Question

In: Biology

Problem 2) Skin colour in dolphins is determined by a single gene showing incomplete dominance giving...


Problem 2) Skin colour in dolphins is determined by a single gene showing incomplete dominance giving rise to 3 phenotypes Brown, Pink & White. A survey of a large population of dolphins was carried out and the phenotypes recorded. However a student who had some of the data stored on their phone fell overboard and unfortunately the data was lost. The only data saved was that the brown homozygous dominants (BB) were 64% of the population .

a) What would be the percentage of pink heterozygotes and white homozygotes?








b) What percentage of the gametes that give rise to the next generation of dolphins in this population will contain the b allele?

Solutions

Expert Solution

An allele is considered dominant if the phenotypic appearance remains same in homozygous as well as heterozygous conditions. Incomplete dominance or epistasis means the allele ‘B’ is not completely dominant over ‘b’, therefore the heterozygous individual shows intermediate colouration (in between brown (BB) and white(bb)), that is, pink.

Answer a) The only data we have is homozygous brown dolphins are 64% of the total population.

Frequency of BB = 64% = 0.64

Let us assume that the frequency of B is p and b is q, and the total numbers of dolphins are 1000.

Genotype BB (p2) = 0.64

p = √(0.64) = 0.8

p+q = 1; therefore q=0.2

Genotype

Frequency

Number of individuals

BB (p2)

0.8 x 0.8 = 0.64

0.64 x 1000 = 640

Bb (2pq)

2 x 0.8 x 0.2 = 0.32

0.32 x 1000 = 320

bb (q2)

0.2 x 0.2 = 0.04

0.04 x 1000 = 40

Therefore, percentage of pink heterozygotes (Bb) = (320/1000) x 100 = 32%

percentage of white homozygotes (bb) = (40/1000) x 100 = 4%

Answer b)

There are two copies of each allele, so the total number of b allele is divided by 2000 (because, 1000 dolphins each carrying 2 alleles = 1000 x 2 = 2000)

Frequency of b allele = [(2 x 40)+320]/2000 = 400/2000

*40 individuals carrying two copies of b allele, and 320 heterozygous individuals carrying one copy of b allele, thus [(2 x 40)+320]

Percentage of the gametes containing b allele = (400/2000) x 100 = 20%


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