Question

Determining the expected frequency of each genotype Considering the same population of cats as in Part A, what is the expected frequency of each genotype (TLTL, TLTS, TSTS ) based on the equation for Hardy-Weinberg equilibrium? Keep in mind that you just learned in Part A that: The allele frequency of TL is 0.4. The allele frequency of TS is 0.6. The equation for Hardy-Weinberg equilibrium states that at a locus with two alleles, as in this cat population, the three genotypes will occur in specific proportions: p2+2pq+q2=1p2+2pq+q2=1 For help applying the Hardy-Weinberg equation to this cat population, see Hints 1 and 2. Genotype Phenotype (tail length) Number of individuals in population TLTL long 60 TLTS medium 40 TSTS short 100 Enter the values for the expected frequency of each genotype, TLTL, TLTS, TSTS, respectively, separated by commas. Enter your answers numerically (as decimals), not as percentages. View Available Hint(s)

Answer 1

The Genotypes of TLTL = 0.16

The Genotypes of TLTS = 0.48

The Genotypes of TSTS = 0.36

This population fully abide by the Hardy-Weinberg equation.

p² + 2pq + q² =1

0.16 + 0.48 + 0.36 = 1

Step-by-step explanation:

According to the question,

Genotype Phenotype No. of individuals in population

TLTL Long 60

TLTS Medium 40

TSTS Short 100

The TL allele count = 60 × 2

= 120

But, TL is also present in TLTS

∴ The total TLTL allele count is = 120 + 40 × 1

= 120 + 40

= 160

Total individual = ( 60+40+100)

= 200

Total allele count = 200 × 2

= 400

∴ p = 160/400

= 0.4

We know that, according to the Hardy-Weinberg equation, the expected frequencies of the genotypes should add up to 1.

p + q = 1

or, 0.4 + q = 1

or, q = 1 - 0.4

∴ q = 0.6

We know that, according to the Hardy-Weinberg equation, the expected frequencies of the genotypes should add up to 1.

p² + 2pq + q² =1

⇒ (0.4)² + 2× (0.4)×(0.6) + (0.6)² = 1

⇒ 0.16 + 0.48 +0.36 = 1