LAB GENETIC PROBLEMS 1. Mrs. Wong and Mrs. White had their babies on the same da

Question

LAB GENETIC PROBLEMS 1. Mrs. Wong and Mrs. White had their babies on the same day. The hospital mixed the offspring up and could not tell them apart. The doctors ordered blood tests on all concemed in order to straighten the situation out. "Two Wongs," they said, "don't make a white." Here are the results for both sets of parents andbabiese Mrs. Wong B Mr. Wong Mrs. White A Mr. White A. Baby girl Baby boy B Decide to whom the babies belong.ifanyonel 2. Color blindness is a sex-linked recessive trait. A man and his wife have normal color vision, but a daughter has red-green color blindness. The man sues bis wife for divorce on grounds of infidelity. Does he have a case? Explain. 3. Right bandedness CR is dominant to left handedness (r. Write the genotype for the following individuals: a. Homozygous right bandedness b. Heterozygous right handedness Left handed d. A man who is left handed marries a woman who is heterozygod for right handedness. What percentage of left handed children can they anticipate? 4. The ability to taste PTC is dominant to the inability to taste. If two parents are beterozygous for the ability to taste, what percent of their children will be aon-tasten?

Explanation / Answer

1. Baby boy may belongs to Mr and Mrs. Wong while Baby girl belongs to Mr. and Mrs. White. As There is no allelle for blood group B in Mr and Mrs White therefore the baby has only probability to have either blood group A or O. In case of Mr and Mrs Wong, can have offspring with blood groups A, B, AB as well as O. If it is sure that one of the baby definetely belong to them then it has to be a child with blood group B.

2. Yes he has a case as there are no chances that this couple has a color blind girl. The possibility of genotype for mother are:

i) Xc Xwhich is known as carrier as the disease is sex linked recessive disease and will not be expresses in heterozygous condition.

ii) X X Which is a normal female.

As the father is having normal vision then the only possible phenotype of father is:

XY

So the two possible crosses are:

First, Xc X  Mother and XY Father

So, in this case the girl will only be carrier of the gene and not have the disease.

In second case, XX mother and XY father there is no disease causing allelle present, so there will be no offspring with disease.

3. a. RR

b. Rr

c. rr

Reason, left handedness is recessive therefore can be expressed only in homozygous condition as in third case only.

d. Genotype for left handed man- rr

Genotype for heterozygous right handed woman- Rr

So, there is 50% probability of there children to be left handed.

4. Genotype for heterozygous taster- Tt

Genotype for non taster- tt. Only 25% probability to have non taster children as it is a recessive trait and can be expressed only in homozygous condition.

X Y Xc XcX XcY X XX XY

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