What factors should we consider when determining why the chicken "loses" to the fox? A chicken can fly and perch on higher place
1 answer:
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Question
Pedigree attached
a. Provide evidence from the pedigree that conclusively shows that the tune deafness allele is autosomal dominant, not autosomal recessive. Explain your reasoning.
b. Identify the genotypes of individuals 5 and 6, and then draw the Punnett square for the cross of these two individuals.
c. Compare the expected percentage of each phenotype of the offspring from the cross in part (b) with the actual percentage of each phenotype observed in the children of individuals 5 and 6.
Answer/Explanation:
a. Autosomal dominant means that even having one copy of the allele T will produce a tune deaf individual. The evidence of this is that individuals II.3 and 4 have children (III. 8 and 9) with normal tune perception. This suggests that they have inherited the normal allele from both parents. That means II.3 and II4 must be heterozygous, in order to pass on the normal allele. This makes sense if tune deafness is dominant (because 3 and 4 will both be Tt and Tt, so can pass on the t alleles). However, if tune deafness was recessive, that would mean that individual II 3 and 4 both have to carry 2 copies, meaning there is no way their children would have normal tune perception.
b. Since individual 5 is unaffected and we know this is the recessive trait, they must be Tt. Since individual 6 is affected, they must be either Tt or tt. However, they have children who are unaffected (11, 12, 13) who must be tt. This means, individual 6 must have the t allele to pass on, and must be Tt. Therefore, the cross between individuals 5 and 6 is tt x Tt:
T t
t Tt tt
t Tt tt
c) The punnet square shows the children from (b) have a 50:50 chance of being tune deaf vs normal tune perception (50% each). In contrast, the actual ratios are 25% tune deaf to 75% normal tune perception. (1:4). This deviates from the expected ratio, but since all of these are due to chance, it is not an unexpected occurrence. Perhaps if there were 10, 50 or 100 children (!) from this cross, the results would be more like the expected ratio
