Chapter 25.2, Problem 23AQP

a.

Summary Introduction

To determine:

The calculation of genotypic and allelic frequency for the population.

Explanation of Solution

Table 1 Represents the frequency for M locus encodes for enzyme malate dehydrogenase are given:

Genotype	Number of individuals in population
${\text{M}}^{\text{1}}{\text{M}}^{\text{1}}$	20
${\text{M}}^{\text{1}}{\text{M}}^{\text{2}}$	45
${\text{M}}^{\text{2}}{\text{M}}^{\text{2}}$	42
${\text{M}}^{\text{1}}{\text{M}}^{\text{3}}$	4
${\text{M}}^{\text{2}}{\text{M}}^{\text{3}}$	8
${\text{M}}^{\text{3}}{\text{M}}^{\text{3}}$	6
TOTAL	125

Frequencies of different genotypes are as FOLOW:

$\begin{array}{c}\text{Genotypic frequency}=\frac{\text{Number of individual with genotype}}{\text{Total number of individual}}\\ \text{f}\left({\text{M}}^{\text{1}}{\text{M}}^{\text{1}}\right)\text{=}\frac{\text{20}}{\text{125}}\\ \text{= 0}\text{.16}\\ \\ \text{f}\left({\text{M}}^{\text{1}}{\text{M}}^{\text{2}}\right)\text{=}\frac{\text{45}}{\text{125}}\\ \text{= 0}\text{.36}\\ \\ \text{f}\left({\text{M}}^{\text{2}}{\text{M}}^{\text{2}}\right)\text{=}\frac{\text{42}}{\text{125}}\\ \text{= 0}\text{.34}\\ \\ \text{f}\left({\text{M}}^{\text{2}}{\text{M}}^{\text{3}}\right)\text{=}\frac{\text{8}}{\text{125}}\\ \text{= 0}\text{.064}\\ \text{f}\left({\text{M}}^{\text{3}}{\text{M}}^{\text{3}}\right)\text{=}\frac{\text{6}}{\text{125}}\\ \text{= 0}\text{.48}\end{array}$

The genotypic frequency of :

$\begin{array}{l}\text{f}\left({\text{M}}^{\text{1}}{\text{M}}^{\text{1}}\right)\text{= 0}\text{.16, }\\ \text{f}\left({\text{M}}^{\text{1}}{\text{M}}^{\text{2}}\right)\text{=0}\text{.36, }\\ \text{f}\left({\text{M}}^{\text{2}}{\text{M}}^{\text{2}}\right)\text{= 0}\text{.34, }\\ \text{f}\left({\text{M}}^{\text{2}}{\text{M}}^{\text{3}}\right)\text{= 0}\text{.064, }\\ \text{and f}\left({\text{M}}^{\text{3}}{\text{M}}^{\text{3}}\right)\text{= 0}\text{.48}\end{array}$

In Hardy-Weinberg equilibrium the numbers of expected individuals are:

$\begin{array}{c}\text{f}\left({\text{M}}^{\text{1}}\right)\text{= F}\left({\text{M}}^{\text{1}}{\text{M}}^{\text{2}}\right)\text{+}\frac{\left({\text{M}}^{\text{1}}{\text{M}}^{\text{2}}\right)}{\text{2}}\text{+}\frac{\left({\text{M}}^{\text{1}}{\text{M}}^{\text{3}}\right)}{\text{2}}\\ \text{=}\left(\text{0}\text{.16}\right)\text{+}\frac{\text{0}\text{.36}}{\text{2}}\text{+}\frac{\text{0}\text{.32}}{\text{2}}\\ \text{= 0}\text{.16+0}\text{.18+0}\text{.016}\\ p\text{ = 0}\text{.356}\end{array}$

$\begin{array}{c}\text{f}\left({\text{M}}^{\text{2}}\right)\text{= F}\left({\text{M}}^{\text{2}}{\text{M}}^{\text{2}}\right)\text{+}\frac{\left({\text{M}}^{\text{1}}{\text{M}}^{\text{2}}\right)}{\text{2}}\text{+}\frac{\left({\text{M}}^{\text{2}}{\text{M}}^{\text{3}}\right)}{\text{2}}\\ \text{=}\left(\text{0}\text{.34}\right)\text{+}\frac{\text{0}\text{.36}}{\text{2}}\text{+}\frac{\text{0}\text{.64}}{\text{2}}\\ \text{= 0}\text{.34+0}\text{.18+0}\text{.032}\\ q\text{ = 0}\text{.552}\end{array}$

$\begin{array}{c}\text{f}\left({\text{M}}^{\text{3}}\right)\text{= F}\left({\text{M}}^{\text{3}}{\text{M}}^{\text{3}}\right)\text{+}\frac{\left({\text{M}}^{\text{1}}{\text{M}}^{\text{3}}\right)}{\text{2}}\text{+}\frac{\left({\text{M}}^{\text{2}}{\text{M}}^{\text{3}}\right)}{\text{2}}\\ \text{=}\left(\text{0}\text{.48}\right)\text{+}\frac{\text{0}\text{.32}}{\text{2}}\text{+}\frac{\text{0}\text{.64}}{\text{2}}\\ \text{= 0}\text{.048+0}\text{.16+0}\text{.032}\\ r\text{ = 0}\text{.096}\end{array}$

The frequencies of alleles are ${\text{M}}^{\text{1}}{\text{, M}}^{\text{2}}\text{, and}{\text{M}}^{\text{3}}$ 0.356, 0.552, and 0.096.

b.

Summary Introduction

To determine:

The expected number of genotype if the population is in Hardy Weinberg equilibrium.

Explanation of Solution

The expected frequency for the genotypes in the population in Hardy Weinberg equilibrium:

$\begin{array}{c}\text{f}\left(M^1{M}^1\right)\text{= }{p}^2\\ \text{=}{\left(\text{0}\text{.356}\right)}^{\text{2}}\\ \text{= 0}\text{.127}\\ \text{Expected number of Leopards with genotype (}{M}^1{M}^1\text{)}=\text{ 0}\text{.127×125}\\ \text{= 16}\\ \\ \text{f}\left(M^1{M}^2\right)\text{= }2pq\\ \text{= 2×}\left(\text{0}\text{.356}\right)\left(\text{0}\text{.552}\right)\\ \text{= 0}\text{.393}\\ \text{Expected number of Leopards with genotype }(M^1{M}^2\text{)= 0}\text{.393×125}\\ \text{= 49}\end{array}$

$\begin{array}{c}\text{f}\left(M^2{M}^2\right)\text{= }{q}^2\\ \text{=}{\left(\text{0}\text{.552}\right)}^{\text{2}}\\ \text{= 0}\text{.305}\\ \text{Expected number of Leopards with genotype (}{M}^2{M}^2\text{)}=\text{ 0}\text{.305×125}\\ \text{= 38}\end{array}$

$\begin{array}{c}\text{f}\left(M^1{M}^3\right)\text{= }2pr\\ \text{= 2}\left(\text{0}\text{.352}\right)\left(\text{0}\text{.096}\right)\\ \text{= 0}\text{.68}\\ \text{Expected number of Leopards with genotype (}{M}^1{M}^3\text{)= 0}\text{.168×125}\\ \text{= 8}\text{.5}\end{array}$

$\begin{array}{c}\text{f}\left(M^2{M}^3\right)\text{= }2qr\\ \text{= 2}\left(\text{0}\text{.552}\right)\left(\text{0}\text{.096}\right)\\ \text{= 0}\text{.106}\\ \text{Expected number of Leopards with genotype}\left(M^2{M}^3\right)\text{= 0}\text{.106×125}\\ \text{= 13}\end{array}$

$\begin{array}{c}\text{f}\left(M^3{M}^3\right)\text{= }{r}^2\\ \text{=}{\left(\text{0}\text{.096}\right)}^{\text{2}}\\ \text{= 0}\text{.009}\\ \text{Expected number of Leopards with genotype (}{M}^3{M}^3\text{)= 0}\text{.009×125}\\ \text{= 1}\end{array}$

Table 1: Represent the chi-square test:

Genotype	Observed number (O)	Expected number (E)	$\left(O-E\right)$	${\left(O-E\right)}^2$	$\frac{{\left(O-E\right)}^{\text{2}}}{\text{E}}$
${\text{M}}^{\text{1}}{\text{M}}^{\text{1}}$	20	16	4	16	1
${\text{M}}^{\text{1}}{\text{M}}^{\text{2}}$	45	49	-4	16	0.33
${\text{M}}^{\text{2}}{\text{M}}^{\text{2}}$	42	38	4	166	0.42
${\text{M}}^{\text{1}}{\text{M}}^{\text{3}}$	4	8	-4	116	2
${\text{M}}^{\text{2}}{\text{M}}^{\text{3}}$	8	13	-5	25	1.9
${\text{M}}^{\text{3}}{\text{M}}^{\text{3}}$	6	1	5	25	25

Calculation for chi-square:

$\begin{array}{c}\text{Chi-square=}\sum \frac{{\left(O-E\right)}^2}{E}\\ =30.\text{65}\\ \text{Degree of freedom}=\text{ Number}\text{of}(\text{genotypes-alleles)}\\ =6\text{-3}\\ =3\end{array}$

From the above table of chi-square it is clear that value of p is less than 0.05. Thus, the population is in hardy –Weinberg equilibrium but locus is rejected.

Conclusion

Population of bear was found to be at Hardy-Weinberg equilibrium.