genetics of populations

1. Genetic
Constitution
of Population

2. Species: The classic definition of a species is a group of related individuals or
populations which are potentially capable of interbreeding and producing fertile
offspring.
Strain: A strain is a group of individuals which share a presumed common
ancestry and have clear-cut physiological but not usually morphological
distinctions (e.g. a strain of winter wheat).
Breed: A breed is a group of animals presumably related by descent from
common ancestors and visibly similar in most characteristics.
A breed is a type of animal that differs from all others of the same species in
some way, has a separate history and breed name, and has been breeding true for
a number of generations.
Variety A variety or type is arbitrarily defined as a group of animals that share a
single physical trait. For example, rats with curly fur belong to the "rex" variety,
rats with low-set ears are called "dumbo," rats without hair are called "hairless."
A variety is therefore a purely descriptive category, it says nothing about the
biological relationship.
Review
of
some
terminologies

3. Variety
A variety or type is arbitrarily defined as a group of animals that share a single
physical trait. For example, rats with curly fur belong to the "rex" variety, rats
with low-set ears are called "dumbo," rats without hair are called "hairless."
A variety is therefore a purely descriptive category, it says nothing about the
biological relationship.
What's the difference between a breed and a strain?
There's obviously some overlap between the two terms: both refer to subgroups of
individuals in a given species which share common traits due to common descent.
However, the term strain trends to refer to physiological differences, while the
term breed tends to refer to morphological differences.
Physiological differences are differences in physical and chemical function, and
therefore are often internal and invisible.
Example: a high resistance to cold or heat, a blood clotting disorder, an
immunodeficiency, high milk yield (cows), strong wool (sheep) are examples of
physiological differencs that may define a strain.
Morphological differences, on the other hand, are differences in form or function,
and therefore tend to be external and visible.
Example: body size, leg length, coat color, coat length, tail length, head shape,
and ear placement are examples of morphological differences that may define a
breed.
Review
of
some
terminologies

4. Population
A population in the genetic sense is not just a group of individuals, but also a
breeding group. The genetics of a population is concerned not only with the
genetic constitution of the individuals but also with the transmission of the
genes from one generation to the next. In the transmission, the genotypes of the
parents are broken down and new set of genotypes are constituted in the
progeny – from the genes transmitted in the gametes. The genes carried by the
population thus have continuity from generation to generation, but the
genotypes in which they appear do not. The genetic constitution of a population,
referring to the genes it carries, is described by the array of gene frequencies,
that is by specification of the alleles present at every locus and the numbers or
proportions of the different alleles at each locus.
Review
of
some
terminologies

5. Population Genetics
Population genetics, is concerned with the statistical consequences of
Mendelism in a group of families or individuals; it studies the hereditary
phenomenon on the population level. The mechanism of heredity is presumed to
be what Mendelian genetics has described.
The life of an individual is limited in length of time, and his genetic make-up is
fixed throughout his life, barring mutation. In contrast, a population is
practically immortal, may be large or small in size, may be distributed over
wide or limited area, may change in genetic process chance, migration,
mutation or of selection. The study of population genetics is thus inevitably
related to that of organic evolution, which, from the genetical view point, is but
a process of cumulative change in the hereditary characteristics in a continuous
population certain principles or laws has emerged and population genetics shall
deal with those that have become well established.
Review
of
some
terminologies

6. Gene/Allele Frequency
The term gene frequency refers to the relative abundance or relative rarity
of a particular gene in a population as compared to its own allele in that
particular population. More precisely, the frequency of a gene is the
proportion of the loci of given allelic series occupied by particular gene.
Or,
Gene frequency indicates the proportion of a gene in a population as
compared to its own alleles of that particular population.
If nP is the number of P alleles in a population, then f (P) = nP/2n
Gene frequency may also be calculated from genotype frequencies.
If f(PP), f(Pp), f(pp) are the frequencies of three genotypes,
then f(P) = f(PP) + 1/2 f(Pp). The sum of all frequencies = 1

7. Genotype/Genotypic Frequency
The term genotype frequency refers to the relative abundance or relative
rarity of a particular genotype in a population as compared to its own
genotype in that particular population.
Or, Genotype frequency indicates the proportion of a genotype in a
population as compared to its own genotype in that particular population.
If the number of individuals in a population of polled and horned cattle with
each genotype is represented by n subscript with genotype, then
nPP = the no. of homozygous polled (PP)
nPp = the no. of heterozygous polled (Pp)
npp = the no. of homozygous horned (pp)
The genotype frequencies can then be represented as follows:
f(PP) = nPP/N, f(Pp) = nPp/N, f(pp) = npp/N; Where, N = nPP+ nPp+ npp
The sum of frequencies of all possible events must equal to 1.

8. Allele Frequencies: Consider an individual locus and a population of
diploid individuals where two different alleles, A and a, can be found at
that locus. If your population consists of 100 individuals, then that
group possesses 200 alleles for this locus (100 individuals x 2 alleles
at that locus per individual). The number of A alleles present in that
population expressed as a fraction of all the alleles (A or a) at that
locus represents the frequency of the A allele in the population.
(cont…)
How to Calculate Gene Frequency

9. 1. To calculate allele frequencies for populations of diploid organisms,
first multiply the number of individuals in the population by 2 to obtain
the total number of alleles at that locus.
2. Select one of the alleles for your first set of calculations. Let’s first
choose the A allele from the example provided above.
a. Individuals homozygous for the A allele will each possess 2 A alleles.
Multiply the number of AA homozygotes by 2 to calculate the number
of A alleles.
b. Heterozygotes will each possess only one A allele.
c. The total number of A alleles in the population = [(the number of Aa
heterozygotes) + (2 x the number of AA homozygotes)]
3. The frequency of the A allele = [(total number of A alleles in the
population) / (total number of alleles in population for that locus)]
4. The frequency of the a allele = (1 - frequency of the A allele)
How to Calculate Gene Frequency (cont.…)

10. Genotype Frequencies: Consider the same population, locus, and
alleles described above. Genotype frequencies represent the
abundance of each genotype within a population as a fraction of the
population size. In other words, the frequency of the AA genotype
represents the fraction of the population homozygous for the A allele.
1. To calculate genotype frequencies for populations of diploid
organisms, first determine the number of individuals with each
genotype present in the population. In the example used above, you
would count the number of individuals with the following genotypes: AA,
Aa, and aa.
2. To determine the frequency of each genotype, divide the number of
individuals with that genotype by the total number of individuals in the
population.
a. Frequency of AA genotype = # AA individuals / population size.
b. Frequency of Aa genotype = # Aa individuals / population size.
c. Frequency of aa genotype = # aa individuals / population size
How to Calculate Genotypic Frequency

11. Gene Counting Method
RR Rr rr Total
Red Roan White
496 x 2 416 x 1, 416 x 1 88 x 2
992 R 416 R, 416 r 176 r 1000 x 2
992 + 416 = 1408 416 + 176 = 592 2000
In a population of 1000 Shorthorn cattle, 496 have the red coat color, 416
are roan and 88 are white. Calculate the gene frequency in that population.
*1000 heads of cattle represented 2000 genes at the locus for R and r genes.
The 496 red cattle represent 2 x 496 “R” genes and 416 roan cattle
represent 416 R gene. Thus, the total number of R genes among the 2000
genes is 2 x 496 + 416 = 1408.
Calculation of Gene Frequency (Example 1)

12. Gene Counting Method (Cont.…)
Coat color
phenotype
Genotype of
individual
Number Number of
“R”
gene
“r” gene Total
Red RR 496 992 -
Roan Rr 416 416 416
White rr 88 - 176
Total 1000 1408 592 2000
The relative frequency of R genes in this population of 1000 cattle is
therefore, f(R) = 2 x 496 + 416/2000 = 1408/2000 = 0.704 = 0.7
The relative frequency of r genes in this population of 1000 cattle is
therefore, f(r) = 2 x 88 + 416/2000 = 592/2000 = 0.296 = 0.3
Calculation of Gene Frequency (Example 1 cont…)

13. In a population of 100 Shorthorn cattle, 50 are red, 40 are roan and remaining
10 are white coat color. What will be the frequency of red and white in that
population?
Coat color
phenotype
Genotype of
individual
Number Number of
“R”
genes
“r”
genes
Total
Red RR 50 100 0
Roan Rr 40 40 40
White rr 10 0 20
Total 100 140 60 200
140 out of 200 genes are red, thus the frequency of red genes is,
f(R) = 140/140+60 = 140/200 = 0.70 and
The frequency of white genes is therefore,
f(r) = 60/140+60 = 60/200 = 0.30
Calculation of Gene Frequency (Example 2)

14. Gene/Allelic Frequencies from Molecular Data
The individual whose DNA is found in lane 1 is homozygous. All of the others are
heterozygous. Individuals 2 and 8 share one allele, but differ at the other allele.
Likewise individuals 3 and 10 share one allele, but are polymorphic at their second
alleles. If a large number of individuals are scored with respect to this locus, then
allelic frequencies for each allele (band) can be determined.

15. Now let’s score the allelic frequencies for a locus recognized by a molecular marker. The human
population represented by the above image contains individuals with either a 6.5 kb fragments or
a 3.0 kb fragment or both fragments. (Remember that those individuals with a single band
actually contain two copies of that band.) The numbers of individuals scored with respect to each
biochemical phenotype are given above the band. The allelic frequency of each band can be
determined in a similar manner as in the previous examples. p is the frequency of the 6.5 kb
fragment in the population. Its value is p = [(2)(30) + 50]/200 = 0.55. The frequency of the 3.0kb
fragment is then q = [(2)(20) + 50]/200 = 0.45
Gene/Allelic Frequencies from Molecular Data

16. Causes of changes of genetic
constitution of a population
(i) Population size
(ii) Differences in fertility and viability
(iii) Migration and mutation
(iv) Mating system

17. Causes of changes of genetic
constitution of a population (contd..)
(i) Population size: The genes passed from one generation to the
next are sample of the genes in the parent generation. Therefore the
gene frequencies are subject to sampling variation between
successive generations and the smaller the number of parents the
greater in the sampling variation. For practical purposes a large
population is one in which the number of adult individuals is in the
hundreds rather than in the ten.

18. (ii) Differences in fertility and viability: The different genotypes
among the parents may have different fertilities. In this way gene
frequency may be changed in the transmission. Further, the
genotypes among newly formed zygotes may have different
survival rates and so the gene frequencies in the new generation
may be changed by the time the individuals are adult and
themselves become parents.
Causes of changes of genetic
constitution of a population (contd..)

19. (iii) Migration and mutation: The gene frequencies in the population
may also be changed by migration and mutation of individuals from
another population.
(iv) Mating system: The genotypes in the progeny are determined by
the union of the gametes in pairs to form zygotes, and the union of
gametes is influenced by the mating of the parents. So the genotype
frequencies in the offspring generation are influenced by the
genotypes of the pairs that mate in the parent generation.
Causes of changes of genetic
constitution of a population (contd..)

20. Factors modifying gene frequency
1. Selection: This is very important factor that may responsible for
changes in gene frequencies. Selection means that individuals
possessing certain desirable traits are caused or allowed to
produce the next generation. This causes an increase in the
frequency of some genes and a decrease in the frequency of
others.

21. Factors modifying gene frequency (contd...)
Example: Assume that a herd of 100 shorthorn cows consisting of 25
red, 50 roan and 25 white:
Coat color
phenotype
Genotype No. of individuals
No. of
R gene r gene
Red RR 25 50 0
Roan Rr 50 50 50
White rr 25 0 50
Total 100 100 100
In this herd, the frequency of the red and white genes would be equal (0.50 each).
The relative frequency of red genes (R) in this population is therefore,
f(R) = 2 x 25 + 50/200 = 100/200 = 0.5
The relative frequency of white genes (r) in this population is therefore,
f(r) = 2 x 25 + 50/200 = 100/200 = 0.5

22. Now if we decide to cull and sell all of the white individuals and leaving
25 red and 50 roan individuals, then after culling all white individuals-
Factors modifying gene frequency (contd...)
Coat color
phenotype
Genotype No. of individuals
No. of
R gene r gene
Red RR 25 50 0
Roan Rr 50 50 50
Total 75 100 50
By culling the white and roan individuals in this herd, the frequency of the white
gene would be reduce to 0 and that of red gene would be increased to 100.
Now, the relative frequency of red genes (R) would be,
f(R) = 2 x 25 + 50/150 = 100/150 = 0.667 (increased)
The relative frequency of white genes (r) would be,
f(r) = 50/150 = 0.333 (decreased)

23. Factors modifying gene frequency (contd...)
2. Mutation: The source of genetic variability may be caused by
mutation. Mutation proposes new genes, and selection dictates
whether these genes are retained or rouged from the population.
Mutation must be recognized for the important part because it plays
in providing new variation over long periods of time, permitting the
population to adopt the changing environmental stress. It is not a
potent force in changing gene frequency in most economically
important species because most mutation rates are low and most
mutations are harmful. Their general recessive nature makes it
difficult to increase their frequency in initial selection.

24. Factors modifying gene frequency (contd...)
3. Migration/A mixture of population: The introduction of new
genes into a population can change gene frequency. Suppose
we have a population of human in which the frequency of the M
blood group is 0.925. In another population some miles away the
frequency of the same gene is 0.251. If the individuals from the
first group migrated to the second and they intermarried, the
frequency of the M gene in the children would average
somewhere between 0.215 and 0.915, depending upon how
much intermarried occurred.

25. Factors modifying gene frequency (contd...)
4. Genetic drift or sampling nature of inheritance: This is
another factor that may be responsible for changes in gene
frequencies and may result in a type of genetic variation within a
population that is not due to natural selection.
Suppose that in large population of human the frequency of M gene
were 0.60 and that of N gene were 0.40. Let us also assume that a
small group of people (say a dozen) left the large group and
migrated to a new country to settle in a remote area where they are
isolated from other humans. It is possible that the frequency of the
M gene in the small group might be very low and that of the N gene
may be very high or vice-versa, just because of change, due to a
small population (sample) coming from large one.

26. The Hardy-Weinberg Law of
Genetic Equilibrium

27. Hardy-Weinberg Law
In 1908, Hardy a British mathematician and Weinberg, a German physician
working independently, published certain fundamental ideas of gene
distribution in populations. These are known as Hardy- Weinberg law.
The law states that in a large random mating population with no selection,
mutation or migration, the gene frequencies and genotype frequencies are
constant from generation to generation and further more, there is a simple
relationship between the gene frequencies and the genotype frequencies.

28. A Large Breeding population: A large breeding population helps to ensure that chance alone
does not disrupt genetic equilibrium. In a small population, only a few copies of a certain allele
may exist. If for some chance/reason the organisms with that allele do not reproduce
successfully, the allelic frequency will change. This random, nonselective change is what happens
in genetic drift or a bottleneck event.

29. Random Mating: In a population at equilibrium, mating must be random. In assortative mating,
individuals tend to choose mates similar to themselves; for example, large blister beetles tend to
choose mates of large size and small blister beetles tend to choose small mates. Though this
does not alter allelic frequencies, it results in fewer heterozygous individuals than you would
expect in a population where mating is random.

30. No Change in Allelic Frequency Due to Mutation: For a population to be at Hardy-
Weinberg equilibrium, there can be no change in allelic frequency due to mutation. Any
mutation in a particular gene would change the balance of alleles in the gene pool.
Mutations may remain hidden in large populations for a number of generations, but may
show more quickly in a small population.

31. No Immigration or Emigration: For the allelic frequency to remain
constant in a population at equilibrium, no new alleles can come into the
population, and no alleles can be lost. Both immigration and emigration
can alter allelic frequency.

32. No Natural Selection: In a population at equilibrium, no alleles are selected
over other alleles. If selection occurs, those alleles that are selected for will
become more common. For example, if resistance to a particular herbicide
allows weeds to live in an environment that has been sprayed with that
herbicide, the allele for resistance may become more frequent in the population.

33. Hardy-Weinberg Equilibrium
A population with constant gene and genotype frequencies is said to be in
Hardy-Weinberg equilibrium. The necessary conditions for equilibrium
are-
 The organisms in question is diploid
 Reproduction is sexual
 Generations are non-overlapping
 Mating is random
 Population size is very large
 Migration is negligible
 Mutation can be ignored
 Natural selection does not affect the locus under consideration

34. Proof of Hardy-Weinberg Law:
The proof of Hardy-Weinberg law involves four steps, which are summarized
in the following table:
Step Deduction from: to Conditions
01. a) Gene frequency in parents 01. Normal gene segregation
02. Equal fertility of parents
03. Equal fertilizing capacity of gametes
04. Large population
05. Random mating
06. Equal gene frequencies in both parents
07. Equal viability
Gene frequency in all gametes
01. b) Gene frequency in gametes forming
zygotes
02. Genotype frequencies in zygotes
03. Genotype frequencies in progeny
04. Gene frequency in progeny
Steps of deduction in proof of Hardy-Weinberg law and condition that must hold

35. Application of Hardy-Weinberg Law:
 Gene frequency of recessive allele: The frequency of recessive gene is
equal to the square root of the frequency or proportion of homozygous
recessive individuals.
 Frequency of carriers: The proportion of heterozygous dominant
individuals (carriers) in a population would be two times the frequency
of the dominant gene.
 Test of Hardy-Weinberg Law: If the sum of the frequency of two
homozygous genotypes equal 1, then the population is said to be in
equilibrium.

36. The Hardy-Weinberg Equation
To estimate the frequency of alleles in a population, we can use
the Hardy-Weinberg equation. According to this equation:
p = the frequency of the dominant allele (represented here by A)
q = the frequency of the recessive allele (represented here by a)
For a population in genetic equilibrium:
p + q = 1.0 (The sum of the frequencies of both alleles is 100%.)
(p + q)2 = 1 so, p2 + 2pq + q2 = 1
The three terms of this binomial expansion indicate the
frequencies of the three genotypes:
p2 = frequency of AA (homozygous dominant)
2pq = frequency of Aa (heterozygous)
q2 = frequency of aa (homozygous recessive)

37. How to Calculate Gene Frequency of Recessive Allele
As we know gene frequencies among a group of individuals can be
determined from their genotypic frequencies p = P + ½ H and q = Q + ½
H, but for this it was necessary to know the frequencies of all three
genotypes.
Consequently, the relationship in equations p = P + ½ H and q = Q + ½
H cannot be applied to the case of a recessive allele, when the heterozygote
is indistinguishable from the dominant homozygote. If the genotypes are
in Hardy-Weinberg proportions, however, we do not need to know the
frequencies of all the three genotypes.
Let a, for example be a recessive gene with a frequency of q: then the
frequency of aa homozygote is q2, and the gene frequency is the square-
root of the homozygote frequency (√q2 = gene frequency).

38. Calculation of Gene Frequency of Recessive Allele (example)
1. In a population of 1000 Angus cattle, 40 were red and remaining
were black. What are the frequency of red and black gene?
2. Blue versus dark eye color in men could be due to a recessive
genetic difference, bb = blue, BB and Bb = dark. If the frequency
of blue individuals in a population is 16%, what is the gene
frequency of blue and dark eye color?

39. How to Calculate Gene Frequency of Carriers
It is often interest to know the frequency of heterozygotes or carriers of recessive
abnormalities and this can be calculated if the gene frequency is known. If
Hardy-Weinberg equilibrium can be assumed, the frequency of heterozygotes
among all individuals, including homozygotes is given by 2q (1-q).
If “q” is known then heterozygotes = 2pq Since
p = (1-q) therefore, 2pq = 2q (1-q) or carrier is = 2q (1-q)
The frequency of heterozygotes among normal individual denoted by H’ is ratio
of genotype frequencies Aa/(AA+Aa), where a is recessive allele, so, when q
is the frequency of “a”,
H’ = Aa/AA+Aa = 2pq/p2+2pq = 2q (1-q)/(1-q)2+2q(1-q) = 2q (1-
q)/(1-q)(1-q)+2q(1-q) = 2q(1-q)/(1-q){1-q+2q} = 2q (1-q)/(1-q+q+q) =
2q (1-q)/(1-q)(1+q) = 2q/(1+q)

40. Calculation of Gene Frequency of Carriers (example)
1. Phenylketonuria (PKU) is a human metabolic disease due to a single
recessive gene. Homozygotes can be detected a few days after birth. There
are five cases in 55,715 babies. How many carriers would you expect to be in
that population?
2. Assume a gene pool of 100 dogs. 16% of the population is PRA affected.
How many would be considered carriers (meaning - what is the gene
frequency of PRA for the breed)?
Hint:
1. q2 = 5/55,715 = 1/11000
q = √1/11000 = √0.000090 = 0.009486 = 0.0095.
2. If the population is 100 and 16% are affected... the probability is that the
carrier rate of the PRA gene would be 48%. The gene frequency would be
40% within the population. Calculate using formulas as appropriate.

41. The Hardy-Weinberg Law of Genetic Equilibrium
Estimating Allelic Frequency： If a trait is controlled by two alternate alleles,
how can we calculate the frequency of each allele? For example, let us look at
a sample population of pigs.
The allele for black coat is recessive to the allele for white coat. Can you
count the number of recessive alleles in this population?

42. Sample Problem 1
In a population of pigs, the allele for black coat is recessive. Using the Hardy-Weinberg
equation can you determine the percent of the pig population that is heterozygous for
white coat?
Hint:
1. Calculate q2
Count the individuals that are homozygous recessive in the illustration above.
Calculate the percent of the total population they represent. This is q2.
2. Find q
Take the square root of q2 to obtain q, the frequency of the recessive allele.
3. Find p
The sum of the frequencies of both alleles = 100%, p + q = l. You know q, so what is p,
the frequency of the dominant allele?
4. Find 2pq
The frequency of the heterozygotes is represented by 2pq. This gives you the percent
of the population that is heterozygous for white coat:

43. In a certain population of 1000 fruit
flies, 640 have red eyes while the
remainder have sepia eyes. The
sepia eye trait is recessive to red
eyes. How many individuals would
you expect to be homozygous for
red eye color?
Hint: The first step is always to
calculate q2! Start by determining the
number of fruit flies that are
homozygous recessive.
Calculations:
q2 for this population is 360/1000 = 0.36
q = 0.6, p = 1 - q = 1 - 0.6 = 0.4
The homozygous dominant frequency = p2 = (0.4)(0.4) = 0.16.
Therefore, you can expect 16% of 1000, or 160 individuals, to be
homozygous dominant.
Sample Problem 2

44. Allelic Frequency vs Genotypic Frequency
Genotypic frequency is the frequency of a genotype - homozygous recessive,
homozygous dominant, or heterozygous - in a population. If you don't know the
frequency of the recessive allele, you can calculate it if you know the
frequency of individuals with the recessive phenotype (their genotype must be
homozygous recessive).
Sample Problem 3
If you observe a population and find that 16% show the recessive trait, you
know the frequency of the aa genotype. This means you know q2. What is
q for this population?

45. In a population of 1000 individuals, you find that the population consists of
90 individuals homozygous for the A allele (AA genotype), 490 individuals
homozygous for the a allele (aa genotype) and 420 individuals
heterozygotes (Aa genotype). Assume that you have no information on the
presence or absence of evolutionary mechanism in this population.
1. Calculate the genotype and allele frequencies for this population.
2. Determine if this population meets Hardy Weinberg Assumptions (in
other words determine if evolutionary mechanisms operate in this
population).
Sample Problem 4

46. Sample Problem 4 (cont…)
Calculation of Allele and Genotype Frequencies
Since you do not know if this population meets Hardy Weinberg Assumptions,
you must calculate both the allele and genotype frequencies using the raw
data.
1. Allele Frequencies: The frequency of the A allele will equal: (total
number of A alleles in the population) / (total number of alleles in population
for locus) = [(90*2) + 420] / (1000*2) = 0.30 The frequency of the a allele will
equal: (1 - 0.30) or (total number of a alleles in the population) / (total
number of alleles in population) = [(490*2) + 420] / (1000*2) = 0.70

47. 2. Genotype frequencies:
Frequency of AA genotype = # AA individuals / population size = 90/1000 = 0.09
Frequency of Aa genotype = # Aa individuals / population size = 420/1000 = 0.42
Frequency of aa genotype = # aa individuals / population size = 490/1000 = 0.49
Hardy-Weinberg Predictions: If no evolutionary mechanisms operate on this locus, then
the Hardy-Weinberg Equilibrium Theory predicts that the genotype frequencies should be
as follows:
Frequency of AA = (frequency of A allele)2 = (0.3)2 = 0.09
Frequency of Aa = 2 x (frequency of A allele) x (frequency of a allele) = 2*0.3*0.7 = 0.42
Frequency of aa = (frequency of a allele)2 = (0.7)2 = 0.49
Conclusion: Since the observed genotype frequencies equal to those predicted by the
Hardy-Weinberg Equilibrium Theory, we conclude that no evolutionary mechanisms
operate on this locus in this population (i.e., the population meets the assumptions of the
Hardy Weinberg Theory).
Sample Problem 4 (cont…)

48. A pair of co-dominant alleles governs coat color in Shorthorn cattle, CRCR
=Red, CRCW =Roan and CWCW =White. A sample of a population
revealed the following phenotypes: 180 red, 240 roan and 80 white. Does
the sample indicate that the population is in equilibrium?
Sample Problem 5

49. In a population 80% are horned cattle and the rest are polled animals, if
polled condition is desirable then how much increase in the frequency of
desired genes can be achieved in subsequent generations?
Sample Problem 6

50. 1. If the frequency of two alleles in a gene pool is 90% A and 10% a, what is
the frequency of individuals in the population with the genotype Aa?
a. 0.81
b. 0.09
c. 0.18
d. 0.01
e. 0.198
The Hardy-Weinberg Law of Genetic Equilibrium (Quiz)
2. In a population that is in Hardy-Weinberg equilibrium, the frequency of the
homozygous recessive genotype is 0.09. What is the frequency of individuals
that are homozygous for the dominant allele?
a. 0.7
b. 0.21
c. 0.42
d. 0.49
e. 0.91

51. 4. In humans, Rh-positive individuals have the Rh antigen on their red blood
cells, while Rh-negative individuals do not. If the Rh-positive phenotype is
produced by a dominant gene (A), and the Rh-negative phenotype is due to
its recessive allele (a), what is the frequency of the Rh-positive allele if 84%
of a population is Rh-positive?
a. 0.04
b. 0.16
c. 0.48
d. 0.60
e. 0.84
3. Which of the following is NOT a condition that must be met for Hardy-
Weinberg equilibrium?
a. Large population size
b. No mutations
c. No immigration or émigration
d. Dominant alleles more frequent than recessive alleles
e. No natural selection
The Hardy-Weinberg Law of Genetic Equilibrium (Quiz)

52. P
A
V
V
population
in
variation
phenotypic
Total
gene
of
effects
additive
to
due
Variation
Definition:
Heritability is defined as the proportion of the total
phenotypic
variance in a trait that is due to the additive effects of
genes, as
compared to dominance or environmental effects.
Heritability
h2
=
=

53. P
G
V
V
population
in
variation
phenotypic
Total
genetics
to
due
Variation
 Heritability is the proportion of the total phenotypic
variance of
a trait that is due to all genetic effects including additive
and
dominance effects.
=
H
In genetics,
 Heritability is the proportion of phenotypic variation in a
population that is attributable to genetic variation among
individuals.
 Variation among individuals may be due to genetic and/
or environmental factors.

54. Estimation of Heritability:
The main basis of estimating heritability of a trait is the
degree of resemblance between relatives.
The degree of resemblance between relatives depends
upon the degree to which relative’s have-
• Genes in common
• Genotype in common
• Environment in common
Scale of heritability:
• Low heritability ranges from: 0-0.11
• Medium heritability ranges from: 0.12-
0.30

55. Relatives Covariance (VA)
Offspring-parents 1/2
Half sibs 1/4
Full sibs 1/2
Cousin 1/8
Double cousin 1/16
The following tables shows the covariance of the relative

56. Methods of Estimating of Heritability:
•Parent-offspring regression method
•Sib analysis or half/ full sib correlation
•Using twin information or twin analysis
•Results of selection experiment
•The knowledge of repeatability estimate
•Restricted Maximum Likelihood (REML)
approach
•Markov Chain Monte Carlo (MCMC) simulation
approach.

57. Relatives Covariance (VA) Regression(b)/correlati
on (r)
Offspring and one
parent
Offspring and mid
parent
½VA
VA
b= ½ h2 or h2= 2b
b= h2
Half sibs ¼ V r = ¼ h2 or h2 = 4r
Parent-offspring regression method:
 The data are obtained in the form of measurement of parent
and their offspring. Then the covariance is calculated from cross-
product of the paired values:

58. Sib correlation Estimates of heritability
Half sib’s, t(HS)
=
= σ2
s/ σ2
t
Sire component; h2= 4 σ2
s/
σ2
t
= 4t
Dam component; h2= 4
σ2
s/ σ2
t
= 4t
Full sib’s, t(FS)
=
Sire + Dam component;
h2= 2 (σ2
S+ σ2
D/ σ2
t)
Variation
Total
sire
to
due
Variation
Variance
Total
variance
Dam
variance
Sire 
Sib analysis or half/ full sib correlation:

59. •Twin analysis:
 The method is widely used in estimating
heritability of human traits and in some extent in
cattle.
Because occurrence of twin in human is more but
in cattle it is 0.3-0.4%.
 The reasons for difference in any trait between
the members of identical twin pair are
environment.
 In the other hand, the reasons for differences
any trait between the members of fraternal twin
pairs are heredity and environment.

60. FT
to
due
variation
heredity
to
due
Variation
P
H
V
V
Identical twin (IT)
Environment
Fraternal twin (FT) Heredity + Environment
Variation due to heredity
VH= VFT –VIT; σ2
G = σ2
FT – σ2
IT
So, heritability in broad sense,
H =
σ2
G/ σ2
FT
Reasons for variation:

61. variation
Total
nt
environmne
to
due
Variation
genetic
to
due
Variation 
P
EP
G
V
V
V 
•The knowledge of repeatability estimation:
Repeatability is the degree of association between
two records
for the same trait measured on the same animal and
in terms of variance components-
Repeatability =
r =

62. Intra-class correlation by analysis of variance
The intraclass correlation is commonly used to quantify the
degree to which individuals with a fixed degree of relatedness (e.g.
full siblings) resemble each other in terms of a quantitative trait
(heritability).
The intraclass correlation is a correlation coefficient
representing the correlation between two observations in the
same class.
This intraclass correlation is equal to σ2
a / σ2
a + σ2
e.
This is the proportion of the total variance among, because the
total variance is σ2
a + σ2
e.

63. Selection differential
The difference between the mean of a population and the
mean of the individuals selected to be parents of the next
generation.
Breeding value the sum of gene effects of a breeding animal as
measured by the performance of its progeny
Genetic gain is the amount of increase in performance that is
achieved through artificial genetic improvement programs.
Selection Index is a method of artificial selection in which several
useful traits are selected simultaneously

64. Utility or Uses of Heritability:
 To estimate progress and set backs in different
traits from different mating
 To estimate the genetic improvement in the
progeny of selected parents
 To decide which animal evaluation method
should be used
 To know the degree of inheritance of that
particular trait
 It can be used in order to plan and execute

65.  To Predict the genetic merit or breeding values of
animals
 Used for estimating the correlation of genetics between
traits
 Used to decide mating strategy or plan in practical
animal breeding
 Used to suggest which general type of breeding
program is optimum for a particular production situation
 To decide which type of records should be used for
prediction of genetic merit of animals