What did you know for sure after one test? two test? three test?
1 answer:
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Answer: The probability of obtaining an individual who is CcmmLLPp 1/32.
Explanation: This can be achieved by crossing similar genes to obtain individual probability as shown in the attached image.
When you cross Cc in the first genotype with CC in the second genotype, the following probabilities will be obtained;
P (CC) = ¾, P(Cc) = ¼.
Similarly, crossing Mm with mm, we get;
P (Mm) = ½, P (mm) = ½
Crossing Ll with Ll, we get
P (LL) = ¼, P (Ll) = ½, P (ll) = ¼
Crossing PP with pp, we get
P (Pp) = 1
Therefore, the probability of individual with genotype CcmmLLPp will be;
P (Cc) x P (mm) x P (LL) x P (Pp)
= ¼ x ½ x ¼ x 1
= 1/32
Answer:
Probability of having a child with genotype Ww is 50%.
Explanation:
This question involves a single gene coding for hairline shape in humans. The allele for widow's peak hairline (W) is dominant over the allele for straight hairline (w). This means that allele W will mask the phenotypic expression of allele w in a heterozygous state.
In a cross between a mother with genotype, Ww and father with genotype, ww, the following gametes will be produced by each parent:
Mother (Ww)- W and w
Father (ww)- w and w
Using these gametes in a punnet square (see attached image), four possible offsprings will be produced with genotypes: Ww, Ww, ww and ww.
Ww(2) : ww(2) or Ww(1) : ww(1)
Hence, 1 out of 2 children produced by the parents will have a genotype Ww, i.e. 1/2. Therefore, the percentage probability is 1/2 × 100= 50%.
