Biology questions I really need help with

Question

1. Consider that the gene pool of a certain animal population contains just 2 alleles for the genes that controls hair color. If the frequency of the H allele is 0.45 what percent of the population is homozygous for H (i.e. HH)?
2. Assume that in a certain population of giraffes, dark brown spots are dominant to light brown spots. If the frequency of dark brown giraffes is 24%, what is the frequency of the dark brown allele?
3. The frequency of the aa genotype is 26 percent what is the frequency of the AA genotype? Assume that there is only 2 alleles (i.e. a and A)?
4. In a non-evolving population of wildflowers, there are 400 individuals with serrated leaves and 454 individuals with smooth ones. If the allele for serrated is dominant to the allele for smooth. What is the expected frequency of the serrated allele?

Answer 1

These questions are asking you to apply the Hardy-Weinberg equation: p² + 2pq + q² = 1. In this equation:

p = freq of the dominant allele,

q = freq of the recessive allele,

p² = freq of the homozygous dominant genotype

2pq = frequency of the heterozygous genotype, and

q² = freq of the homozygous recessive genotype.

The Hardy-Weinberg equation assumes that the population is at equilibrium (no natural selection, no genetic drift, no non-random mating, and no gene flow, etc.). If true, you can then use allele frequencies to calculate genotype frequencies, or genotype frequencies to calculate allele frequencies, using the equation.

Question 1. The question indicates that p=0.45, so q=0.55 (since p+q=1 in any two allele system). Using Hardy-Weinberg, the frequency of HH is p², the frequency of Hh is 2pq, and the frequency of hh = q².

Question 2. This is a trickier question. The question indicates that p² + 2pq = 0.24 (dark is dominant, dark phenotypes can be HH or Hh, so p² (freq of HH) + 2pq (freq of Hh) = 0.24). By extension (p² + 2pq + q² must equal 1), the frequency of the homozygous recessive genotype (hh = q²) must be 0.76. Because q² = 0.76, we can determine q by taking the square root of 0.76 which = 0.87. Since we now know q, we can calculate p as 1-0.87. (You have to take this convoluted route because you were given the frequency of HH + Hh. If you were given the frequency of HH by itself, then p is simply the square root of the frequency of HH.)

Question 3. The answer here uses the same logic as question 2. If the frequency of aa is 0.26, then the frequency of a (q in the standard notation) is the square root of 0.26 = 0.51. The frequency of A (p in the standard notation) must then be 1-0.51 = 0.49. The frequency of the AA genotype is p².

Question 4. The answer here also uses the same logic as question 2, but you first need to calculate genotype frequencies from the numbers. If 400 of 854 individuals have serrated leaves, then the frequency of the dominant phenotype is 400/854 = 0.47. So p² + 2pq = 0.47. The frequency of recessive genotype (q²) must then be 0.53 (1-0.47). Determine q using the square root of 0.53 = 0.73. Once you have q you can calculate p.

If something here doesn't make sense, feel free to post a follow-up question.