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
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