Ans. (B). Mutated gametes (sex cells) fuse to create offspring.
Genetic mutations or germline mutations can be defined as permanent change in the DNA sequence of germ cells (cells, which form ovum and sperm). These mutations can pass from one generation to other, when a mutated sperm or mutated oocyte (gametes) get fertilized with other oocyte or sperm.
A germline mutation, present in a single-celled zygote will present in all cells of a muticellular organism, as single-celled zygote divides and produce all of the cells in that organism.
Thus, the correct answer is option (B).
Answer:
Natural selection is the process through which populations of living organisms adapt and change. Individuals in a population are naturally variable, meaning that they are all different in some ways. ... Individuals with adaptive traits—traits that give them some advantage—are more likely to survive and reproduce.
Copper deficiency is a rare disease that leads to hematological disorders. It is an essential trace element. It is an essential part of some enzymes (c0-factor) that plays role in cellular respiration. Copper is also an anti-oxidant that scavenges free radicals. In case of cellular respiration copper is responsible for electron transfer of oxygen.
the answer is ATP. ATP is the bi- product that cells can easily absorb 4 energy. the body digests the glucose first, whatever is left after that is ATP..
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.
