I've actually learned something similar recently so your best bet is to go with B.
The last <span>process to occur in the epiphyseal plate is ossification</span>
Answer:
Daughter cells with abnormal number of chromosomes will be produced (aneuploidy)
Explanation:
Ideally, homologous pair of chromosomes are meant to separate into opposite poles during the anaphase I stage of meiosis. However, a failure to separate, a process called MEIOTIC NON-DISJUNCTION, will cause some daughter cells to have more than the required/normal chromosome number while some daughter cells will have less than the normal or no chromosome.
This abnormality in chromosome number caused by failure of chromosome to separate is called ANEUPLOIDY. In a nutshell, when this occurs, the daughter cells will possess abnormalities in chromosome number i.e trisomy, monosomy etc.
I think there should be options to choose. Anyway I know the correct answer. The cells of diplomonads and <span>parabasalids </span>have modified mitochondria.
