Allele, Genetic Drift, and Artificial Selection

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Exercise 1 Assessment
Question 1: What is the probability that any one individual in the lab group possesses all 6
traits listed in your trait table?
Solution:
∙ This sum is equal to one: (p+q) 2 = p 2 + 2pq + q 2 = 1.
∙ "p" is the frequency of the dominant allele, and "q" is the frequency of the recessive allele. The "2pq" in the equation shows the frequency of heterozygotes in the population.
∙ According to question, the frequency of dominant allele (p) = 0.7
∙ If p+q=1, then q=1 - p => q = 1 - 0.7 =0.3
∙ Hence, the frequency of recessive phenotype = q 2 = (0.3) 2 =0.09.
Question 2: If a mother has free-hanging earlobes and a father has free-hanging earlobes, what is the probability that their child will have attached earlobes? Assume heterozygosity for the parents & explain your answer.solution:
25% chance that their next child will have attached earlobes. The only way two people with unattached earlobes can produce a baby with attached earlobes is if they are both heterozygous.
Question 3: Chondrodystrophic dwarfism¹ is the result of the inheritance of a dominant allele (D). A person with chondrodystrophic dwarfism, whose father also has dwarfism but whose mother does not, has a baby of average stature. What is the probability or frequency that they will have a child with chondrodystrophic dwarfism? Explain your answer
Solution:
(p+q) 2 = p 2 + 2pq + q 2 = 1.
∙ "p" is the frequency of the dominant allele, and "q" is the frequency of the recessive allele. The "2pq" in the equation shows the frequency of heterozygotes in the population.] According to the question, if it is a recessive disease, then “p" is the frequency of normal dominant allele, "q" will be the frequency of disease-causing recessive allele, and "2pq" will be the frequency of carrier individual.
∙ In the question the frequency of disease-causing recessive allele is given i.e., q=80% or o.8
∙ If p+q = 1 then, p = q - 1 => 1 - 0.8 = 0.2 or 20%
∙ Thus, the probability will be = 2 X p X q => 2 X 0.8 X 0.2 = 0.32 or 32%
Exercise 2 Assessment
Problem 1: If the allele frequency of B is 0.3, what is the allele frequency of b?
solution
p + q = 1
p represents the frequency of the dominant allele (B in your case), and q represents the frequency of the recessive allele (b in your case).
We get the following result when we plug in B's frequency of 0.3 into the equation:
0.3+ q =1
q= 1 – 0.3
q= 0.7
So, t...