Answer: 1/16, or approximately 6.25% (see explanation below)
Explanation:
Answering this question requires two steps.
First, we need to figure out the probability that this couple will have a child with albinism in the first place. We know the following:
- Both parents are unaffected.
- The couple has already had one affected child.
- Albinism follows an autosomal recessive inheritance pattern.
Let ( M = normal gene ) and ( m = mutated gene ). Since the condition is recessive, the affected child can be assumed to have a “mm” genotype. Barring the possibility of a de novo mutation (which are assumed to be rare), the affected child must have inherited one ”m” allele from each parent. Since both of them are unaffected, however, we can assume that they are both carriers (genotype “Mm”). In conclusion, 1/4 of their offspring (25%) <em>for any given pregnancy</em> may be expected to have albinism. See the resulting Punnett square:
<u> | M | m </u>
<u>M | MM | Mm </u>
<u>m | Mm | mm </u>
Note that the question asks about the probability that not one but two consecutive births result in affected children. Since it can be assumed that both events are independent (meaning: the outcome of a pregnancy does not influence the outcome of following ones), we may apply the rule of multiplication for probabilities. The final answer is therefore 1/4 * 1/4 = 1/16.
The physiological process which explains the principles is mitosis b.
The polluted bacteria and algae interacts with fish and other aquatic wildlife, so they will be negatively impacted by that relationship in this case.
Answer:
81/256
Explanation:
By doing a test cross of albinism and sickle cell anemia you'll see that the probability that one child is unaffected by either conditions will be 9/16. Both children would be 9/16 * 9/16.
Since fraternal twins are developed from two separate fertilization events the probability chance will be independent.
Answer: The average speed of the zebra during time between 0s and 40s is 34 seconds.
