Question

Question 6:

Calculate the genotypic frequencies that you would expect if the population is in Hardy-Weinberg equilibrium.

(Remember, that we are preparing to use the Chi-square goodness-of-fit test to determine if this population is in fact in Hardy-Weinberg equilibrium).

Round your answer to three decimal places and submit your answer in the correct format (e.g. 0.062).

(a) The frequency of the SS genotype is:

(b) The frequency of the Ss genotype is:

(c) The frequency of the ss genotype is:

The total number of leopards in this population is 284.

For reference:

Phenotype	Genotype	Number of leopards
brown pigmentation	SS	194
brown pigmentation	Ss	88
red pigmentation (strawberry leopard)	ss	2

S = dominant allele for brown pigmentation

s = recessive allele for red pigmentation (erythrism)

the frequency of the S (dominant) allele in this population: 0.916

the frequency of the s (recessive) allele in this population: 0.084

Answer 1

Answer-

According to the given question-

Here we have a population of about 284 leopards having two-color of pigment, one brown which is dominant and another is red which is recessive. we have to find the genotypic frequencies when the population is in Hardy-Weinberg equilibrium.

the dominant allele is for brown color pigment = S

the recessive allele is for red color pigment or erythrism = s

Brown pigmentation = genotype SS = 194 leopards

Brown pigmentation = genotype Ss = 88 leopards

Red pigmentation = genotype ss = 2 leopards

Total number of leopards = 194 + 88 + 2 = 284

We know that according to the Hardy Weinberg equilibrium-

p + q = 1

and p² + 2pq + q² = 1

where

p = frequency of dominant allele

q = frequency of recessive allele

p² = Frequency of individuals having dominant genotype

2pq = Frequency of heterozygous individuals

q² =Frequency of individuals having dominant genotype

The frequency of Recessive allele = q = (Number of homozygous recessive 2) + Number of heterozygous Total population 2 = (2   2) + 88   284 2

= 92 568 = 0.1619

q = 0.1619

So p + 0.1619 = 1 , p = 1 - 0.1619 = 0.8381

Thus p = 0.8381 and q = 0.1619

p² = 0.7024 and q² = 0.0262

p² + 2pq + q² = 1

0.7024 + 2pq + 0.0262 = 1

2pq = 1 - 0.7286 = 0.2714

2pq = 0.2714

frequency of dominant allele (S) in this population = p = 0.8381

frequency of the recessive allele(s) in this population = q = 0.1619

(a) The frequency of the SS genotype = p² = 0.7024

(b) The frequency of the Ss genotype = 2pq = 0.2714

(c) The frequency of the ss genotype = q² = 0.0262

The number of homozygous dominant = p² number of population = 0.7024 284 = 200

The number of homozygous recessive = q² number of population = 0.0262 284 = 7

The number of heterozygous = 2pq number of population = 0.2714 284 = 77