Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

Round all answers to 2 decimal places. 1. A population of rabbits may be brown (

ID: 150187 • Letter: R

Question

Round all answers to 2 decimal places. 1. A population of rabbits may be brown (the dominant phenotype) or white (the recessive phenotype). Brown rabbits have the genotype BB or Bb. White rabbits have the genotype bb. The frequency of the BB genotype is .4. 1. 2. 3. What is the frequency of heterozygous rabbits? What is the frequency of the B allele? What is the frequency of the b allele? 2. A hypothetical population of 12,000 humans has 7840 individuals with the blood type AA (Homozygous for blood type A) 3860 individuals with blood type A1 (Heterozygous) and 300 individuals with the blood type lB1B (Homozygous for blood type B) What is the frequency of each genotype in this population? 1. 2. 3. What is the frequency of the A allele? What is the frequency of the B allele? If the next generation contained 30,000 individuals, how many individuals would have blood type BB, assuming the population is in Hardy-Weinberg equilibrium?

Explanation / Answer

Answer:

1).

1. The frequency of heterozygous rabbits = 0.47

2. The frequency of B allele = 0.63

3. The frequency of b allele = 0.37

Explanation:

The frequency of BB genotype = 0.4

The frequency of B allele = SQRT of BB = 0.63

As p+q=1

The frequency of b allele = 1-0.63 = 0.37

The frequency of heterozygous rabbits = 2pq = 2 * 0.63 * 0.37 = 0.47

2).

1. The frequency of IA alleles = 0.81

2. The frequency of IB alleles = 0.19

3. Number of IBIB individuals out of 30000 = 0.0361 * 30000 = 1083

Explanation:

IAIA = 7840

IA = 7840 +7840 = 15680

IBIB = 300

IB = 300+300 = 600

IAIB = 3860

IA = 3860 & IB = 3860

Total IA alleles = 15680 +3860 = 19540

Total IB alleles = 600+3860 = 4460

Total alleles = 19540 + 4460 = 24000

The frequency of IA alleles = 19540 / 24000 = 0.81

The frequency of IB alleles = 4460 / 24000 = 0.19

IBIB genotype frequency in new next generation = 0.19 * 0.19 = 0.0361

Number of IBIB individuals out of 30000 = 0.0361 * 30000 = 1083