2. Assume a population of pea plants consists of 300 TT and 700 tt with T (tall)
ID: 196812 • Letter: 2
Question
2. Assume a population of pea plants consists of 300 TT and 700 tt with T (tall) dominant to t (short) What are the phenotypic, genotypic, and allele frequencies? a. Further assume that you allow this population to undergo one generation of random mating with each other and you know that mutation is very rare, there is no selection happening, and the population is large enough that drift is negligible. You count the number of tall plants to be 382 and the number of short plants to be 368. What is the genotypic frequency of this generation? What is the allele Frequency of this generation? c. Was the original population in HW equilibrium? d. Did any evolution occur from one generation to the nextExplanation / Answer
a.
According to Hardy-Weinberg principle,
p2 +2pq+ q2 = 1, p + q= 1, where
p2 is the percentage of homozygous dominant individual
q2 is the percentage of homozygous recessive individual
2pq is the percentage of heterozygous individual
p is the frequency of the dominant allele in the population
q is the frequency of the recessive allele in the population
This population of pea plants consist of 300 (TT) and 700 (tt) plants. T represents tall (dominant) and t represents short (recessive)
Thus, homozygous dominant percentage is = 300/ (300+700) = 0.3 or 30% =p2
Homozygous recessive percentage is = 700/ (300+700) = 0.7 or 70%=q2
Then, the phenotypic frequency is 30% tall (TT) plants and 70% short (tt) plants. As there is no heterozygous population exist, then the genotypic frequency is same as phenotypic frequency.
If p2 =0.3, then p= 0.3 = 0.54 or 54%
As p+q =1, then q= 1- 0.54= 0.46 or 46%
Thus, allelic frequency: frequency of the dominant allele in the population is 54% (T) and the frequency of the recessive allele in the population is 46% (t).
B. After one generation by random mating, the number of tall plants is 382, and the number of short plants is 368
Here, p2 +2pq+ q2 = 1,
thus, the phenotypic frequency of tall plants is p2 + 2pq = 382 /(382+368) = 0.51 or 51%, and
the phenotypic frequency of short plants is q2= 368 / (382+368) = 0.49 or 49%.
Thus, the homozygous recessive genotype (tt) is 49%, which raise only the short plants.
Then the frequency recessive allele t (q) is 0.49 = 0.7 or 70%
As p+q = 1, then p = 1- 0.7 = 0.3 or 30% is the frequency of dominant allele T (p) in the population
Thus the frequency of homozygous dominant genotype is p2 = (0.3)2 = 0.09 or 9%
And the frequency of heterozygous dominant genotype is 2pq = 2* 0.3 *0.7 = 0.42 or 42%.
C. If the allele frequencies after one round of random mating change at all from the original frequencies, the population is not in Hardy-Weinberg equilibrium and evolution has occurred within the population.
Thus, in the problem, after one round of random mating, the allele frequencies, that is the frequency of dominant allele T (p) and recessive allele t (q) is changed, as
Before and after mating the frequency of dominant allele (T) is 54% and 30% respectively.
Before and after mating the frequency of recessive allele (t) is 46% and 70% respectively.
Thus, the original population does not maintain Hardy-Weinberg population equilibrium.
D. As the population doe not maintain Hardy-Weinberg equilibrium, then evolution occurs from one generation to another.
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