Genotypes and Phenotypes

Question

Consider a certain species of grasshopper with a body that exhibits two phenotypes for pattern: 1) light green spots and 2) solid color with no spots. 

The solid allele is dominant and the green-spotted allele is recessive. For a total of 487 grasshoppers, 445 are solid and 42 have green spots.

a. List and describe all possible genotypes and phenotypes.

b. What is the phenotypic frequency of the grasshoppers that have no spots? (Round answer to two decimal places.)

c. What is the phenotypic frequency of the green-spotted grasshoppers? (Round answer to two decimal places.)

d. Calculate the frequency of the recessive allele in this population. (Round answer to two decimal places.)

e. Calculate the frequency of the dominant allele in this population.

f. Explain why the phenotypic frequency and genotypic frequency of the green spotted grasshoppers is the same value.

g. Explain why the phenotypic frequency and genotypic frequency of the solid grasshoppers IS NOT the same value.

Answer 1

1. Possible genotypes and phenotypes
2. HH (homozygous dominant) = solid
3. Hh (heterozygous) = solid
4. hh (homozygous recessive) = green spots
5. Phenotypic frequency of grasshoppers with no spots = 445/487 = 0.9138 (2 decimal places = 0.91)
6. Phenotypic frequency of grasshoppers with green spots = 42/487 = 0.08624 (2 decimal places = 0.09)
7. Frequency of recessive allele = √(genotypic frequency homozygous recessives) = √0.08624 = 0.29
8. Hardy Weinberg Equation: p² + 2pq + q² = 1, and p + q = 1
9. p = frequency of dominant allele
10. q = frequency of recessive allele
11. p² = genotypic frequency of homozygous dominant
12. 2pq = genotypic frequency of heterozygous
13. q² = genotypic frequency of homozygous recessive
14. q² is equal to the phenotypic frequency of grasshoppers with green spots (i.e. homozygous recessives), or 0.08624 as calculated in question C
15. Therefore q, the recessive allele frequency = √q² = √0.08624 = 0.29 (rounded to 2 decimal places)
16. Frequency of dominant allele = 0.71
17. Per the Hardy Weinberg equation, p + q = 1
18. The allele frequencies of the dominant and recessive alleles must equal 1
19. We calculated the frequency of the recessive allele to be 0.29
20. Therefore, the frequency of the dominant allele (or q) is 1-0.29 = 0.71
21. The phenotypic frequency of the green spotted grasshoppers is the same as the genotypic frequency because there is only ONE GENOTYPE (hh) for green spotted grasshoppers. Recall that this corresponds to the value of q² that we used in question D to calculate the recessive allele frequency.
22. The phenotypic frequency of the solid grasshoppers is NOT the same as the genotypic frequency because there is MORE THAN ONE GENOTYPE (Hh and HH) for solid grasshoppers.