A population of English peas has genotypic frequencies of PP 0.25, Pp 0.50 and p

Question

A population of English peas has genotypic frequencies of PP 0.25, Pp 0.50 and pp- 0.25. The P allele codes for purple flowers, the p allele for white flowers What color are the flowers of heterozygous plants? Is the population in H-W equilibrium? 6. 7. A blue crab population has genotypic frequencies of KiK 0.25, K K2-0.01 and kik, -0.74. How many phenotypes are expressed among these three genotypes? 1s the population in H-W equilibrium? What are the genotypic frequencies in the crab population above after one generation of random mating? a. You discover a new population of cotton rats on Bermuda Island that have immigrated from the mainland on ships from different ports. You collect the following genetic data for the enzyme lactate dehydrogenase (Ldh), shown in the table below. Use these data 8. to answer the following questions. What are the allelic frequencies for the new population? What are the population's genotypic frequencies? Is this population in H-W equilibrium? Offer an explanation. Rat i Sex dh genotype LiLs LaL LiL 10 31

Explanation / Answer

6.

P codes for purple colour and p codes for white colour

Genotype of heterozygous plant = Pp

Phenotype = Purple (because there is one copy of dominant allele P determining purple colour)

Frequency of homozygous dominant (PP) = p2 =0.25

Frequency of Heterozygous (Pp) = 2pq = 0.5

Frequency of homozygous recessive (pp) = q2 = 0.25

If the population is in H-W equation, then,

p2 + 2pq + q2 = 1

since, 0.25 + 0.5 + 0.25 = 1, therefore the population is in H-W equilibrium.

7.

Since both K1 and K2 are dominant alleles, there will be three phenotypes as K1K1 , K1K2 and K1K2 will exhibit separate phenotype

Frequency of K1K1 = p2 =0.25

Frequency of K1K2 = 2pq = 0.74

Frequency of K2K2= q2 = 0.01

If the population is in H-W equation, then,

p2 + 2pq + q2 = 1

since, 0.25 + 0.74 + 0.01 = 1, therefore, the population is in H-W equilibrium.

8. Frequency of allele x is given by f(x) = (2(no of homozygous) + no of heterozygous)/2(total population)

Total number of L1L1 = 4

Total number of L1L2 = 2

Total number of L2L2 = 4

Allelic frequency

f(L1) = p =   ((2 x 4) + 2)/ (2 x 10) = 0.5

and f(L2) = q =   ((2 x 4) + 2)/ (2 x 10) = 0.5

Genotypic frequency

f(L1L1) = p2 = 0.5 * 0.5 =0.25

f(L1L2) = 2pq = 2 * 0.5 * *0.5 = 0.5

f(L2L2) = q2 = *0.5 * *0.5 = 0.25

multiplying each frequency with population no, the expected genotype should be

2.5 (2 or 3) L1L1

5 L1L2

2.5 (2 or 3) L2L2

The expected and actual values are not same, therefore, the population is not in H-W equilibrium.

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