Question

8)Determine the value of a test statistic to test the null hypothesis that a population with the following number of individuals is in Hardy-Weinberg equilibrium:
Number of TT individuals = 1324
Number of Tt individuals = 658
Number of tt individuals = 198

Question 8 options:

	1) 0.433
	2) 0.0312
	3) 5.1
	4) 3.84
	5) 68.05

9)Based on your result from the question above, would this population be in Hardy Weinberg Equilibrium (with a 5% test)?

Question 9 options:

	1) No
	2) Yes

Q 10)

You are told that in a population of 10,000 individuals, there were 1,764 matings recorded between heterozygous individuals and 882 matings recorded between homozygous dominant and homozygous recessive individuals. Is this population in Hardy-Weinberg equilibrium?

Question 10 options:

	1) Yes
	2) No

Answer 1

8.
The first step is to calculate the allele frequencies for M and N.
            2*(# of TT) + (# of Tt)         2*(1324) + (658)
freq(T) = ---------------------------- =  -----------------
           2*(total # of individuals)     2*(1324 + 658 + 198)
freq(M) = 3306/4360 = 0.758
freq(N) = 1 - freq(M) = 1 - 0.758 = 0.242

The next step is to calculate the expected frequencies of each of
the genotypes given that the population is in Hardy-Weinberg
proportions.
freq(TT) = 0.758^2 = 0.575
freq(Tt) = 2*0.758*0.242 = 0.367
freq(tt) = 0.242^2 = 0.058

The expected numbers of each genotype are determined by
multiplying the expected frequency of each genotype by the number
of individuals in the sample.
expected number of TT = 0.575 * 2180 = 1253.5 = 1253
expected number of Tt = 0.367 * 2180 = 800.66 = 801
expected number of tt =  0.058 * 2180 = 126.44 = 126

The final step is to compare the expected and observed
distributions of numbers with chi-square analysis.  This will test
the null hypothesis that the population is in Hardy-Weinberg
proportions.                        
                       TT            Tt          tt             
observed numbers       1324          658         198
expected numbers       1253          801         126
((obs-exp)^2)/exp      4.02          25.53       41.14

chi-square value (Test Statistics) = 4.02 + 25.53 + 41.14 = 70.69

So closes option is 5.

9.
The degrees of freedom for the critical value is 1
(number of phenotypes - number of alleles) = (3-2) = 1
Thus the critical value to reject is 3.814 at confidence level of 95%.
As the calculated value is greater than the critical value, we
reject the null hypothesis that our population is in Hardy-Weinberg
proportions.  

So, Option 1 is correct i.e. NO

10.

Violations of the Hardy–Weinberg assumptions can cause deviations from expectation. How this affects the population depends on the assumptions that are violated.

• Random mating: The HWP states the population will have the given genotypic frequencies (called Hardy–Weinberg proportions) after a single generation of random mating within the population. When the random mating assumption is violated, the population will not have Hardy–Weinberg proportions. A common cause of non-random mating is inbreeding, which causes an increase in homozygosity for all genes.

Since in this example, a lot of inbreeding is happening (1,764 matings recorded between heterozygous individuals as compared to 882 matings recorded between homozygous dominant and homozygous recessive individuals) Random mating assumption is violated and thus the population will not be in Hardy-Weinberg equilibrium.