Answer:
The probability that the child will have type blood B equals <u>3/16</u>.
Explanation:
<u>Available data:</u>
- Individuals with the rare Bombay blood phenotype lack both the A and B antigens in individuals and/or are of hh genotype.
- Cross between two parents that are both of I A I B Hh genotype
Cross: IAIB Hh x IAIB Hh
Gametes) IAH, IAh, IBH, IBh
IAH, IAh, IBH, IBh
Punnett square) IAH IAh IBH IBh
IAH IAIAHH IAIAHh IAIBHH IAIBHh
IAh IAIAHh IAIAhh IAIBHh IAIBhh
IBH IAIBHH IAIBHh IBIBHH IBIBHh
IBh IAIBHh IAIBhh IBIBHh IBIBhh
F1) Genotype
- 1/16 IAIA HH
- 2/16 IAIAHh
- 1/16 IAIAhh
- 2/16 IAIBHH
- 4/16 IAIBHh
- 2/16 IAIBhh
- 1/16 IBIBHH
- 2/16 IBIBHh
- 1/16 IBIBhh
Phenotype
- 3/16 Blood type A
- 6/16 Blood type AB
- 3/16 Blood type B
- 3/16 Blood type 0
Answer:
D. A silent variant near the 5' end of the TBX1 gene.
Explanation:
TBX1 gene is wild type human being. It gives instructions for making protein called T-box 1. It plays an important role in tissue formation and organs during embryonic development.
It is expected to see in the offspring of a woman who has DMD and a man who does not have the disease that all of their sons and none of their daughters will have the disease
Option A.
<h3><u>
Explanation:</u></h3>
Duchene muscular dystrophy, also known as DMD is a recessive, x-linked disease. It occurs when there is a mutation in the dystrophin gene. This mutation further affects the muscles of the body leading it to degenerate and ultimately death.
When a woman with DMD and a man without DMD have an offspring, their daughters and the sons will get the mutated gene signature from the mother as she is the carrier. But since the father has normal genes, the daughters will become the carriers and will not be afflicted by it. This is because the normal X chromosome from the father will be passed on to them. The sons on the other hand will have the disease as they will get one of their mother's X chromosome which would be carrying the mutated gene.
D is the correct answer ur welcome