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:
Somatic hypermutation is a process in which point mutations build up in the antibody V-regions of both the heavy and light chains.
This process occurs at rates that are about 106-fold higher than the background mutation rates observed in other genes.
It allows B cells to mutate the genes that they use to produce antibodies. This then ensures the B cells to produce antibodies that are better able to bind to bacteria, viruses and other infections.
What r the options if u want me to solve it I need options
Answer:
Parasitism
Explanation:
This is because parasitism is a type of relationship that exist between organisms where one which is the predator feed on the prey thereby causing harm or injury or death to the prey. In the question, small frogs are the predators and insects are the prey, therefore the predators cause harm to the prey, meaning they only benefits from the farm they cause the prey.
A trait is the organism's feature. So, a trait would be eye color. A phenotype would be hazel. An allele is a gene. It can be recessive or dominant.
