Question

In need of assistance with questions 9 and 10 please

Hardy Weinberg Lab

When you have finished copy and past your answers into the submission box in the Hardy Weinberg Lab assignment.

Equations: p + q = 1 p 2 + 2pq + q2 = 1

1. what is the hardy weinberg symbol for the frequency of the dominant allele? ________

2. what is the hardy weinberg symbol for the frequency of the recessive allele __________

3. what is the hardy weinberg symbol for the frequency of the heterozygous genotype ________

4. what is the hardy weinberg symbol for the frequency of the homozygous dominant genotype _____

5. what is the hardy weinberg symbol for the frequency of the homozygous recessive genotype ______

6. If the allele frequency of "r" is .75 what is the allele frequency of "R" ______________

7. If the allele frequency of "R" is .15, what is the allele frequency of "r" _________________

8. In a population the homozygous dominant individuals make up 81% of the population, heterozygous individuals make up 18% and homozygous recessive, what is the frequency of the "p" allele?

9. Tay-Sachs disease is caused by a recessive allele. The frequency of this allele is 0.1 in a population of 3600.

A. what is the frequency of the dominant allele?

B. How many people in this population will have Tay-Sachs?

C. How many people in this population are Tay-Sachs carriers?

10. If a population is at Hardy Weinberg equilibrim then evolution _______________ happening.

Answer 1

9) Let the recessive allele be a, and dominant allele be A. According to the eqation:

p2+2pq+q2=1 ; p= frequency of A & q= frequency of a

A) So here it is given that the frequency of a is 0.1, have therefore the frequency of A or dominant allele is:

1-0.1= 0.9 ( as p+q = 1), so the frequency of the dominant allele is 0.9

B) So from the equation we can calculate the frequency of people that have Tay-Sachs, which is:

p2+2pq+q2=1 ; p= frequency of A & q= frequency of a

so q2 is the frequency of people that have Tay-Sachs, which is (0.1)2 = 0.01, that means 1% of the population have Tay-Sachs. So, we can calculate (1/100)*3600 people have Tay-Sachs = 36 people have Tay-Sachs

C) Tay-Sachs carriers means those people who are heterozygous for that gene, so from the eqation we can calculate that as 2pq, which is equal to 2*0.9*0.1 = 0.18, that is 18% of the population are carriers. Sowe can say that (18/100)*3600 = 648 people are carriers.

10) If a population is at Hardy Weinberg equilibrim then evolution is not happening. If a population is in a state called Hardy-Weinberg equilibrium, the frequencies of alleles, or gene versions, and genotypes, or sets of alleles, in that population will stay the same over generations (and will also satisfy the Hardy-Weinberg equation). Formally, evolution is a change in allele frequencies in a population over time, so a population in Hardy-Weinberg equilibrium is not evolving.