARACTERISTICS
Se limited characters are characters that are more
pronounced in one sex than the other.
Though both sexes may carry genes responsible for
these characteristics, pronounced expression is
strictly limited to one of the two sexes. They are
usually carried on autosomes but may largely be
influenced by the level of sex hormone in the body.
Examples include;
Facial hair, deep voice, baldness etc in males
Breasts, lactation, widening of hip bones, high
pitched sound etc.

34 |ENRICHED ADVANCED LEVEL GENETICS NOTES 2019 EDITED BY KING SOYEKWO ROGERS
0701211966

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