Question

Use this partial DNA sequence (the original sequence with no mutations) to answer the following questions.

5' - C A A - 3'
3' - G T T - 5'

Suppose that the transient rare guanine tautomer shifted back to the common guanine tautomer prior to a second round of replication.

Which DNA sequence(s) would be present in the sister chromatids after this second round of replication? Select all that apply.

A.) 5' - T A A - 3'

3' - A T T - 5'

B.) 5' - T A A - 3'
3' - G T T - 5'

C.) 5' - U A A - 3'
3' - A U U - 5'

D.) 5' - C G G - 3'
3' - G C C - 5'

E.) 5' - C A A - 3'
3' - G T T - 5'

F.) none of these

Answer 1

A.) 5' - T A A - 3'

3' - A T T - 5'

E.) 5' - C A A - 3'

3' - G T T - 5'

Step-by-step explanation:

It's worthy to note that

G is the rare guanine tautomer.

The sister chromatids DNA sequence after first round of DNA replication will be:

5' - CAA - 3'

3' - GTT - 5'

5' - TAA - 3'

3' - G*TT - 5'

Before the second round of replication, transient rare guanine tautomer will reverse to the common guanine tautomer

So,

the sister chromatids will have a DNA sequence

5' - TAA - 3'

3' - ATT - 5'

5' - CAA - 3'

3' - GTT - 5'

(Take a look at the sister chromatid

It is present after the first r abd second replication)

A tautomeric shift has result in mutation of one of the sister chromatids