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31 March, 14:55

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

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Answers (1)
  1. 31 March, 15:46
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    A.) 5' - T A A - 3'

    3' - A T T - 5'

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

    3' - G T T - 5'

    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
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