GENETICS MENDELIANN EPISTATIS UNIVERSITY LECTURE NOTES DIHYBRID

1. BASIC READING LIST FOR BASIC GENETICS
1. Burns GW (1980) The Science of Genetics: an introduction to
Heredity. Macmillan Publishing Company Inc. New York, 4th
edition.
2. Fansworth MW (1978) Genetics. Harper international edition.
Harper and Row Publishers Inc.
3. Green NPO, Stout GW, Taylor DJ (1998) Biological Science Vol.
I & II. Cambridge University Press, United Kingdom.
4. Hartl DL and Jones EW (2001) Genetics: analysis of genes and
genomes. Jones and Bartlett Publishers, United States, 5th
edition.
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5. Muschel LH (1966) Blood Groups, Disease, and Selection.
Bacteriological Reviews, 30: 427‐441.
6. Newman SA (2005) The Pre‐Mendelian, Pre‐Darwinian world:
Shifting relations between genetic and epigenetic mechanisms
in early multicellular evolution; Journal of Biosciences 30: 75–
85
7. Strickberger MW (1968) Genetics. The Macmillan Company,
NewYork.
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DEFINITION OF GENETICS
►Can be defined as a branch of biology that deals with the
mechanisms of hereditary transmission and variation of
inherited characteristics among similar or related organisms
►It seeks to find answers to pertinent questions regarding
heredity suchas:
Why do animal and plant species almost always produce
offspring similar to the parents and only with members of the
same generalgroup?
Why is it that no exact copies are made, but there is always
some variation between parents and off springs and between
off springs themselves?
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2. What controlling processes guide the determination of physical
characteristics?
Where do significant abnormalities / aberrations come from?
►Various theories have been proposed to explain the
mechanisms of inheritance, some of them dating as far back as
several centuries before the birth of Christ
►This course will focus on those theories proposed before the
era of Gregor Mendel (Pre‐Mendelian)
►It will also examine the contributions of Gregor Mendel to the
understanding of inheritance AND finally consider inheritance
patterns that do not conform to Mendelian Inheritance
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Pre‐Mendelian GGeenneettiiccss
❖ Objectives of this topic
➢Review of theories of inheritance that existed
before the era of Gregor Mendel.
➢Compare and contrast the different
hypotheses/theories
➢Examine the strengths and weaknesses of each of
the theories.
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EARLY THEORIES OF INHERITANCE
❖ The theory of vapors andfluids
➢Proposed by the Greek philosopher and mathematician
Pythagoras (≈ 550 B.C).
➢He postulated that human male semen was created
from vapors and fluids collected from the entire body
➢that there was a complete being preformed in the
semen that was transferred intact to the female
➢Thus the male parent played the dominant role in
determining the form of the child.
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➢The mother served as the receptacle for the embryo
formed entirely from male material.
➢Do you see an immediate weakness of this theory?
➢Empedocles (≈ 453 B.C) proposed a modification to
cater for blending between the male and female sexual
material in the production of the embryo.
➢Another Greek philosopher (Aristotle) postulated that
semen was purified blood.
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3. ➢According to his doctrine, both parents contributed purified
blood to the embryo, but the male semen was more purified
than the female menstrual fluid.
➢That the male parent contributes the vitality and the
blueprint (source of life), the female the building blocks.
➢ Do you notice any male chauvinism here??
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➢Could this be the basis of the belief in most male
dominated societies that children in human marriages
belong to the fathers??.
➢This concept persisted for over 2000 years and lasted
although there was no available systematic anatomical
evidence to confirm or disprove it.
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THE THEORY OF PREFORMATION
❖ proposed by Jan Swammerdam (1637‐1680)
➢postulated that each sperm contained,
in miniature form, a complete human
being (Homonculus)
➢That the mother provided only the location
and nourishment for the growth of the
embryo, since the male sperm contributed
everything else
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➢Believers in this school of thought were generally known as
‘spermatists’.
➢The spermatists therefore also tended to ignore the role of
the female in contributing genetic material to the offspring.
➢Charles Bonnet (1720‐1793) suggested the reverse; that
the potential embryos were in the eggs, not the sperms.
➢believers in this school of thought were known as the
‘ovists’.
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4. ➢They believed that each succeeding generation was
similarly housed within the present generation thus:
(encapsulation theory)
Question: what would happen when the last Homunculus
wasreached??
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➢ Do you see any basic weakness of such a theory?
➢It failed to explain how both parents could transmit
their traits to the offspring
➢Pierre Louis Moureau de Maupertuis (1698‐1759)
opposed ovists and spermatists by proposing a blending of
maternal and paternal particles in the embryo
➢He suggested that particles from one parent could exert
more influence than those from the other (a remote
suggestion of dominance and recessiveness??)
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THE THEORY OF EPIGENESIS
➢ proposed by Kaspar FFrriieeddrriicchhWolff (1733‐1794)
➢healso postulated that reproductive cells do not contain
preformed embryos but rather particles which have the
power to guide the development of the body parts in the
growing embryo.
➢He proposed a vital force that operated within the growth
period of the embryo and controlled the development of the
offspring.
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➢However, he failed to explain how this force worked
➢BUT his theory did emphasize the importance of some
material "encoded" inside the fertilized egg.
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5. THE THOERY OF PANGENESIS
➢ proposed by Charles Darwin.
➢He postulated that each part of the body produced
minute particles (pangenes)
➢these particles moved in the blood stream to the
reproductive organs and formed similar organs in the
developing embryo.
➢had the concept of blending
➢characteristics acquired in life could affect the pangenes
and therefore the offspring.
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➢Thus tended to support the concept of the inheritance of
acquired characteristics.
➢disproved by August Weismann (1834‐1914)
➢He demonstrated that the life experience of one
generation had no influence upon the biological structures
of thenext.
➢He cut off tails from rats for 22 generations and yet all off
springs had tails!
Through these experiments, was Weissman disproving
inheritance of acquired characteristics?
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THE THEORY OF THE GERMPLASM
➢proposed by August Weismann (1834‐1914) in 1884 as an
alternative to the theory of Pangenesis.
➢Postulated that sex cells (germplasm) endure from one
generation to the next and are not affected by the bodily
experiences of the organism.
➢BUT it does not explain the existence of variation between
parents and offspring
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MENDELIAN INHERITANCE
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6. OBJECTIVES OF THIS TOPIC
➢ By the end of this topic, you should be able to:
➢ enumerate and explain characteristics of good genetic
organisms
➢ explain why Mendel was more successful than his
predecessors
➢ explain the meaning of basic standard genetic
vocabulary
➢ predict results from a given monohybrid cross
➢ restate and explain Mendel’s first law of inheritance
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BBIIOOGGRRAAPPHHYYOF MENDEL
➢Born in a poor family in Austria on 22nd
July 1822; died on 6th January 1884
➢1843: began studying at the St. Thomas
Monastery
➢ ordained as a priest in August 1847
➢ 1849: assigned to a secondary school
in the city of Znaim.
➢ he then took the qualifying exams for
teacher certification and failed.
" he lacks insight and the requisite clarity of
knowledge".
BIOGRAPHY OF MENDELcontd.
➢ 1851: entered the University of
Vienna to train to be a teacher
of Mathematics and Biology
➢ 1854: returned to teaching
in Brunn
➢ 1856: attempted the state
certification examination but
withdrew.
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➢1856: returned to Brunn
& continued to teach part‐
time
➢1868: promoted in the
monastery to an Abbot.
Site of Gregor Mendel’s experimental
garden in the Czech Republic
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7. MENDEL’S PIONEERING WORK
➢He did groundbreaking work leading to the current
theories of heredity
➢more than any other scientist, he synthesized the basic
principles of heredity into a body of knowledge that has
formed the very core of modern genetics, a reason why he is
called the father of modern genetics.
➢He combined the ideas put forth by other scientists into a
definite set of working principles that are acceptable even
today
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➢ Mendel's ideas on the mechanism of inheritance were
published in 1866 in the Journal of the Brno Natural
History Society but had absolutely no impact
➢ His ideas and principles temporality died with him in
1884 and the new Abbot who replaced him burnt all his
papers
➢ However, his ideas were rediscovered 34 years later by
three scientists, Hugo De Vries (Dutch), Tschermarck
(Austrian) and Correns (German) when they obtained the
same results in the experiments conducted by each one of
them independently in 1900
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➢ The basic ideas and the conclusions drawn by scientists
after this rediscovery came to be known as Mendelian
Genetics
➢ he used the garden pea (Pisum sativum) for his
hybridization experiments
➢ He crossed plants of the same species, obtained fertile
offspring & crossed them again to study subsequent
generations
➢ The garden pea possessed a number of characteristics
that made it a good genetic organism
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GENERAL CHARACTERISTICS OF GOOD
GENETIC ORGANISMS
➢ Possession of sufficient discontinuous variation
➢ ability to undergo genetic recombination through sexual
reproduction
➢ ability to carry out / be subjected to controlled mating
➢ short life cycles
➢ production of a large number of offspring
➢ easiness and convenience of handling
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8. ADVANTAGES OF GARDEN PEAS OVER OTHER
SPECIES
➢ Several varietieswith distinct and discontinuous
characteristics were available (Mendel selected 7 such
characteristics)
➢ The plants were easy to cultivate.
➢ Artificial and controlled cross breeding between con‐specific
varieties was possible by transferring pollen grains from one
plant toanother
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➢ Pea flower morphology makes selfing possible facilitating
controlledmating
➢ Pea flowers exhibit a condition known as cleistogamy; i.e.
their petals remain closed under natural conditions thus
facilitatingself‐pollination
➢ This trait is very important in establishing pure breeding lines
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Emasculation illustrated
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9. 33
OTHER REASONS FOR MENDEL’S SUCCESS
➢ He chose the garden pea which was an ideal organism for
investigating inheritance through controlled mating
experiments (he was able to observe inheritance patterns
in up to two generations a year).
➢ He initially established pure‐breeding lines of varieties
for each of the characteristics he investigated (spent 2
years doing justthis!)
➢ carried out preliminary investigations (necessary for fine‐
tuning experimental techniques)
➢ Considered only one variable at a time (makes analysis
and interpretation of data easy)
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➢ took extra care to prevent contamination (enabled him to
make correct deductions)
➢ kept accurate and meticulous records of all the
experiments and the results obtained (enabled him to
observe consistent trends in his experiments)
➢ He pooled the data obtained from similar experiments for
different characteristics and analyzed the results by using
statistical methods and applying the law of probability.
➢ (This enabled him to obtain sufficient data for analysis
and hence minimized his errors and improved the
statistical significance and reliability of his results)
➢ He was fortunate because the genes responsible for the
7 characteristics that he considered are located on
separate chromosomes, i.e. not linked (linkage does not
allow independent assortment that he discovered)
➢ He was also fortunate in that the characteristics he
investigated in the pea plant did not show any complex
forms of gene interaction or linkage (they were
qualitative traits whose phenotypes were not influenced
by environmental factors)
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MENDELIAN MONOHYBRID INHERITANCE
➢ a type of inheritance where transmission of only one
characteristic is considered at a time.
➢ pure breeding parents with contrasting expression of the
same characteristic are crossed.
➢ Mendel conducted his experiments in three distinct
phases
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10. THE 3 PHASES OF MENDEL’S EXPERIMENTS
➢ Establishment of pure breeding lines through continuous
inbreeding for a particular trait
➢ Crossing of pure breeding parents with contrasting
expression of a trait to produce a first filial (F1)
generation.
➢ Selfing of the F1 progeny to produce a second filial (F2)
generation
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MENDEL’S OBSERVATIONS
➢ F1 progeny resembled one of the parents and not the
other.
➢ Selfing of the F1 progeny gave a second‐generation
progeny (F2) expressing both characteristics of the
original parents.
➢ careful mathematical analysis of the F2 progeny
phenotypes showed it to segregate in an approximate 3 :
1 phenotypic ratio
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A few questions to ponder
1. Is the F1 progeny identical to the parent they
resemble?
2. What can you say about the effect of one
characteristic on the appearance of its alternative
expression in the F1?
3. What can you say about the reappearance of the
second character in the F2?
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Pea plant characters studied by Mendel
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11. Pea plant characters studied by Mendel contd.
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Ratios of dominant to Recessive in Mendel’s Plants
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INFERENCES FROM MENDEL’S RESULTS
➢ Phenotypic characteristics are under the control of factors
➢ These factors are the ones that are passed on from one
generation to another.
➢ Each parent possesses two factors controlling a given
characteristic
➢ Each offspring inherited one factor from each parent.
➢ the factors are transmitted to offspring via gametes
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➢ In the gametes these factors are located in the nucleus.
➢ Within the nucleus the factors are located on the
chromosomes.
➢ The egg and the sperm make equal contributions to the
zygote (inferred from the fact that reciprocal crosses give
the sameresults)
➢ Each factor is transmitted from generation to generation
as a discrete unchanging unit.
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12. DEFINITION OF A RECIPROCAL CROSS
➢ a reciprocal cross is a second cross of the first one
whereby the characteristics of the parents are
interchanged.
➢ For example, if the first cross is between a tall male
parent and a dwarf female, the reciprocal cross is the one
between a dwarf male parent and a tall female parent.
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MENDEL’S 1ST LAW OF INHERITANCE
➢ The observations and deductions led Mendel to formulate
his first law of inheritance (the principle of segregation)
➢ ‘The characteristics of an organism are determined by
internal factors which occur in pairs and only one of a pair
of such factors can be present in a gamete’
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SYMBOLIC REPRESENTATION
➢ Choose capital letter symbols to represent dominant
genes and lower case letters to represent recessive genes
➢ Take the example of height:
%Let T represent gene for tallness and t the gene for
shortness
%Then pure breeding tall parent = TT & pure breeding
short parent = tt
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PARENTAL CROSS
t t
Pure breeding
parents (2n)
TT ttX
T T
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Meiosis
Random
fertilization Tt Tt Tt Tt
All F1 Progeny Heterozygous tall (Tt)
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13. T
t
T
t
SELFING F1
X
T t T t
F1 PROGENY
parents (2n)
Meiosis
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Random
fertilization T
T
T
t
T
t
tt
3 Tall 1 dwarf
SUMMARY OF MONOHYBRID CROSS
➢ The F2 generation segregates into a
➢ 3T‐ : 1tt Phenotypic ratio
➢ 1TT : 2Tt : 1tt genotypic ratio
Question: How can we differentiate between a
homozygous and heterozygous dominant individual?
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This can be achieved by carrying
out what is known as a TEST CROSS
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MONOHYBRID TEST CROSS & ITS USE
➢ Is a cross between an individual of a dominant phenotype
and a homozygous recessive phenotype. e.g. T‐ x tt.
➢ It is used to differentiate between homozygous dominant
and heterozygous dominant individuals
➢ If the dominant individual is homozygous, all the progeny
will show the dominant phenotype
➢ If the dominant individual is homozygous, all the progeny
will be dominant
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14. ➢If the dominant individual is heterozygous, half of the
progeny from the testcross will show the dominant
phenotype and half will show the recessive phenotype
TT X
tt
Tt X
tt
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All are Tt Half are Tt Half are tt
Mendelian monohybrid 1:1
test cross ratio
The Mother of all questions is: Did Mendel cheat??
OBJECTIVES OF THIS TOPIC
❖ By the end of the topic, you should be able to:
❖ List and explain the different factors that modify
Mendelian monohybrid genotypic and phenotypic
ratios.
❖ Explain the concept of pleiotropy
❖ Compute modified ratios from given crosses with genes
showing different modes of inheritance.
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MODIFICATION OF MENDELIAN
MONOHYBRID PHENOTYPIC AND
GENOTYPIC RATIOS
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RECAP OF CHARACTERISTIC FEATURES OF A
MENDELIAN MONOHYBRID CROSS
➢ A character is controlled by only two alleles
➢ One of the alleles shows complete dominance over the
other
➢ No allele exerts a lethal effect on the bearer either in the
homozygous or heterozygous condition
➢ Pure breeding parents with contrasting expression of
the same characteristic produce a 3:1 phenotypic and a
1:2:1 genotypic ratio in the F2 generation.
➢ Reciprocal crosses produce the same results
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15. INCOMPLETE DOMINANCE
➢ Pure breeding phenotypically contrasting parents produce
progeny in the F1 which are phenotypically intermediate
between the two parents
➢ Classic example is the inheritance of flower color in the
snapdragon (Antirrhinum majus)
➢ A red flowered plant crossed with a white flowered plant
produce a progeny which is pink‐flowered
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Pure Breeding
red floweredplant
Pure Breeding
White floweredplant
X
F1
All PinkProgeny
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F1 x F1 Pink x Pink
1 Red 2 Pink 1White
Note: the F1 does not resemble one of the parents; 3 phenotypes
appear in the F2 instead of 2; the 3:1 ratio is modified into 1:2:1
SYMBOLICREPRESENTATION
➢ Incompletely dominant alleles are represented by small
letters with subscripts e.g. r1, r2 ; a1, a2
Pure Breeding
White flowered plant
XPure Breeding
red flowered plant
r1 r1 r2 r2
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1White
F1 All PinkProgeny
r1 r2
F1 x F1 Pink x Pink
r1 r2 r1 r2
1 Red 2Pink
r1 r1 r1 r2 r2 r2
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CODOMINANCE
➢ Pure breeding phenotypically contrasting parents
produce progeny in the F1 which are phenotypically a
mixture of the two parental phenotypes
➢ Both alleles at a locus equally & fully express themselves
in a heterozygous condition.
➢ Selfing of the F1 progeny produces an F2 progeny that
segregates into three phenotypic classes in a ratio of 1:2:1
➢ Conventionally, codominant alleles at a locus are
represented by different upper case letters
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16. EXAMPLES OF INHERITANCE THAT SHOW
CODOMINANCE
➢Coat color incattle
➢A and B blood groups in man.
➢Normal and sickle cell haemoglobin inman
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Coat color inheritance in shorthorn cattle
X
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Red short horn (RR) White short horn (WW)
Roan short horn (RW)
Roan = Red & white hairs mixed together
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Selfing the F1 progeny produces an F2 generation
segregating into a 1 Red : 2 Roan : 1 white ratio
RW RW
R W
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R W
RR RW RW WW
1 Red : 2 Roan : 1 white
1:2:1 Monohybrid
Phenotypic ratio
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17. A & B BLOOD GROUP INHERITANCE
➢the alleles for blood group A and blood group B are co‐
dominant.
➢Whereas persons with genotype AA produce antigen A &
BB individuals produce antigen B on the surfaces of their red
blood cells, heterozygotes (AB) produce both A and B
antigens.
➢** Attempt to symbolically illustrate the inheritance of A
and B blood groups in man.
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INHERITANCE OF NORMAL AND SICKLECELL
HAEMOGLOBIN IN MAN
➢The genes responsible for the two types of
hemoglobin are HbA and HbS respectively.
Most persons belong to the
genotype HbA HbA; their red blood
cells are biconcave disk‐shaped
and contain hemoglobin A
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People suffering from sickle
cell anemia have the genotype
HbS HbS & all their red blood cells
are distorted to assume a sickle
Shape .
In the heterozygotes, HbA HbS, some of the red blood cells
contain Hemoglobin A while others contain hemoglobin S.
** Symbolically illustrate the inheritance of heamoglobin
inman
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LETHALGENES
➢ genes whose phenotypic effects are sufficiently drastic
to kill their bearers
➢can manifest themselves at any stage during the life time
of an individual
➢can show gametic lethality or zygotic lethality
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18. ➢ Lethal genes may be:
➢ Dominant e.g. the gene for Huntington’s chorea in
man (an inherited degenerative disorder of the CNS)
➢ Co‐dominant e.g. the gene for sickle cell anemia
➢ Recessive e.g. the gene for cystic fibrosis in man or
the gene for yellow coat color in mice
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➢ Cystic fibrosis is an autosomal recessive hereditary
disease
➢ Affects the mucous glands of the lungs, liver, pancreas
and intestines
➢ Characterized by production of thick mucous and
frequent lung infections, diminished pancreatic enzyme
production leading to early death
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A typical CF treatment using a Nebbuulliisseeraanndda ThAIRapy vveest
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If a completely normal person marries a person with cystic
Fibrosis (CF), the following will happen
Let C stand for the Normal allele and c stand for the allele
for the CF condition
Normal Cystic fibrosis sufferer
CC x cc
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Normal offspring
Cc
Carrier for the cystic fibrosis gene
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19. If 2 carriers married each other, they would produce progeny
That would segregate as follows:
Normal but carrier x Normal butcarrier
Cc Cc
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CC Cc cc
Normal Normal but carriers CF sufferers
1 2 1
Eventually diesoff
Genotypic ratio converted from 1:2:1 to 1:2 ratio and the
number of phenotypic classes reduced from 2 to 1
COAT COLOR INHERITANCE IN MICE
➢ The black color always breeds true i.e. black x black
always produces black.
➢ Crosses between two yellow individuals always
produce yellow and black individuals in a ratio ≈ 2:1.
➢ Crosses between yellow and black mice always
produce yellow and black progeny in the ratio ≈ 1:1.
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SYMBOLIC REPRESENTATION OF COATCOLOR
INHERITANCE
Let Y represent yellow
coat color & y
Represent black coat
color Y y Y y
XYellow
(Yy)
Yellow
(Yy)
YY Yy Yy yy
homozygous heterozygous homozygous
Yellow Yellow black
(dieearly)
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➢homozygous yellow die off in early stages of embryo
development.
➢The resultant progeny therefore segregates into a 2:1
ratio instead of the expected 3:1 ratio.
➢gene Y behaves as a dominant gene as far as coat color
determination is concerned and as a recessive gene as far as
the viability of the embryos is concerned.
➢genes with multiple effects on an individual are said to be
Pleiotropic.
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20. QUESTIONS TO PONDER
➢ Why is it that recessive lethal genes can persist in a
population for a longer time than dominant lethal
genes?
➢ under what circumstances can a dominant lethal gene
persist in a population despite the fact that it kills the
bearer both in the heterozygous and homozygous
conditions?
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MULTIPLE ALLELES
➢ A situation whereby a characteristic is controlled by a
series of more than two alleles of which only two can be
present at the locus at any time e.g. blood groups in
man
➢ A B O blood group system is controlled by three alleles
➢ Alleles A and B are codominant and both are completely
dominant to O
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POSSIBLE BLOOD PHENOTYPES & GENOTYPES
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Phenotype Possible genotypes
Blood group A AA, AO
Blood group B BB, BO
Blood group AB AB
Blood group O OO
ACTIVITY
Predict the possible genotypes of offspring and in what
ratios they occurs for marriages between persons with the
following blood group genotypes
1. AO XBO
2. AB XAB
3. AB XBO
4. AB XAO
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21. 21
CYTOPLASMIC / MATERNAL INHERITANCE
➢ occurs when genes controlling a characteristic are located
in the cytoplasm e.g. chloroplast or mitochondrial genes.
➢ Uniparentally transmitted by maternal parent to all
progeny without exception
➢ Reciprocal crosses involving cytoplasmic genes do not
yield identical results.
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EXAMPLES OF UNIPARENTAL / MATERNAL
INHERITANCE
➢ Shoot variegation in the four o’clock plant: (Mirabilis
jalapa)
➢Shoot color is determined by pigment produced in
chloroplasts.
➢Some of the chloroplasts contain genes producing
normal, green pigment
➢Some chloroplasts contain defective genes producing
no pigment (Leukoplasts).
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➢ As mitosis and meiosis divide up each cell’s cytoplasm,
any one cell line can either inherit
➢ only functional chloroplasts (green),
➢ only non‐functional chloroplasts (white),
➢ or any mixture of the two (variegated).
➢ Only progeny cell lines from variegated tissues can
create all three types.
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Thus, a shoot (stem, leaves and flowers) may be
➢white, green, or variegated.
➢When self‐fertilized, flowers from different
colored shoots produce the following
outcomes:
♀Parent Shoot Color Progeny Shoot Color
Green Green
White White
variegated green, variegated, white
(Irrespective of the paternal phenotype)
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22. 85
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➢ Direction of shell coiling in the snail Limnaea ppeerreeggrraa:
➢The shells of snails coil either to the right (dextral) or to
the left (sinistral).
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DEXTRAL SINISTRAL
➢ A dextral female crossed with a sinistral male will
produce dextral progeny
➢A sinistral female crossed with a dextral male will
produce sinistral progeny.
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23. A human pedigree illustrating maternal
inheritance
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♂ ♀
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Review Questions
➢Two curly‐winged fruit flies (Drosophila) are mated; the
progeny consists of 341 curly‐winged and 162 normal
winged individuals. Explain fully.
➢Compare and contrast the effects of incomplete , co‐
dominance, lethal genes and multiple alleles on the
standard Mendelian monohybrid phenotypic and genotypic
ratios
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24. MENDELIAN DIHYBRID INHERITANCE
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OBJECTIVES OF THE LECTURE
➢ By the end of the topic, you should be able to:
➢Recall the basic concepts of the summation and
product rules of probability
➢Apply the probability rules to predict ratios in standard
dihybrid crosses.
➢Restate and explain Mendel’s second law of
inheritance
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DEFINITION OF DIHYBRID INHERITANCE
➢ is a type of inheritance whereby transmission of two
characteristics is considered simultaneously
➢ is called dihybrid inheritance because a cross of doubly
homozygous parents differing in the expression of two
characteristics produces progeny that is heterozygous for
the two pairs of genes.
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➢Example of one of Mendel’s crosses:
➢ he considered the simultaneous inheritance
of seed shape and seed color
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Pure breeding parents
with round and yellow
seeds
Pure breedingparents
with wrinkled and
greenseeds
All the F1 progeny that was all round and yellow.
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25. ➢selfing the F1 produced an F2 generation which segregated
into four phenotypic categories as follows:
➢315 round and yellow
➢101 wrinkled and yellow
➢108 round and green
➢32 wrinkled and green.
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➢ Detailed analysis of the results revealed that:
➢ Two new combinations of characteristics appeared
in the F2; round and green & wrinkled and yellow.
➢ The four combinations of characteristics in the F2
segregated in a ratio that approximated to a 9:3:3:1
ratio.
➢The ratios of each pair of allelomorphic characteristics
(phenotypes determined by different alleles at the same
locus) independently segregated into a monohybrid ratio
of3:1
98
➢ For the shape of the seeds
(irrespective of the color of the seeds)
we have
(315 + 108) round : (101 + 32) wrinkled
➢ For the color of the seeds we
(irrespective of the shape of the
seeds have
(315 + 101) yellow : (108 + 32)green.
➢ Therefore, the dihybrid cross can be treated as
being composed of two separate and independent
monohybrid crosses
99
SYMBOLIC REPRESENTATION OF A DIHYBRID
CROSS
Choose suitable symbols:
Let R = round seeds; r = wrinkled seeds
Y = yellow seeds; y = green seeds
100
Parental genotypes:
Pure breeding round
and yellow seeds
Pure breeding wrinkled
and green seeds
RRYY
rryy
25

26. RRYY rryy
Pure breeding
Round & yellow
Pure breeding
Wrinkled & green
Gametogenesis
101
ryRY
RrYy Dihybrid
All F1 Progeny round and yellow
➢When the F1 progeny above are selfed, the genes for
these two characteristics assort independently into different
gametes and their union in the zygote is random. i.e.
RrYy RrYy
102
X
Gametogenesis
RY Ry
rY ry
RY Ry
rY ry
NB: All the four different types of gametes
are produced in equal proportions
The outcome of the random union of these gametes can be
illustrated using a Punnett square as follows
♂
103
RY Ry rY ry
RY RRYY RRYy RrYY RrYy
Ry RRYy RRyy RrYy Rryy
rY RrYY RrYy rrYY rrYy
ry RrYy Rryy rrYy rryy
♀
Round
&
Yellow
9
Round Wrinkled
& &
Green Yellow
3 3
wrinkle
d &
Green
1
➢ It can also be observed that the following genotypes are
produced with the following frequencies:
104
➢RRYY(1)
➢RRYy (2)
➢RRyy (1)
RrYY(2)
RrYy (4)
Rryy (2)
rrYY(1)
rrYy(2)
rryy(1)
➢ This gives the classic Mendelian dihybrid
ratio of 1 : 2 : 1 : 2 : 4 : 2 : 1: 2 : 1
26

27. ➢ These observations formed the basis of Mendel’s second
law of inheritance (the principle of independent
assortment)
➢ “Any one of a pair of characteristics may combine with
either one of another pair during the process of gamete
formation”.
➢ formed the basis of classical breeding experiments aimed
at combining desirable characteristics from different
individuals / varieties into one individual.
105
➢ These Mendelian ddiihhyybbrriiddpphheennoottyyppiiccand genotypic
ratios can only be obtained if:
➢The characters in question are controlled by genes
located on different chromosomes (i.e. they are not
linked).
➢The alleles controlling the characteristics show
complete dominance
106
DIHYBRID TESTCROSS
➢ If a doubly heterozygous individual (RrYy) is crossed with
a doubly homozygous individual (rryy), the following
results are obtained.
RrYy rryy
107
Gametogenesis
RY Ry
rY ry
ry
This gives the classic 1:1:1:1 dihybrid phenotypic testcross ratio
1 RrYy
1Rryy
1rrYy
1rryy
108
EVALUATION QUESTION
The table below gives results from three breeding
experiments on tomato plants involving the characteristics
of height and flower color
Offspring Phenotype
Parental phenotypes Tall & Tall & short & short &
purple green purple green
Tall purple x short green 301 287 293 302
Tall green x short purple 521 0 0 0
Tall green x short purple 87 91 77 83
1. Determine which phenotype in each case is dominant and recessive
2. Determine the probable genotypes of the parents in each cross
27

28. RULES OF PROBABILITY AND HOW THEY APPLY
TO MENDELIAN CROSSES
Q The relevant rules which apply to Mendelian crosses are
the sum rule (regarding mutually exclusive events) and the
product rule (regarding independent events)
Q Two events are mutually exclusive if they cannot
occur at the same time. i.e. the probability of
their simultaneous occurrence = 0. P (A and B) = 0
Q If two events are mutually exclusive, then the
probability of either occurring is the sum of the
probabilities of each occurring.
i.e. P (A or B) = P (A) + P (B)
109
GENETIC EXAMPLE OF THE SUM RULE
Q Consider a monohybrid cross GG x gg:
Q F1 genotype is Gg
Q F2 progeny is ¼ GG: ½ Gg: ¼ gg (1:2:1 ratio)
Question: what is the probability that an F2 offspring has a
dominant phenotype (is either GG or Gg)
Answer: P[dominant phenotype] = P[GG] + P[Gg]
= ¼ + ½ =¾
110
GENETIC EXAMPLE OF THE PRODUCTRULE
Q Two eventsare independent if the occurrence of one does
not change / affect the probability of the other occurring.
Q An example would be rolling a 2 on a die and flipping a head
on a coin. Rolling the 2 does not affect the probability of
flipping thehead.
Q If events are independent, then the probability of them
both occurring is the product of the probabilities of each
occurring. (can you restate this in a different way?)
Q i.e. P(A and B) = P(A) x P(B)
111
112
Q Genetic example:
Q Parental genotypes: GGWW x ggww
Q F1 : GgWw
Q F2: 9/16 G-W- : 3/16 G-ww : 3/16 ggW- : 1/16 ggww
Q Question: what is the probability that an F2 offspring of
this cross will have at least one dominant allele for
each trait (i.e. that it will be G-W-)
28

29. Q Solution:
Q First note that the segregation of the G locus is
independent of the segregation at the W locus so that
P[G‐W‐] = P[G‐] x P[W‐]
Q Also remember that GG and Gg are mutually exclusive
(an individual can either be GG or Gg but not both)
Q Likewise WW and Ww are also mutually exclusive
113
Q Therefore:
Q P[G‐] = P[GG or Gg] = P[GG] + P[Gg]
= ¼ + ½
= ¾
Q P[W‐] = P[WW or Ww] = P[WW] + P[Ww]
= ¼ + ½
= ¾
AND
Q P[G‐W‐] = P[G‐] x P[W‐] = ¾ x ¾ =9/16
114
SOLVING COMPLEX GENETIC PROBLEMS
Q Consider the following cross:
Q AaBbCcDdEeFf x AaBbCcDdEeFf
Q Question: What proportion of the F2 progeny will be
➢ AAbbCcDDeeFf
➢ AabbCcDdEeff
➢ Aabbccddeeff
Q How many different phenotypes and genotypes would
be expected in the progeny of the above cross?
115
TESTING THE GOODNESS OF FIT OF GENETIC
DATA
Q Suppose the F2 generation of a monohybrid cross has a phenotypic
ratio close to 3:1 but not exactly 3:1
Q How do we tell if the deviations from expected proportions in a
genetic experiment are due to chance or due to the fact that our
genetic hypothesis is wrong?
Q A: repeat the experiment many times e.g. x 100 and see how
consistently you approximate to the expected outcome
Q Suppose a fair coin is tossed 100 times and you count how many of
the throws are heads and how many are tails,
116
29

30. Ratios of dominant to Recessive in Mendel’s Plants
117
What is the most likely outcome of a single experiment?
A:
If you replicated the expt. 100 times, will all the 100 experiments
have the same outcome?
A:
118Such a situation is a likely outcome of such an experiment
Q Supposing a cross of Aa to an individual of unknown genotype
yields 100progeny.
Q Assume that our hypothesis is that the unknown individual
has the genotypeaa
Q What would be the expected result based on our hypothesis?
Q If the actual progeny contains 65 of the dominant phenotype,
and 35 of the recessive phenotype, can we reject our
hypothesis that the unknown parent was aa and that the cross
was therefore not Aa x aa ?
119
30
Q In other words “how likely are we to get a result this different
from the expected result if our hypothesis is true?”
Q How can we measure / quantify the difference between the
expected and observed results?
120

31. THE CHI-SQUARE TEST (χ2-TEST)
Q This is a test commonly used to measure the goodness of fit of
experimental data to a given null hypothesis (used to determine
whether there is a significant difference between the expected
frequencies and the observed frequencies in one or more categories.
Q i.e. Do the number of individuals or objects that fall in each
category differ significantly from the number you would expect?
Q Is this difference between the expected and observed due to
sampling error, or is it a real difference?
Q The observed number (O) in each category is compared to the
expected number (E) under the null hypothesis.
121
Q We use the square of the difference between
observed and expected numbers, and standardize by
dividing by the expected number in a category i.e. (O
––EE))2/ E
122
Q We then sum the values over each category or
“class” of the data i.e.
χ2 = Σ (O – E)2 / E
Going back to our test cross:
class Observed Expected (O-E) (O-E)2 (O-E)2/E
Aa
aa
65
35
50
50
15
-15
225
225
4.5
4.5
χ2 = 9.0; d.f = # classes - 1
Q Assuming the hypothesis (null hypothesis) we
proposed is correct, if we repeated our genetic
experiment many times (counting 100 progeny each
time) and calculated a χ2 for each, we could obtain a
distribution of chi-square values.
Q In a certain number of these repeated experiments, we
would get 65 or more Aa offspring out of 100
outcomes
123
INTERPRETATION OF THE RESULTS
Q Compare the computed χ2 to a theoretical value
from a chi-square table that has the same
degrees of freedom such as the one below.
124
31

32. Q the χ2 value associated with these cases would be
greater than or equal to the χ2 value we obtained
above (9.0).
Q The proportion of times that this occurs, is the
probability of getting a value of 9 or greater, when
you have 1 degree of freedom.
Q In this case the likelihood associated with a χ2 value
of 9 is less than 1%. i.e. 0.001 < P <0.01 (less than 1%
chance of occurrence)
125
DEGREES OF FREEDOM
Q refers to integer number that is generally equal to
the number of categories in the data, minus 1 i.e.
(n-1)
Q It is the number of phenotypic classes that are
independent of each other
Q if there are two categories, there is only one
degree of freedom
126
Q The ccoonnssttrraaiinnttof ((nn--11))ooccccuurrssbbeeccaauussee,,given the
total number of 100 progeny and the number in one
class (65), the number in the other class (35) is
fixed.
Q Only one class can vary freely, the other is
constrained.
Q The Critical Values of the χ 2 distribution are
exceeded by chance only rarely, and that rarity is
indicated by the probability value.
127
Q The common practice in genetics is to set the
critical value at 5%. (P = 0.05)
Q This means that if an event occurs by chance
alone less than 5% of the time, it is considered
statistically significant and thus justifies the
rejection of the null hypothesis
Q Therefore the standard convention for rejecting the
null hypothesis is if P < 0.05
Q If the difference between observed and expected is
too big to be due to chance alone, then you reject
your proposed genetic hypothesis
128
32

33. Chi‐Square TestRequirements
✓there must be quantitative data toanalyse.
✓there must be one or more categories to
compare.
✓There must be independentobservations.
✓Sample size in each category must be adequate
(at least10).
✓Simple random samplepreferable.
✓Data should be presented in frequencyform.
✓All observations must beused.
129
Summary of the steps to follow in using the chi
squaretest
✓Write the observed frequencies in column O
✓Figure the expected frequencies and write them in
column E.
✓Use the formula to find the chi‐square value
✓Find the degrees of freedom (df = n‐1)
✓Find the tabulated χ2 value at P = 0.05
130
✓If your calculated χ2 value is equal to or greater than
the tabulated value, reject the null hypothesis i.e. the
differences observed in your data set are not due to
chance alone (i.e. there is a significant difference
between your observed and your expected values)
✓If your calculated χ2 value is less than the tabulated
value, accept the null hypothesis i.e. the differences
observed in your data set are not due to mere
sampling error (no significant differences exist
between your observed and expected values)
131
Q Review question:
Q Assume you have crossed two pea plants with
purple flowers. Your hypothesis is that both
plants are heterozygous for a dominant allele at
a single locus controlling flower color. Out of 166
progeny, 110 had purple flowers and 56 had
white flowers. Test the goodness of fit of these
results to the Mennddeelliiaannmmonohybrid phenotypic
ratio
Q Test the goodness of fit of the data in slide 98to
the Mennddeelliiaann ddiihhyybbrriiddpphheennoottyyppiicc ratio
132
33

34. (i) In one family, the four children each have a different blood group.
Their mother is group A and their father is group B. Complete the
genetic diagram to show how this is possible.
Blood group A
————————––
————————––
Blood groupB
————————––
————————––
133
Parental phenotypes
Pareennttaallggeennoottypeess
Gameteess
Offspring ggeennoottypes
Offspring pphheennoottypes
————––——————————————————–
—————––———————————————
(3marks)
(ii) In many such families where the parents could have produced
children of all four blood groups, a total number of 120 children
were genotyped and each of the blood groups was found to be
represented as follows: Blood group A = 26; Blood group B = 31;
Blood group AB = 39; Blood group O = 24
iii) Carry out a χ2 test to find out how best these results fit the
expected values in a freely mating population.
Suggest a suitable null hypothesis for this test.
...................................................................................................................
...................................................................................................................
...................................................................................................................
...................................................................................................................
(1mark)
Commpplleetteetthhettaablebbeellooww to calculate tthhevvaalluueffoorχ2 for thhese results..
134
iv) Using the extract from the χ2 tables below, Explain what the
calculated value of χ2 tells us about these results.
...................................................................................................................
...................................................................................................................
...................................................................................................................
...................................................................................................................
...................................................................................................................
...................................................................................................................
..................................................................................................................
(3marks)
135
Modification of dihybrid
phenotypic and genotypic
ratio
136
34

35. OBJECTIVES OF THE LECTURE
Q By the end of the topic, you should be able to:
Q State and explain the effects of incomplete
dominance, codominance, lethal genes and
epistasis on the Mendelian ratios.
Q Solve problems involving dihybrid crosses where
one or more of the above factors are operating.
137
INCOMPLETE DOMINANCE
Q has the effect of increasing the number of
phenotypic classes but has no effect on the
genotypic classes.
Q One or both pairs of genes in the cross may show
incomplete dominance.
Q If one pair of genes shows incomplete dominance,
the number of phenotypic classes increases from
four to six.
138
Q If both pairs of genes show incomplete dominance,
you get nine phenotypic classes.
Q This expectation will only hold if the two loci under
consideration assort independently
Q when one pair of alleles shows incomplete
dominance and the second pair shows complete
dominance, we have Aab1b2 x Aab1b2
139
Q consider the genes in the above dihybrid cross to
be assorting independently and the cross to be
made up of two independent monohybrid crosses.
140
35

36. 141
Q When both loci show incomplete dominance, we
have a1a2b1b2 x a1a2b1b2.
Q The progeny from this cross will segregate into nine
phenotypic classes in the ratio of 1:2:1:2:4:2:1:2:1
instead of the standard 9:3:3:1.
Q Show the working how these ratios are obtained
142
LETHAL GENES
Q Most lethal genes show recessive lethality i.e. they
exert their effect only in a homozygous recessive
condition.
Q if you have two pairs of genes Aa and Bb showing
complete dominance, a cross AaBb x AAaaBBbbis
expected to produce four phenotypes in the ratio 9 A-
B-- :: 3 A- bb : 3 aaaaBB--:: 1aaaabbbb..
Q if gene a shows recessive lethality, phenotypes 3 aaaa
B-- aanndd1 aabb will be eliminated from the progeny
143
Q the phenotypic classes will be reduced from four to
two and the phenotypic ratio will be modified from a
9:3:3:1 ratio to a 9:3 (3:1) ratio.
Q determine the oouuttccoommeeof the cross AAaaBBbbx AAaaBBbbiiff
both genes a & b show recessive lethality.
Q What would happen to the genotypic ratios when
➢ one locus shows recessive lethality?
➢ both loci show recessive lethality?
144
36

37. CODOMINANCE
Q increases the number of phenotypic classes in
exactly the same way like incomplete dominance. It
may be shown by one or both pairs of genes.
Q Apply the same reasoning as in incomplete
dominance to determine the phenotypic and
genotypic ratios in a dihybrid cross where i) one
and ii) two pairs of genes are showing
codominance.
145
EPISTASIS
Q is a phenomenon of gene interaction whereby the
phenotypic effects of either one or both members
of one pair of alleles are masked by a gene of a
different pair at another locus
Q The masking gene is said to be epistatic while the
masked gene is said to be hypostatic.
146
Q Epistasis differs from dominance in that it is shown
by genes in different pairs and it can also be shown
by recessive genes.
Q In epistasis, two different genes interact to regulate
a single phenotype in an individual. Intermediate
phenotypes do not appear.
Q It has the effect of reducing the number of
phenotypic classes.
147
SINGLE DOMINANT EPISTASIS (A>B, b):
Q Sometimes known as ‘Dominant eeppiissttaassiissttyyppee II’’..
Q presence of a dominant allele of one gene masks the
genotype of a different gene irrespective of whether
the second gene is dominant or recessive i.e. A>BB,,bb..
Q The second gene can only manifest itself
phenotypically when the epistatic gene is in a
homozygous recessive condition.
148
37

38. EXAMPLE OF DOMINANT EPISTASIS
Q Genes controlling grain color inheritance in
oats show this type of gene interaction
Q Results of one of the experiments:
149
Pure breeding X
white grains
pure breeding
black grains
Black
Q Selfing of the F1 produced an F2 progeny of 560
individuals that segregated into three phenotypes as
follows; 418 black, 106 gray and 36 white.
Q This approximates to a 12:3:1 phenotypic ratio.
150
GENETICEXPLANATION
Q Note that the F2 segregates in a ratio that adds up
to16.
Q This implies that there are two pairs of dominant
genes controlling grain color (one responsible for
the black color and one responsible for the gray
color)
Q However, the gene for black color is epistatic to the
gene for gray color
151
Q This implies that the gene for gray can only
manifest itself when the gene for black is in a
homozygous recessive condition.
Q When both genes are in a homozygous recessive
condition, the white color is obtained.
152
38

39. SYMBOLIC REPRESENTATION
Q Let the 2 loci be A & B
Q Pure breeding black genotype = AABB
Q Pure breeding white genotype = aabb
153
12 : 3 : 1
RECESSIVE EPISTASIS
Q Symbolically represented as (a>B,b)
Q two pairs of genes are involved and homozygosity
for a recessive allele with respect to one gene masks
the expression of the genotype of a different gene.
Q This type of gene interaction produces a 9:3:4
phenotypic ratio in the F2.
154
Q Exemplified by genes controlling coat color in
rodents e.g. laboratory mice
Q In these rodents, 3 coat colors exist; the agouti or
wild type (most common color in the natural
populations) the albino and the solid black.
Q The albino phenotype is white with pink eyes and it
is true breeding
Q Agouti is dominant to black and both Agouti and
black are dominant to albino.
155
Q Agouti is dominant to black and both Agouti and
black are dominant to albino.
Q The Agouti phenotype is due to a dominant allele
A, and in aa animals the coat color is black.
Q A second dominant gene C is necessary for the
formation of hair pigments of any kind and cc
animals are albino (white)
156
39

40. Q A cross between true--bbrreeeeddiinnggAgouti (AACC) and
albino (aacc) produces a uniform F1 of Agouti
(Heterozygote).
Q Crosses between F1 males and females produce F2
progeny which segregate into phenotypic
proportions of 9 Agouti (A--CC--): 3 bbllaacckk((aaaaCC--): 4
albino (A--cc & aaaacccc))..
157
158
159
160
40

41. 161
DUPLICATE RECESSIVE EPISTASIS
Q Symbolically represented as (a=b > B,A)
Q observed when a homozygous recessive mutation
in either or both of two different genes causes the
organism to show the same phenotype
Q Leads to the production of a 9:7 ratio in the F2
generation
162
Q Occurs when 2 pairs of genes control the same
characteristic by participating in the same chain
reaction
Q Recessive homozygosity aatteeiitthheerror both of the
loci interrupts the reaction and behaves as
epistatic to the genes that follow in the reaction
Q Exemplified by genes controlling hydrocyanic acid
(HCN) production in the clover (Trifolium rreeppeennss).
163
Q Some strains of T. repens have high
concentrations of HCN while others do not produce
HCN
Q A cross between a plant positive for HCN and a plant
negative for HCN results in an F1 testing uniformly
positive for HCN.
Q What would be your immediate logical deduction
fromthis?
164
41

42. Q If your deduction were correct, what would you
then expect the progeny resulting from the selfing
of the F1 to be in the F2?.
Q However, Sullivan and Atwood (1943) obtained an
F2 progeny of 607 individuals which segregated
into 351 HCN positive : 256 HCN negative.
Q Apply the χ 2 test to check if these results conform
to a 3:1 monohybrid phenotypic ratio
165
Q Assuming HCN production is controlled by two
pairs of genes, compute the ratio to which these
results conform and test for the goodness of the
results to this ratio.
Q Explanation:
The formation of HCN involves a chain reaction
whereby two enzymes participate:
* one enzyme converts a precursor substance
into cyanogenic glucoside
* the second enzyme converts the cyanogenic
glucoside into HCN
166
enzyme A
Precursor Cyanogenic glucoside
enzyme B
HCN
This can be illustrated as follows:
167
List reasons for the production / no production
of HCN
When compared to the classic dihybrid phenotypic
ratio, we have:
9/16 A – B -; 3/16 A – bb; 3/16 aaB –; 1/16 aabb
HCN +ve HCN -ve
Q If you crossed two individuals AaBb x AaBb
determine the ratios of the progeny showing the
following traits:
➢ cyanogenic glucoside +ve : cyanogenic glucoside
–ve
➢ HCN +ve : cyanogenic glucoside +ve : precursor
only +ve
➢ Enzyme A & B +ve : Enzyme A only +ve : Enzyme B
only +ve : neither enzyme A nor B +ve.
168
42

43. AABB AABb AaBB AaBbAB
Punnett square illustration of duplicate
Recessive Epistasis (AaBb X AaBb)
AB Ab aB ab
169
AABb AAbb AaBb Aabb
AaBb Aabb aaBb aabb
A
b
aB AaBB AaBb aaBB aaBb
a
b
9 : 7
DOMINANT AND RECESSIVE EPISTASIS
Q sometimes known as ‘Dominant eeppiissttaassiissttyyppeeIIII’’
or dominant suppression
Q type of eeppiissttaassiisswhheerreebbyya ddoommiinnaannttgene at one
locus is epistatic to another gene at another locus
while the second gene, when in a homozygous
recessive condition, is epistatic to the first one.
Q Denoted as a=B > B,A
170
Q Exemplified by genes controlling feather color in
two breeds of chicken; the white leghorn and the
white Wyandotte
Q Feather color is controlled by two independent loci,
C & I
Q C responsible for colored feathers; I inhibits color
production (epistatic to C and leads to production of
white feathers)
Q White leghorn is CCII & white Wyandotte is ccii
171
Pure Breeding
White Leghorn
(CCII)
Pure Breeding
White Wyandotte
(ccii)
X
White (CcIi)F1
172
F2 9 C- I-
White
: 3C- ii : 3cc I- : 1ccii
colored white white
13 white : 3 colored
selfing
43

44. SEMI EPISTASIS
Q A character is controlled by 2 loci such that recessive
homozyggoossiittyyaatteither of the loci but not the ootthheerr
makes individuals phenotypically indistinguishable
but different from either hhoommoozzyygotes
Q Exemplified by genes controlling coat color in Duroc-
Jersey pigs (loci R &S)
173
Q Red coat color requires the presence of both loci in
a dominant form (R--SS--)
Q Double recessive individuals (rrss) are white
Q R-- ssssand rr SS--individuals have a sandy coat color
174
Pure Breeding
Red coat
(RRSS)
Pure Breeding
White coat
(rrss)
X
Red (RrSs)F1
175
F2 9 R- S-
Red
: 3R- ss
sandy
: 3rrS- :
1rrss sandy
white
9 red : 6 sandy : 1 white
selfing
ISO-EPISTASIS
Q A characteristic is controlled by two pairs of genes
in such a way that presence of one or both pairs of
genes in a dominant state produces the full
phenotype.
Q This type of interaction modifies the 9 : 3 : 3 : 1 ratio
into a 15:1 ratio.
Q shown by the genes that control grain color in wheat
176
44

45. Q Production of colored kernels is controlled by
two enzymes which participate in alternate
metabolic pathways
EnzymeA
177
Precursor Produc
t
Enzyme B
Q a functional enzyme A or B can produce aproduct
from a common precursor.
Pure Breeding
Colored grains
(AABB)
X Pure Breeding
white grains
(aabb)
178
Because either of the genes can provide the wild type
phenotype, this interaction is also called duplicate gene
action.
SUMMARY OF EPISTATIC INTERACTIONS
179
A- B- A- bb aa B- aabb
Complete dominance &
noepistasis
9 3 3 1
Dominant epistasis
12 3 1
Recessive epistasis
9 3 4
Duplicate recessive
epistasis
9 7
Semiepistasis
9 6 1
Iso -epistasis
15 1
Dominant and recessive
epistasis
13 3
RECIPROCAL/COMPLEMENTARY GENE ACTION
Q Dominant alleles of two or more genes are required
to generate a particular trait.
{compare with epistasis whereby the action of an
allele at one gene locus can hide traits normally
caused by the expression of alleles at another
gene locus}.
Q interaction is shown by the genes controlling
inheritance of comb shape in chicken.
180
45

46. Q Four different types of comb shape exist:
Rose comb
e.g. in
the Wyandottes
Pea comb
e.g. in
the Brahmas
single comb
e.g. in
the Leghorns
181
Q A cross between the Wyandottes (Rose comb)
and the Brahmas (Pea comb) produces hybrid
chicken with a walnut comb, a phenotype not
expressed in either parent.
182
Q Selfing oof tthheeF1 produces an F2 progeny having all
the 4 phenotypes in a characteristic 9:3:3:1 ratio; a
classic example of complementary gene action.
Q The ratio suggests that 2 loci control comb shape;
one for pea comb and one for rose comb
Q When both loci are dominant, a walnut comb is
produced and when both are homozygous
recessive, the single comb is produced.
183
Pure Breeding
Rose comb
(ppRR)
Pure Breeding
Pea comb
(PPrr)
184
Walnu
t
PpRr
46

47. Q Selfing the F1 generation:
Walnut
PpRr
Walnut
PpRr
X
185
9 Walnut : 3 Pea : 3 Rose : 1 single
Determine the probable genotypes of the four
different phenotypes above.
186
187
OTHER FORMS OF GENEINTERACTION
Q The concepts of Penetrance and expressivity:
Q These two concepts are different and yetrelated
and are often confused for one another
188
Q Penetrance simply refers to whether or not a given
gene or gene combination expressed
phenotypically in a population
Q refers to the statistical regularity with which a
gene produces its effect when present in the
requisite homozygous (or heterozygous) state (%
of individuals with the same alleles in a population
that express the associated phenotype)
47

48. Q Penetrance can either be complete or
incomplete
Some genes in homozygous as well as in heterozygous conditions
fail to fully express themselves phenotypically e.g. genes for
Diabetes mellitus in man
189
Q Expressivity refers to the extent to which
individuals with the same alleles express a
phenotype
Q We can have either full expressivity or variable
190
expressivity e.g. Gene for baldness inman.
Can range from thinning hair
to complete lack of hair.
Q Both the degree of penetrance (complete or
incomplete) and expressivity (invariant or variable)
are influenced by modifier genes and the
environment e.g. conditional mutations
Q Dark vs. light coloration in Siamese cats show the
effect of environment on the link between
genotype and phenotype.
191
TAKE HOME MESSAGE
Penetrance is a qualitative concept while expressivity is
a quantitative concept
Expressivity is dependent on penetrance because it is
not possible to measure the degree of expression if
the genotype is not expressed in the phenotype
Penetrance describes statistical variability among a
population of genotypes while expressivity describes
individual variability
192
48

49. The terms penetrance and expressivity quantify the
modification of gene expression by varying
environment and genetic background; they measure
respectively the percentage of cases in which the gene
is expressed and the level of expression.
193
MULTIPLE ALLELIC
INHERITANCE
194
OBJECTIVES OF THE LECTURE
Q By the end of the topic, you should be able to:
Q Explain what multiple alleles are.
Q Enumerate different characteristics under the
control of multiple alleles
Q Explain the basis of compatibility of different
blood groups in transfusion in man
195
Q Explain the iinnhheerriittaanncceeof sseellff--iinnccoommppaattiibbiilliittyyaalllleelleess
and the basis of sseellff--iinnccoommppaattiibbiilliittyyiinnssoommeespecies
of plants
Q Predict outcomes of progeny blood groups from
specific matings
Q Explain the pattern of inheritance of the Rh antigen
and consequences of pregnancies between
incompatible parents
196
49

50. DEFINITION
Q A characteristic is said to under the control of
multiple alleles if:
➢ its phenotypic expression is controlled by more
than two alternative alleles (BUT)
➢ only two of the alleles can be present at a locus at
any given time (one on each of the homologous
chromosomes passed on from the maternal and
paternal parents respectively).
197
Q when the entire population of a species is
considered, more than two alleles may be
phenotypically manifested in the entire
population at such a locus
Q Thus, multiple allele inheritance at a locus has
the effect of increasing the number of
phenotypic classes at that locus.
198
EXAMPLES
Q The A B O blood groups in man:
Q Controlled by three alleles, namely:
➢IA ; this one produces the antigen A on the
surfaces of red blood cells,
➢IB ; this one produces the antigen B on the
surfaces of red blood cells,
➢IO ; this is a recessive allele and produces
neither antigen A or B.
199
Q The alleles show the following dominance
relationships: (IA = IB) > IO.
Q Four possible blood group phenotypes exist (A,
B, AB and O) with the following characteristics.
200
Bloodgroup RBC antigen
Serum
antibodies
Possible
genotypes
A A Anti-b IAIA, IAIO
B B Anti-a
IBIB, IBIO
AB A &B None IAIB
O None
Both anti-a&
anti-b
IOIO
50

51. PRACTICAL IMPORTANCE OF BLOOD GROUP
TYPING
Q Cheap & easy tool for resolving paternity
disputes and baby mix-ups in hospital nurseries
(an exclusive test).
The following offspring are not possible from the
following matings
201
A B AB O
A AB & B
B None AB &A
AB O O O
O AB & B AB &A AB & O A, B, &AB
Q Determining compatibilities in blood
transfusion:
Q it is important that blood types be matched
before blood transfusions take place
Q If any two different blood types are mixed
together, the blood cells may begin to clump
together in the blood vessels (agglutination),
causing a potentially fatal situation.
202
Q agglutination is caused by a reaction between
the antigens on the surface of the red blood
cells of the donor and the antibodies in the
blood serum of the recipient
Recipient
203
Donor
A
Anti-b
B
Anti-a
AB
No
antibodies
O
Both
antibodies
A √ X √ X
B X √ √ X
AB X X √ X
O √ √ √ √
Q There are two types of ABO blood typing,namely:
➢Front or forward typing: here commercial anttii-A
and anti--Bare uusseedd
➢Here, the commercial antibodies are added to
the blood test sample in a test tube and then
centrifuged
➢The RBC’s are then examined for clumping
204
51

52. Anti A Anti B Anti A Anti B
A B
205
Anti A Anti B Anti A Anti B
OAB
Back or reverse typing: here commercially available
RBCs coated with Antigens A and B are added to two
tubes of plasma as shown below
A A BA BA B
AB
B
B A O
206
Q Persons with blood group O are called universal
donors and persons with blood group AB are
called universal recipients.
Q blood group O can only receive blood from a
group O donor and blood group AB can only
donate blood to group AB.
207
THE RHESUS FACTOR (ANTIGEN D)
Q when you are told your blood type, it is usually
expressed as a letter followed by either a
positive (+) or negative (-).
Q +/- refers to the rhesus system. One can be
rh+ve or rh-ve
Q The Rhesus factor (Rh-factor) was first
discovered in rhesus monkey blood
Q It is clinically the most important blood group
after the ABO system
208
52

53. Q frequency of this antigen is about 85% amongst the
whites and 92% amongst blacks and most Orientals.
Q The blood serum does not naturally contain antibodies
for the Rhesus antigen; they can only be produced
after the rhesus antigen has been introduced into the
blood.
Q an Rh- individual cannot receive Rh+ blood. Otherwise
he would become sensitized to produce anti-Rh
antibodies on the first transfusion such that on any
subsequent transfusion of Rh+ blood, agglutination
would occur.
209
210
COMPREHENSIVE ABO Rh COMPATIBILITY TABLE
Donor
R
e
c
i
p
i
e
n
t
O- O+ A- A+ B- B+ AB- AB+
O- √
O+ √ √
A- √ √
A+ √ √ √ √
B- √ √
B+ √ √ √ √
AB- √ √ √ √
AB+ √ √ √ √ √ √ √ √
RH FACTOR & PREGNANCY
Q Before, or at birth, it is common for some of the
baby's and mother's blood to mix
Q If the mother is Rh-ve & the baby Rh+ve, the
mother may be sensitized to produce Rh
antibodies against the Rh factor in the fetus
Q the first baby is not affected as they are
produced in her blood after this baby is born.
211
Q If subsequent pregnancies are Rh+, accumulated
antibodies can pass into the baby's circulation
across the placenta and destroy the baby's red
blood cells.
Q Medical condition is called erythroblastosis
foetalis
Q The baby either suffers from jaundice, anemia,
enlargement of the liver and spleen, dyspnea,
severe edema or quite often it is stillborn.
212
53

54. ANY PREVENTION?
Q The drug known as RhoGam is used (Rho(D)
immune globulin).
213
Q The Rh- mother is given an injection containing
Rh+ antiserum (serum containing Rh
antibodies ) after the delivery of her first child
Q These antibodies attach to Rh+ blood cells that
have passed from fetal circulation and
neutralize them, thus preventing the production
of anti-Rh+ antibodies.
PRECAUTIONS TO TAKE
Q Women should have their blood tested during
the first pregnancy to find out whether they
withRh-ve
Q b) The Rh+ antiserum must be given within 72
hours of the delivery of the first child.
Q c) Such an injection must not only be given
after the delivery of a child but also after an
abortion
214
Questions to ponder
54

55. Can universal blood be created from other types?
%The answer isyes
%In 2007 it was reported that types A, B, AB blood types
can be converted into type O
%This conversion is done using a glycosidase enzyme
from bacteria
%They cleave the blood group antigens from the RBC
%However, this does not address the problem of Rh
antigen in Rh+ individuals
%Only Blood from Rh- donors must be used
217
Q Coat color inheritance in rabbits:
Q Controlled by four alleles, namely:
➢C (for normal or wild-type / agouti coat
color),
➢Cch (responsible for the chinchilla coat
color),
➢Ch (for the himalayan coat color),
➢c (responsible for the albino phenotype).
218
Q The four alleles show graded dominance as
follows: C > Cch > Ch > c.
Agouti coat
(Dark brown)
Chinchilla coat
(Light brown)
220
Himalayan (Pure white
with black extremities)
Albino (Pure white
with pink eyes)
55

56. Q Ten different possible genotypes exist in the
population as follows:
221
Q Crosses between different phenotypes can produce
the following progeny depending on the genotypes
of the parents.
222
Questions to Ponder
223
224
Q Self incompatibility in flowering plants:
Q many plants show adaptations that encourage
cross--ppoolllliinnaattiioonnbut this does not necessarily insure
against inbreeding
Q this is because seeds and pollen arenormally
dispersed within a limited area
Q thus neighboring plants that are likely to fertilize
each other are often closely related to each other
Q In some species extra mechanisms exist that
prevent inbreeding between closely related
individuals
56

57. 225
Q One such mechanism is self incompatibility
Q It is usually controlled by a single gene locus S at
which there are several different alleles e.g. s1, s2, s3
etc.
Q two types of sseellff--incompatibilityeexxiisstt::ggaammeettoopphhyytic
and ssppoorroopphhyytic sseellff--iinnccoommppaattiibbiilliittyy
Q it is important to note that in order for successful
pollination to take place, pollen grains must land on
the stigma of a flower, germinate, grow through the
style of the stigma and reach the ovules in the ovary
where fertilization takes place
Q Gametophytic sseellff--iinnccoommppaattiibbiilliittyy:
Q it is always important to keep in mind that the
pollen genotype is always haploid and the stigma
genotype is always diploid
Q in such a system if pollen lands on a stigma with
a similar allele in its genotype, the pollen tube fails
to reach the ovule and so fertilization does not
occur
Q it is the most commontype
Q this occurs in plants such as the forage legume
Trifolium pprraatteennssee and tobacco Nicotiana ttaabbaaccuumm.226
227
Q Sporophyttiicc sseellff--iinnccoommppaattiibbiilliittyy:
Q here, the incompatibility phenotype of the pollen
grain is determined by the genotype of the parent
plant ((tthheessppoorroopphhyyttee)which produces the pollen.
Q in such a system it is possible for one allele to be
dominant. e.g. let s1 be dominant to s2
Q the dominance effect operates on the pollen while
it is forming so all pollen produced by s1 s2 plants
has the s1 phenotype regardless of its own
genotype
228
57

58. Thus, s2 pollen produced by an s1s2 plant will not grow
on any stigma that contains s1
229
Q This type of incompatibility is found in the
cabbage family (cauliflowers, broccoli, cabbages,
Brussels sprouts)
230
SampleQuestion
The prevention of self pollination in tobacco is controlled by a series of
multiple alleles known as S1, S2 and S3. A stigma of genotype S1S2
allows fertilization by pollen carrying S3 but not by pollen carrying
S1 or S2; i.e. fertile pollen must be of a different genotype to the
stigma.
i) Two tobacco plants of genotypes S1S3 and S2S3 are crossed
reciprocally. What are the genotypes of the F1 plants in each case?
ii) If all the F1 plants from the above crosses are pollinated with pollen
from S1S2 plant, what proportion of them will fail to set seed? (note
that pollen genotypes are haploid and stigma genotypes are
diploid).
GENE LINKAGE
231
OBJECTIVES OF THE LECTURE
Q By the end of the topic, you should be able to:
Q Define the terms linkage, crossing over and
cross over value.
Q Differentiate between partial and complete
linkage.
Q Explain the effect of linkage on test cross
ratios.
232
58

59. Q Explain and demonstrate the effect of crossing
over on phenotypic ratios in dihybrid crosses.
Q explain the importance of crossing over in a
sexually reproducing population.
Q Compute recombination frequencies between
linked genes
233
CONCEPT OF LINKAGE
Q Was arrived at based on deductive reasoning
that Mendel used to explain the mechanism of
inheritance in his monohybrid crosses
Q i.e. since the number of genes per species by
far exceeds its number of chromosome pairs &
since genes are located on chromosomes,
Q Then several genes occur on the same
chromosome
234
Q The genes located on the same chromosome
are said to be linked and they form one linkage
group.
Q The number of different linkage groups in
sexually reproducing higher organisms is
usually equal to the haploid number of
chromosomes
Q Can you think of any exceptions to this rule?
235
Q Linked genes are expected to be inherited as a
single block and hence they do not show
independent assortment.
Q genes that are always transmitted together are
said to show complete linkage
Q However, linked genes are occasionally
separated into different gametes by the
process of crossing over.
236
59

60. Q Such genes are said to show partial linkage.
Q Crossing over is the reciprocal exchange of
genetic material between non-sister chromatids
of homologous chromosomes during the
pachytene stage of Prophase I of meiosis.
Q During pachytene the chromatids of the
homologous chromosomes are criss-crossed
at points known as chiasmata.
237
Q Chromosomes exchange segments of DNA at the
chiasmata giving rise to individual chromosomes
that have a combination of DNA originally derived
from two different parents
238
ADVANTAGES OF CROSSING OVER
Q Increases genetic variability within a sexually
reproducing species, thus increasing adaptability
to changing environments over an evolutionary
timescale.
Q facilitates rapid spread of advantageous mutations
in a population.
239
DISADVANTAGE OF CROSSING OVER
Q it sometimes separates advantageous gene
combinations developed over a long
evolutionary time thereby reducing the overall
fitness of the population.
240
60

61. DETECTION OF LINKAGE
Q Recap the characteristic features of
independentassortment:
Q Genes located on different chromosomes
Q Gametes produced in equal proportions Q
Production of a characteristic 1:1:1:1 test
cross ratio
241
Q Discovered in D. melanogaster by Hunt Morgan
through non--iinnddeeppeennddeennttassortment of loci
Morgan’s expts.
Q Considered two characteristics: body color & wing shape
242
Dominan
t
Recessiv
e
He obtained a uniform normal F1 generation; test
crossed it with a double mutant; got 4 phenotypes
(total of 2300)
243
Each time he carried out this test cross he obtained the
following results instead of the expected 1:1:1:1 test
cross ratio
244
61

62. 245
5.22 : 5.10 : 1.11 : 1 (Ratio)
41.5% : 41.5% : 8.5% : 8.5% (%ages)
Q Note:
Q Some phenotypes are oveerr--represseenntteeddwhile
others aarreeuunnddeerr--rreepprreesseenntteeddin the test cross
progeny (this is always an indicator of linkage
between loci)
246
Q The uunnddeerr--rreepprreesseenntteeddgroups are the
recombinants due to crossing over and the over-
represented are the parental combinations.
Q Ratios such as 4:1:1:4; 7:1:1:7;5:1:1:5 etc are a
tale--ttaallee sign of partial linkage between loci.
Q Do you now realize the second use of a test cross?
… testing for linkage
SYMBOLIC REPRESENTATION
Q Choose symbols:
Q G represents gray body color & g represent
black body color
Q N represents normal wings and n represent
vestigial wings
Q Assign parental genotypes:
Q True breeding wild type parent: GGNN
Q Double mutant parent ggnn
247
Parental cross
248
62

63. Test cross
249
After random fertilization, the following progeny
genotypes were produced in the following
proportions
The proportion of the recombinant phenotypes to the total
progeny of the test cross when expressed as a % is known
as the recombination frequency or Cross over value.
COV = { (206 + 185) / (965 + 944 + 206 +185) }
250
Attempt this Please
Predict what would happen to the expected
dihybrid phenotypic ratio (9:3:3:1) and the
expected dihybrid test cross ratio (1:1:1:1) if
the loci for body color and shape of the wings
were completely linked?
251
63