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Furkat [3]
3 years ago
5

Imagine that a newly discovered, recessively inherited disease is expressed only in individuals with type O blood, although the

disease and blood group are independently inherited. A normal man with type A blood and a normal woman with type B blood have already had one child with the disease. The woman is now pregnant for a second time. Assuming that both parents are heterozygous for the gene that causes the disease? Assume that both parents are heterozygous for the gene that causes the disease. Enter your answer as a fraction - e.g. 1/2 - and do not put anything else in the box.
Biology
1 answer:
erastova [34]3 years ago
3 0

Complete question:

Imagine that a newly discovered, recessively inherited disease is expressed only in individuals with type O blood, although the disease and blood group are independently inherited.

A normal man with type A blood and a normal woman with type B blood have already had one child with the disease. The woman is now pregnant for a second time.

Assuming that both parents are heterozygous for the gene that causes the disease, what is the probability that the second child will also have the disease? Express your answer as a fraction using the slash symbol and no spaces (for example, 1/2).

Answer:

The probability that the second child will also have the disease is 1/16.

Explanation:

<u>Available data:</u>

  • Two genes independently inherited: one for blood type, the other for disease
  • Man with type A blood  x  Woman with type B blood
  • Both parents are heterozygous for the gene that causes the disease; Dd

If the man has A blood, and the woman has B blood, and they already have an affected child, this means that they must be heterozygous for blood type too.

Cross:  

Parentals)           AiDd         x      BiDd

Gametes)  AD  Ad  iD  id       BD  Bd  iD  id

Punnett square)      AD       Ad       iD       id

                     BD   ABDD  ABDd  BiDD BiDd

                     Bd   ABDd   ABdd   BiDd Bidd

                     iD    AiDD    AiDd    iiDD   iiDd

                     id    AiDd     Aidd     iiDd    iidd

F1) <u>Genotype</u>:

1/16 ABDD

2/16 ABDd

1/16 ABdd

1/16 AiDD

1/16 BiDD

2/16 AiDd

2/16 BiDd

1/16 Aidd

1/16 Bidd

1/16 iiDD

2/16 iiDd

1/16 iidd

<u>Phenotype:</u>

3/16 A/B normal

4/16 A normal

4/16 B normal

3/16 0 normal

1/16 0 affected by the disease.

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