Correct question:
if a nondisjunction occurs at anaphase I of the first meiotic division, what will the proportion of abnormal gametes (for the chromosomes involved in the nondisjunction)?
Answer:
100%
Explanation:
Nondisjunction at meiosis-I means that two homologous chromosomes of at least one homologous pair fail to separate from each other during anaphase-I. This would result in the formation of one cell with one extra chromosome and the other with one less chromosome by the end of meiosis-I. Meiosis-II in these two cells would maintain this chromosome number in the daughter cells. Therefore, out of the total four gametes formed by the end of the meiosis, two would have one extra chromosome and would be denoted as "n+1". The rest of the two gametes would have one less chromosome and would be denoted as "n-1".
I believe the Answer is B but I could be wrong
Answer:
This may help!
Explanation:
In the lytic cycle, a phage acts like a typical virus: it hijacks its host cell and uses the cell's resources to make lots of new phages, causing the cell to lyse (burst) and die in the process. Entry: The phage injects its double-stranded DNA genome into the cytoplasm of the bacterium.