Question

Question1 Not yet graded/1 pts In humans, having six fingers is dominant to five fingers. If all the assumptions of the Hardy- Weinberg Law are met, and 49% of the population has five fingers, what is the frequency of the heterozygotes in the population? Your Answer: Question 2 Not yet graded/1 pts A population of 400 Dachshunds includes 364 individuals with short hair. Short hair is dominant to long hair, and all of the assumptions of the Hardy-Weinberg Law are met in this population. How many individuals in this population are homozygous dominant? Write your answer as a whole number. Your Answer:

Answer 1

1) Consider the dominant allele for six fingers to be A and the recessive allele for five fingers to be a. Frequency of the recessive genotype (aa)= 49%

f(aa)= 49%

Hence, q²= 49/100=0.49

q= square root of 0.49= 0.7

p= frequency of dominant allele A

p+q=1

p= 1-q= 1-0.7= 0.3

According to Hardy-Weinberg equation

p²+2pq+q²=1

2 pq= Frequency of heterozygotes Aa

f(2pq)= 2 * 0.3*0.7= 0.42

Frequency of heterozygotes is 42%.

2) Homozygous dominant will have a genotype AA, where A is dominant allele. The recessive allele is ‘a’.

Short hair is dominant to long hair. Hence, frequency of dominant allele is p

p= 364/400=0.91

q=Frequency of recessive allele= 1-p= 1-0.91= 0.09

According to Hardy-Weinberg equation

p²+2pq+q²=1

p2 is the frequency of homozygous dominant genotypes= 0.91*0.91=0.8281

Hence, Number of homozygous individual= 0.8281*400= 331.24=331

Right option is 331.