Population Genetics Problems In a population of Linanthus pa

Population Genetics Problems In a population of Linanthus parryae, a small annual plant of the Mojave Desert, 70% of the flowers are blue and the remainders are white. Given that the flower color is controlled by alternate alleles at a single locus and that blue is dominant to white, answer the following questions. Using B for the dominant (blue) allele and b for the recessive (white) alleles what flower color phenotypes correspond to the genotypes BB, Bb, bb? Using your answer to a, and the information provided above, what is the frequency of genotype bb? Letting p be the frequency of the B allele and q be the frequency of the b allele, and assuming that the population is mating at random, write down a formula for the expected frequency of the bb homozygote. Using your answers to parts b and c, calculate q.

Solution

Blue color flowers 70% (BB and Bb)

White color flowers 30% (only bb)

According to Hardy-Weinberg law

p² + 2pq + q² = 1 and p + q = 1

p = frequency of the dominant allele in the population
q = frequency of the recessive allele in the population
p² = percentage of homozygous dominant individuals
q² = percentage of homozygous recessive individuals
2pq = percentage of heterozygous individuals

q² = percentage of homozygous recessive individuals
q²=30%=30/100=0.3; q= 0.3=0.548
p+q=1; p=1-q or p=1-0.548=0.452
p=0.452
q=0.548
BB=p²=(0.452) 2=0.204
Bb=q²=(0.548) 2=0.300
Bb=2pq=2*0.452*0.548=0.495
(p+q=1; 0.452+0.548=1)
(p² + 2pq + q² = 1 : 0.204+0.300+0.495=1)

(a)B is the dominant allele results in blue color

.b is the recessive allele, so homozygous recessive alleles give white color

BB= Blue color flowers (homozygous dominant)

Bb= Blue color flowers (heterozygous)

.bb=White color flowers (homozygous recessive)

(b)Homozygous recessive genotype frequency bb=q²=0.3

(c)If mating is random, then by considering the probability of two alleles combining in the next generation we can calculate their frequencies going to be obtained.

B (p) 0.452

.b(q) 0.548

B (p) 0.452

BB

0.452*0.452=0.204

Bb

0.452*0.548=0.248

.b(q) 0.548

Bb

0.452*0.548=0.248

bb

0.548*0.548=0.3

(d) q = frequency of the recessive allele in the population

(q²)=0.3; q=0.3=0.548

0.548 is the frequency of the recessive allele in the population

	B (p) 0.452	.b(q) 0.548
B (p) 0.452	BB 0.452*0.452=0.204	Bb 0.452*0.548=0.248
.b(q) 0.548	Bb 0.452*0.548=0.248	bb 0.548*0.548=0.3