Question

1) You are screening a population of 350 island sheep for coat color, and you count 100 black sheep and 250 white sheep. What is p, the frequency of the “black” allele of the TYRP1 gene? Crossing true breeding black sheep with true breeding white sheep always results in 100% black sheep.

a.		p = 0.2857
b.		p = 0.7143
c.		p = 0.8452
d.		p = 0.1548

2) In the same population, if it is under Hardy-Weinberg equilibrium, how many individuals do you expect to be heterozygous at TYRP1?

a.		92
b.		100
c.		8
d.		75

3) You come back after the population has gone through one generation, and count 445 white sheep and 208 black sheep. What is the expected number of heterozygotes in the new generation, if this population is in Hardy-Weinberg equilibrium?

a.		171
b.		187
c.		11
d.		208

4) What is the observed number of heterozygotes in the new generation?

a.		535
b.		118
c.		188
d.		156

5) Using the same logic, you get the expected and observed number of homozygotes in the new generation, and run a χ² test. You find your p-value is 0.8377. What can you say about this population?

a.		This population is in Hardy-Weinberg equilibrium at the TYRP1 locus: the TYRP1 is not experiencing any force of evolution.
b.		This population is in Hardy-Weinberg equilibrium at the TYRP1 locus: therefore this population is not undergoing any evolution at all.
c.		This population deviates from HWE at the TYRP1 locus: the TYRP1 locus is causing nonrandom mating
d.		None of the above

Answer 1

Answer=

1) a] 0.2857

Explaination=

population of = 350 island sheep for coat color

black sheep =100

and white sheep = 250

p, the frequency of the “black” allele=?

the frequency of the “black” allele p= no of black sheeps /Total number of sheeps

= 100 /350

= 0.2857

P = 0.2857

2) In the same population, if it is under Hardy-Weinberg equilibrium, how many individuals do you expect to be heterozygous at TYRP1?

The population, is under Hardy-Weinberg equilibrium,

then

frequency of heterozygous individual = 2pq

where p=frequency of black allele = 0.2857

q= frequency of black allele

we know that when

The population, is under Hardy-Weinberg equilibrium,

then p+q =1

q = 1- p

= 1-0.2857

=0.7143

q = 0.7143

frequency of heterozygous individual = 2pq = 2 X 0.2857 X 0.7143

2 pq =0.408151

the number of heterozygotes in a population of 350 sheeps = 0.4081 X 350

= 142

3) You come back after the population has gone through one generation, and count 445 white sheep and 208 black sheep. What is the expected number of heterozygotes in the new generation, if this population is in Hardy-Weinberg equilibrium?