Question

1). 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)     Where there is no mutation, migration or selection what percentage of the next generation would be homozygous dominants, heterozygotes and homozygous recessives?                                                         

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

1) a) Let's P be the allelic frequency of dominant allele and q be the allelic frequency of the recessive allele.

So, qq is the genotype frequency of homozygous recessive allele.

qq = 142/1856 = 0.076

So, q = √0.076 = 0.276

So, p = 1-0.276 = 0.724

So, 2pq = 2*0.276*0.724 = 0.4

So, the proportiono of the grouse population would you expect to be heterozygous at the short-tail locus is 0.4

b) Homozygous dominant = 0.724*0.724 * 100% = 52.41%

Homozygous receive = 0.276*0.276 *100% = 7.62%

Heterozygous = 0.4*100% = 40%

C) Number of grouse killed = 2861-2369 = 492

As all of them were short tailed grouse, thus genotype frequency of homozygous short tailed grouse = 492/2861 = 0.17

So, allelic frequency of q in 2003 = √0.172 = 0.414

So allelic frequency of p in 2004 = 1-0.414 = 0.586

So, genotype frequency of pp in 2003 = 0.343

Genotype frequency of heterozygote in 2003 = 2*0.414*0.586 = 0.485

In 2004, 2369 grouse are lived and all are long tailed.

So, qq is zero in 2004. Relative genotype frequency of pp in 2004 is = 0.343/(0.343+0.485) = 0.414

Relative genotype frequency of heterozygote in 2004 = 0.485/(0.343+0.485) = 0.586

So, p in 2004 = 0.414+(0.586/2) = 0.707

q in 2004 = 0 + (0.586/2) = 0.293

So, it is evident that killing causes drop in allelic frequency of q from 0.414 in 2003 to 0.293 in 2004.

d) Homozygous dominant = 0.707*0.707*100% = 50%

Homozygous recessive = 0.293*0.293*100% = 8.58%

Heterozygous = 2*0.707*0.293*100% = 41.43%