Answer:
E) Either anaphase I or II
Explanation:
Failure of segregation of homologous chromosomes during anaphase I or failure of segregation of sister chromatids during anaphase II leads to the presence of the abnormal number of chromosomes in resultant gametes. In the given example, the egg mother cell with 48 chromosomes (24 pairs) would enter meiosis I but the failure of one pair of homologous chromosomes to segregate from each other followed by normal meiosis II would result in the formation of two gametes with one extra chromosome and two gametes with one less chromosome.
On the other hand, if the nondisjunction occurs at anaphase II of meiosis II, two normal gametes, one gamete with one extra chromosome and one gamete with one less chromosome will be formed. Therefore, nondisjunction at anaphase I or anaphase II would have resulted in the production of eggs with one extra chromosome.
Answer:
0.7
Explanation:
Using Hardy-Weinberg equation of genetic variation being constant when disturbing factors such has mutation and others are removed.
p² + pq + q² = 1 and p + q = 1
where p² is the frequency of the homozygous dominant genotype (RR) and q² is the frequency of the homozygous recessive genotype (rr) and 2pq is the frequency of heterozygous genotype (Rr). p represent the frequency of "R" and q represent "g". since the coefficient against the green/green homozygote is 0.30 then
the fitness of the green/green homozygote = 1 - 0.3 = 0.7
The<em> cell membrane</em> controls the movement of substances in and out of cells and organelles. In this way, it is selectively permeable to ions and organic molecules.
The answer is C hope this helps
