Question

Grouse in Russia show a recessive mutation “short-tail” that causes problems in the control of direction when flying. In 2002 a survey of a large, freely interbreeding, population of 1,856 grouse revealed 142 with short tail feathers.

a) What proportion of the grouse population would you expect to be heterozygous at the short-tail locus

b) What proportion of the grouse population would you expect to be heterozygous at the short-tail locus

c) Due to rising affluence there was an increase in recreational grouse hunting. In a later survey of the grouse over some years it was found that a significant decrease had taken place in the population. In 2003 there were 2,861 grouse and in 2004 there were only 2,369 and all the surviving grouse had long tails. Support by calculation that this change would reflect greater vulnerability of short tail grouse to killing by hunters.

d)After further random mating of survivors what would be the percentage of homozygous dominants, heterozygotes and homozygous recessives expected in the next generation in the absence of any further hunting?

Answer 1

Ans:As given in the question that the mutation is recessive type which means only homozygous recessive type will show shot tail and homozgous dominant as well as heterozygous grouse will show wild type phenotype. Now as given in the question that the number of individuals having short tail is 142 out of 1856. So the allelic frequency of homozygous recessive individuals is 142/1856 and according to Hardy Weinberg equation p²+q²+2pq=1 where p² is the allelic frequency of homozygous dominant and q² is the allelic frequency of homozygous recessive and 2pq is the allelic frequency of heterozygous individuals, also (p+q)²=1.

a) So q² is 142/1856 and hence q is equal to 0.277 and p is equal to 0.723 and so the heterozygous population is equal to 2*p*q which is equal to 0.40.

b) The proportion of the grouse population we would be expect to be heterozygous at the short-tail locus is equal to 0.40.

c) As given in the question that in 2003 there were 2,861 grouse and 7.6% of this population are homozygous recessive for the trait which we can calculate from the above question so 7.6% of 2861 is 220 which means approximately 220 individuals are homozygous recessive for this trait and as given in the question that in 2004 here were only 2,369 and all the surviving grouse had long tails so from this data we can analyze that individuals with short tails are more vulnerable to be killed by the hunters.

Ans 4: So as described above that the allelic frequency of A is 0.727 and allelic frequency of a is 0.277 and also it is given that the remaining individuals are randomly mating so this means that population is in Hardy Weinberg equilibrium. Now using punnette square for calculating the frequency:

Female Male	A	a
A	AA	Aa
a	Aa	aa

So the percentage of homozygous dominant individuals= 0.727*0.727*100 = 52.85%.

Similarly percentage of homozygous recessive individuals = 0.273*0.273*100 = 7.45%

Percentage of heterozygous individuals = 2*0.727*0.273*100= 39.69%.