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 bacteria was not killed off. In those three days symptoms had stop because the bacteria was dormant. They must have been activated because Jim provided a suitable environment for their growth
Answer:
d. consumers, heterotrophs
Explanation:
