Question

Please show WORK!

In one species of rabbits, fur color is (hypothetically) controlled by a single gene, with dark fur being dominant to light fur. You study a wild population of these rabbits, and determine that 84% have dark fur.

1. If the population is in Hardy-Weinberg equillibrium, what would be the percentages of each genotype? (Homozygous dominant, Heterozygous, and Homozygous recessive)

2. The following spring, you examine 500 newly born babies, and observe that 390 of these have dark fur. Perform a Chi-Squared test (use p=0.05) to determine whether this generation remains in Hardy-Weinberg equillibrium. Be sure to make a conclusion based on these results (and phenotypes that you observe).

- What is your Chi-Squared value

- How many degrees of freedom

- What is your critical value (p=0.05)

- Is this population still in Hardy-Weinberg equillibrium? Why or why not?

3. If the population had 84% dark fur and had significant inbreeding (F= 0.2), what percentage of the next generation would you have expected to be heterozygous?

Answer 1

The Phenotype expressing dominant D is 84% of total population.

The phenotype expressing recessive d is the remaining population i.e 16%.(100-84)

The phenotype expressing dominant D is the sum of homozygous dominant (DD) and heterozygous dominant (Dd) population.

for easy calculation lets assume population of 100.if this population is in hardy weinberg equilbrium

(D+d)2 = 100

DD+2Dd= 84 (84% *100) (given)

(D+d)2 = D2+2Dd+d2 =100

84+d2 =100

d=4

(D+d)2 =100

(D+d)=10

Sub d=4 we get D= 6

So in a population in Hardy weinberg equilbrium

The Homozygote Dominant is = D2 = 36

The Heterozygote Dominant = 2Dd=48

Homozygote Dominant = d2=16

Since total population is 100 we can consider the above values in percentage itself (36/100*100)

2) Among 500 new born, 390 of them express dominant phenotype (D)

For a population in Hardy weinberg equilbrium phenotypes expressing dominant phenotype is 84 % of total population.In this case total population is 500 .so the expected number= (84/100 *500) = 420

To understand whether the population is following Hardy weinberg ,

we have to test the goodness of fit (Χ2) for the expected (E) and observed (O) population

Χ2 = (O-E)2 / E = (390-420)2/ 420 = 2.14

if the expected and observed values are equal the Χ2= 0 ( Note: The probability of this happening for all practical purpose is zero or p=0)

So we assume a significant p value to see the closeness of two population (p=0.05,this is accepted level of significance for large population)

Now we need to know the Χ2 value for p=0.05.This value is found in the Χ2 table . that is given below .We can see that we need to find a value degree of freedom to get the Χ2 value at p= 0.05 .

The degree of freedom is nothing but value  = no of expected categories -1 .The no of expected categories or genotypes is  2 (DD or Dd) .So degree of freedom =1(2-1)

Chi-Square (?) Distribution Area to the Right of Critical Value 0.95 0.90 0.10 0.05 Degrees of Freedom 0.99 0.975 0.025 0.01 2.706 4.605 6.251 0.020 0.115 0.297 0.554 0.872 1.239 1.646 2.088 2.558 3.053 3.571 4.107 4.660 5.229 5.812 6.408 7.015 7.633 8.260 8.897 9.542 10.196 10.856 11.524 12.198 12.879 13.565 14.257 14.954 0.001 0.051 0.216 0.484 0.831 1.237 1.690 2.180 2.700 3.247 3.816 4.404 5.009 5.629 6.262 6.908 7.564 8.231 8.907 9.591 10.283 10.982 11.689 12.401 13.120 13.844 14.573 15.308 16.047 16.791 0.004 0.103 0.352 0.711 1.145 1.635 2.167 2.733 3.325 3.940 4.575 5.226 5.892 6.571 7.261 7.962 8.672 9.390 10.117 10.851 11.591 12.338 13.091 13.848 14.611 15.379 16.151 16.928 17.708 18.493 0.016 0.211 0.584 1.064 1.610 2.204 2.833 3.490 4.168 4.865 5.578 6.304 7.042 7.790 8.547 9.312 10.085 10.865 11.651 12.443 13.240 14.042 14.848 15.659 16.473 17.292 18.114 18.939 19.768 20.599 7.779 9.236 10.645 12.017 13.362 14.684 15.987 17.275 18.549 19.812 21.064 22.307 23.542 24.769 25.989 27.204 28.412 29.615 30.813 32.007 33.196 34.382 35.563 36.741 37.916 39.087 40.256 3.841 5.991 7.815 9.488 11.071 12.592 14.067 15.507 16.919 18.307 19.675 21.026 22.362 23.685 24.996 26.296 27.587 28.869 30.144 31.410 32.671 33.924 35.172 36.415 37.652 38.885 40.113 41.337 42.557 43.773 5.024 7.378 9.348 11.143 12.833 14.449 16.013 17.535 19.023 20.483 21.920 23.337 24.736 26.119 27.488 28.845 30.191 31.526 32.852 34.170 35.479 36.781 38.076 39.364 40.646 41.923 43.194 44.461 45.722 46.979 6.635 9.210 11.345 13.277 15.086 16.812 18.475 20.090 21.666 23.209 24.725 26.217 27.688 29.141 30.578 32.000 33.409 34.805 36.191 37.566 38.932 40.289 41.638 42.980 44.314 45.642 46.963 48.278 49.588 50.892

From the above table Χ2 value with degree of freedom (1) and at (p=0.05)

Χ2 = 3.841

The two are same only if calculated Χ2 value < Χ2 value at p=0.05 or Χ2 Critical value

the calculated Χ2 value is 2.14 and  Critical Χ2 value is 3.84

Since it is less than critical value , the two population are similar and follow Hardy Weinberg Equilbrium

3)

Inbreeding always causes a decrease in heterozygote population This is computed from the F coefficent for inbreeding.

Hin = Hwb (1-F)

For the population in hardy weinberg equilbrium Heterozygote(Dd) population is 48 (Calculation shown in 1st problem)

Hin = 48 (1-0.2) =38.4

38.4 % of the next generation of population (inbreeding coefficent =0.2) will be heterozygotes.