Question

As the Covid-19 pandemic continues around the globe, one area of active research is to identify a plausible origin of this novel coronavirus (also called SARS-CoV-2). Early genomic studies suggest that the evolutionary precursor to SARS-CoV-2 may reside in bats since the bat coronavirus and SARS-CoV-2 have many similar sequences. However, there have been no documented cases of direct bat-human transmission, which suggests that an intermediate host was likely involved between bats and humans.

There are many different species of bats, and the Chinese horseshoe bat is just one that may be linked to the Covid-19 pandemic. However, much research is still needed to support or disprove this hypothesis.

The diploid number for a horseshoe bat is 62 (i.e., 2n = 62).

Give the number of each of the following that would be present in a cell in a horseshoe bat during each of the following stages of either mitosis or meiosis.

A. Anaphase of mitosis:

I) chromosomes:

ii) chromatids:

iii) DNA molecules:

iv) telomeres:

B. Anaphase of meiosis I

I) chromosomes:

ii) chromatids:

iii) DNA molecules:

iv) telomeres:

C. Anaphase of meiosis II

I) chromosomes:

ii) chromatids:

iii) DNA molecules:

iv) telomeres:

Answer 1

Ans :

A. Anaphase of mitosis:

During anaphase, sister chromatids separate that each chromatid is considered a separate, individual chromosome. So the chromosome number is doubled and chromatid number remains the same. Therefore, for a horseshoe bat (i.e., 2n = 62), we have

I) chromosomes: 62 X 2 = 124

ii) chromatids : 124

iii) DNA molecules: For each chromosome, 2 DNA molecules are present. Therefore, 124 X 2 = 248

iv) Telomeres: Telomeres exist each end of the two identical chromatids in a chromosome, so the answer is 4 telomeres per chromosome, Therefore 124 X 4 = 496.

B. Anaphase of meiosis I

In meiosis I, homologous pairs of chromosomes line up together.  Since anaphase I only separates the homologous chromosomes, neither the chromosome number nor the chromatid number changes during anaphase. Therefore, for a horseshoe bat (i.e., 2n = 62), we have

I) chromosomes: 62

ii) chromatids: 124

iii) DNA molecules: 2 X no. chromosomes = 2 X 62 = 124

iv) telomeres: 4 X no.of chromosomes = 4 X 62 = 248

C. Anaphase of meiosis II

The second division of meiosis (meiosis II) appears similar to mitosis, with the only difference being that there are now half as many chromosomes as before due to the reduction division of meiosis I. When anaphase II begins, the sister chromatids split apart, which once again doubles the chromosome number, Therefore, for a horseshoe bat (i.e., 2n = 62), we have

I) chromosomes: 31 X 2 = 62

ii) chromatids: 31 X 2 = 62

iii) DNA molecules: 2 X no. of chromosomes = 2 X 62 = 124

iv) telomeres: 4 X no. of chromosomes = 4 X 62 = 248