Question

In 1814, 15 British colonists founded a settlement on Tristan da Cunha, a group of small islands in the Atlantic Ocean, midway between Africa and South America. One of the early colonists apparently carried a recessive allele for retinitis pigmentosa, a progressive form of blindness that afflicts homozygous individuals. Of the founding colonists' 240 descendants on the island in the late 1960s, 4 had retinitis pigmentosa (rr). The frequency of the allele that causes this disease is ten times higher on Tristan da Cunha than in the populations from which the founders came. Calculate the frequency of the r allele in the original population of 15 colonists and in the 240 descendants. Please show calculations, Thanks!

Answer 1

Answer:

Frequency of the allele "r"causing  the disease on Tristan da Cunha$= 1.67$ %

Frequency of the allele "r"causing  the disease in the original population of 15 colonists $= 16.7$%

Explanation:

Frequency of the allele "r"causing  the disease on Tristan da Cunha

Given -

Out of 240 descendants on the island, 4 had retinitis pigmentosa (rr).

As per Hardy Weinberg's equllibrium equation

The frequecy of recessive individual in a given population is represented by $q^2$

And q represents the frequency of allele r

So, in this case q is equal to

$\frac{4}{240} * 100$

$= 1.67$ %

Frequency of the allele "r"causing  the disease in the original population of 15 colonists

As it is given in the question statement , the frequency of allele "r"causing  the disease in the original population of 15 colonists is ten times the frequency of the allele "r"causing  the disease on Tristan da Cunha

i.e

$1.67 * 10$

$= 16.7$%