Cats have recently migrated from Malaysia to an uninhabited island in the South
ID: 142857 • Letter: C
Question
Cats have recently migrated from Malaysia to an uninhabited island in the South China Sea. The gene pool of the migrants includes both alleles for straight (T) and stub (t) tails. You study selection pressures on these cats in the wild, finding the following survivorship in a cohort of 650 kittens.
126
a) Calculate survivorship and relative fitness for each genotype
b) Calculate the selection coefficient (s) and degree of dominance (h)
c) Is the population of kittens in Hardy-Weinberg Equilibrium? [Assume that freq. (T) = p and freq. (t) = q].
d) Calculate the average fitness of the population of kittens and of the population of adults.
e) Calculate allele frequencies before (kittens) and after (adults) selection in the next generation. [Assume that freq. (T) = p and freq. (t) = q].
f) Which of the two alleles is deleterious? Justify your answer.
Phenotype Straight Bent Stub Genotype TT Tt tt Number of Kittens 100 250 300 Surviving adults 70 161126
Explanation / Answer
a) Survival Rate is the measure of the percentage of individuals that survive upto the reproductive rate.
In this case,
For genotypes
TT , survival rate = 70/100 = 0.7 [ adult surviving individual of each genotype/ total number before adulthood]
Tt= 161/250 =0.64
tt = 126/300 = 0.42
RELATIVE FITNESS (w)
It is the measure or ratio of survival and/or reproductive rate of a genotype to the maximum survival and/or reproductive rate of other genotypes in the given population.
[ It is calculated for each genotype by dividing each genotypes survival or reproductive rate by the highest survival or reproductive rate among the 3 genotypes. ]
For genotypes
wTT , = 0.7/0.7 = 1
wTt= 0.64/0.7 =0.91
wtt = 0.42/0.7 = 0.6
b) Calculating selection coefficient(S)
Selection coefficient is a measure of the relative strength of selection acting against a genotype.
s= 1-w
sTT = 1-1 =0
sTt = 1-0.91 = 0.09
stt = 1-0.6 =0.4
Now , If s = 0 means that genotype TT is not being selected against.
That is, although they are dying, the TT individuals on average are dying less or produce more offspring than the other genotypes in the same population.
Calculating h (degree of dominance)
The parameter h determines the fitness of heterozygote Tt
w= 1-hs
here, wTt= 1-hsTt
0.91= 1-h. 0.09
h=1
now if h=1, the fitness of the heterozygote Tt is same as the homozygote tt.
c) For a population to be in HW equilibrium the number of given observations should match with the expected values.
considering, T=p and t = q
frequency of T , f(T) = (100x2 +250)/1300 = 0.346
f(t) =(300x2+250)/1300 = 0.653
expected genotype frequencies
Multiplying each of these genotype frequencies with the total population number, we find that there should be TT= 100
Tt=250
tt=300
But on multiplying the genotype frequencies with the total population, the observations are:
TT= 77
Tt=293
tt=276
Therefore the population is not in HW equilibrium.
d) Average fitness = W bar
p2 (wTT) + 2pq (wTt) + q2 (wtt) = w-bar
For kittens,
0.119x1+ 0.451x0.91+ 0.426x0.6
=0.78
For adults,
p= 70/357 = 0.196
q= 1-p
= 1-0.196
=0.80
p2= 0.038
q2=0.64
2pq=0.3136
Therefore w bar
p2 (wTT) + 2pq (wTt) + q2 (wtt) = w-bar
(0.038x1)+(0.3136x0.91)+(0.64x0.6)
=0.707.
e) new frequencies for kittens
(dividing each frequency by average fitness)
new TT =(0.119x1) / 0.78
=1.16
new Tt=0.451x0.91/0.78
=0.52
new tt = 0.426x0.6/ 0.78
=0.32
For adults,
new TT=(0.038x1)/ 0.70
=0.054
new Tt = (0.3136x0.91)/ 0.70
=0.40
new tt =(0.64x0.6)/ 0.70
=0.54
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