Question

Please answer and show work :)

2. Let’s assume that fin length in clownfish is a case of incomplete dominance regulated by one gene with two possible alleles, a long-tail allele and a short-tail allele, that are incompletely dominant with each other. If 24% of a clownfish population on a reef have long tails, then what percentage of this population is expected to have medium-length tails (assume the population is in H/W equilibrium)?(remember to show ALL work for credit; round to nearest hundredth)

Answer 1

Let us denote the dominant allele for fin long length to be "L" and the recessive allele for short fins be "l".

Then individuals with genotype "LL" will have long fins and individuals with genotype "ll" will have short fins. But individuals with heterozygous phenotypes "Ll", instead of having long fins will have intermediate length fins as the alleles are incompletely dominant.

If 24% of clownfish population has long tails, it indicates that 24% of the population has "LL" type genotype. According to H/W equation,

p + q = 1

Here p²= 0.24 ( $\because$ 24% = 24/100)  

$\therefore p = \sqrt{0.24} = 0.49$

Then

$0.49 + q = 1 \Rightarrow q = 0.51$

The expected frequency of heterozygous individuals in populations that are in H/W equilibrium is 2pq.

$\therefore 2pq = 2 \times 0.49 \times 0.51 \Rightarrow 2pq = 0.4998$

Hence, $49.98% \approx 50%$ $\approx 50%$ % of the population will have medium length tails.

You can cross check the result by putting the values of p², q², and 2pq in H/W equation

p² + 2pq + q² = 1

KINDLY RATE IF THIS WAS HELPFUL.