Answer:
9/16 Kinked tail, normal growth
3/16 Kinked tail, obese growth
3/16 normal tail, normal growth
1/16 normal tail, obese growth
Explanation:
Kinked tail (K) is dominant over normal tail (k) and normal growth (N) is dominant over obese growth (n). Two mice that are heterozygous for both traits (KkNn) are crossed:
KkNn x KkNn
offspring:
<em>9/16</em><em> K_N_ - Kinked tail, normal growth</em>
<em>3/16</em><em> K_nn - Kinked tail, obese growth</em>
<em>3/16</em><em> kkN_ - normal tail, normal growth</em>
<em>1/16</em><em> kknn - normal tail, obese growth</em>
(See the attached image for the Punnet's square analysis)
Answer:
The answer is positive interference
Explanation:
<h2>
The correct answer is explained below:</h2>
Explanation:
- According to the question, free ear lobes are dominant over attached ear lobes.
- Let the allele for free ear lobes be represented by F.
- Let the allele for attached ear lobes be represented by f.
- The genotype of the homozygous recessive woman (mother) having attached ear lobes is : ff.
- The gamete produced by the woman is: f.
- The genotype of the heterozygous man (father) with free earlobes is: Ff.
- The gametes produced by the man are: F and f.
- Mating the man with the woman we get the following offspring,
F f
f Ff ff
Phenotype Free ear lobe Attached ear lobe
- The genotype and the phenotype of the probable children obtained is represented in the Punnett square above.
- The probability that the parents would produce children with attached ear lobes is =
.