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:
Excretion helps maintain Homeostasis by excreting wastes from the body, which eliminates harmful toxins housing themselves in your body, which would increase the chance of getting an illness. ... When you exhale, carbon dioxide and some water are removed from the body by the lungs.
<span>The balanced equation for ammonia (NH3) is 3H2 + N2 ď 2NH3. This equation starts from H2 + N2 ď NH3. We have 1 N and 3 H on the right side (the product side) of the equation, so we multiply each by 2 to get 2 N and 6 H. We do this so that we don’t have an odd number of H atoms. We then balance the left side (the reactant side) of the equation with the right side, N is already balanced, but we have to multiply H by 3 to get 6 H atoms (to match the right side of the equation.</span>
