Dna/rna Genetics Worksheet With Answers

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1. The base composition of a virus was found to be 11% A, 32% G, 18% U and 39% C.
It this a DNA or RNA virus? Is it single-stranded or double-stranded?
The presence of Uracil shows that it is RNA. The base composition is unequal, so it must
be single stranded.
2. Which of these DNA fragments will have a higher melting temperature? (circle one)
A) GCATTGACCGGAGGGACT
B) GGATTTCAATTACTTAAT
CGTAACTGGCCTCCCTGA
CCTAAAGTTAATGAATTA
A has a higher melting temperature. The triple hydrogen bonds between G and C are
harder to break, so fragments with higher GC content will have a higher melting
temperature.
3. Here is a very short chromosome of a eukaryotic cell that lacks telomerase.
Replication starts near x. One strand of the DNA has been labeled with heavy (15) N,
hence the capital letters, but all newly synthesized DNA will have normal N.
5’ aaaggg . . . . . . . . x . . . . . . . ccctttggg 3’
3’ TTTCCC . . . . . . . . X . . . . . . . GGGAAACCC 5’
That cell divides to make two daughters, which in turn divide to make two granddaughter
cells.
Draw the cell pedigree, showing this chromosome in the two daughter and four
granddaughter cells. (For the purposes of this question, assume that replication uses a
primer that is only 3 bases long. Also, real chromosomes would have specific sequences
at the ends- I just used these because they are easy to write down.)
This question required you to put together two different concepts: semi-conservative
replication and the function of telomerase.
Semi-conservative replication
means
that heavy N would be present in only one of the
four granddaughter chromosomes.
The lack of telomerase meant that the new strands would be shorter at the 5’ end each
generation, leaving a 3’ overhang. Because it is semi-conservative, the original strand is
unchanged, however. So at the end you have a mix of short and long strands.
Assuming the primer is 3 bases long, the chromosomes will get shorter in multiples of 3.
Daughters
5’ aaaggg . . . . . . . . x . . . . . . . ccctttggg 3’
3’ tttccc . . . . . . . . x . . . . . . . gggaaa
5’
5’
ggg . . . . . . . . x . . . . . . . ccctttggg 3’
3’ TTTCCC . . . . . . . . X . . . . . . . GGGAAACCC 5’

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