I\'m pretty sure that 7. i and ii is inconsistent. But I\'m not sure what ii is
ID: 188399 • Letter: I
Question
I'm pretty sure that 7. i and ii is inconsistent. But I'm not sure what ii is and 7b
7. The following pedigree shows inheritance of the inability to smell Pseudomonas bacteria and the inability to roll the tongue into a U-shape Inability to roll the tongue is an autosomal recessive trait. Use A/a for smelling/non-smelling and R/r for rolling/non-rolling inability to roll tongue 10 Consider the inability to smell Pseudomonas and complete the following conclusions (ignoree X-inactivation) (0) The data are consistent/inconsistent with an autosomal dominant mode of inheritance because (ii) The data are consistent/inconsistent with an X-linked recessive mode of inheritance because (ii) The data are consistent/inconsistent with an X-linked dominant mode of inheritance because [4 pts] (b) If the two genes are assorting independently, what is the probability that the first child born to III-4 and III-5 will be a non-smeller, non-roller?Explanation / Answer
III) The data are inconsistent with an X-linked dominant mode of inheritance as well.If it were the case of X-linked dominant inheritance then at least one of the parents of the affected girl should have possessed the trait of the inability of smelling pseudomonas bacteria, while in case of affected male their mother should have possessed this trait.
7b. the probability of having a non-smeller and non-roller child depends on the probability of the presence of the defective gene in their parents. both the trait is autosomal recessive.
genetic formation of III-4 is r/r as he cannot role his tongue whereas III-5 will have to be the carrier(R/r) to pass on this defect to his child
therefore the probability of III-5 to be carrier = 2/3
now the probability of having r/r for child = 1/2 ( father will pass only r gene while the mother can pass either R or r)
probability of child to be non-roller = 1/2 x 2/3 = 1/3
III-4 is and III-5 will have to be carrier to pass on defective gene (A/a) (non-smeller) to his child
therefore the probability of III-4 and III-5 to be carrier = 2/3
ow probability of having a/a for child = 1/4 ( father and mother both can pass either A or a)
probability of child to be noo.smeller (a/a) = 1/4 x 2/3 x 2/3= 1/9
therefore probability of child being non-roller and non-smeller = 1/3 x 1/9 =1/27