1. Is it possible to know how many purines and pyrimidines there would be in a 2
ID: 6591 • Letter: 1
Question
1. Is it possible to know how many purines and pyrimidines there would be in a 200bp DNAmolecule in humans. Explain (in short). (4 points)
2. If you were given a sample of chromosomal DNA and asked to determine if it is bacterial
or eukaryotic, what experiment would you perform, and what would be your expected
results? [Think about the differences between prokaryotic and eukaryotic DNA, and the
use of enzymes to determine chromosome structure.] (3 points)
3. In a diploid cell in which 2n=14, how many chromosomes and telomeres are there during
the following phases of the cell cycle: (8 points)
a) G1 of mitosis
b) G2 of mitosis
c) Mitotic prophase
d) Mitotic telophase
4. Why are primers needed in DNA replication, and at what cell cycle stage(s) are they
used? (2 points)
5. If species A has more DNA per nucleus than species B, does A necessarily have more
genes than B? Explain (in short). (4 points)
6. In a certain 1kb DNA molecule the G+C content in 60%. How many hydrogen bonds hold
the two strands of this molecule together? (2 points)
7. A DNA molecule of composition [5’ – AAAAAAAAAAA – 3’] is replicated in a solution of
adenine nucleoside triphosphate with all its phosphorus atoms in the form of the
radioactive isotope 32P. Will both the daughter molecules be radioactive? Explain (in
short).
Then repeat the question for the DNA molecule of composition [5’ – ATATATATATAT – 3’].
(2 points
Explanation / Answer
1). In a DNA molecule the number bases depends on the Chargaff’s rule. According to it the purines and pyramidines will be in 1:1 ratio.
The number of bases can be predicted only when any one nucleotide quantity is known. But in the given data, nucleotide percentage is not mentioned. In that case it is difficult to calculate the number of purines and pyramidines.
2). Bacterial and eukaryotic chromosomal DNA show some basic differences. Depending on them they can be determined.
Bacterial chromosomal DNA has single origin of replication and histone proteins are absent.
Eukaryotic chromosomes have histone proteins and they are intervened by the introns.
If histone specific proteolysis is done then we can differentiate the prokaryotic and eukaryotic chromosomes and also based on the intron specificity.
Restriction enzymes which are sequence specific can be used to determine the chromosomal structure.
3). Diploid cell with chromosome 2n = 14 will have the following chromosomes in each stage
a)G1 of mitosis = 7
b) G2 of mitosis = 7
c) Mitotic prophase = 7
d) Mitotic telophase = 14.
The number of telomeres for each chromosome will be 2. Each one present, at the ends of the chromosome. In that way the number of telomeres at each stage are
a)G1 of mitosis = 14
b) G2 of mitosis = 14
c) Mitotic prophase = 14
d) Mitotic telophase = 28.
4). Primers are the short nucleotide sequences. They are used in the process of replication as initiators for the synthesis of lagging strand. They provide the 3’ hydroxyl group. After its use they are degraded by the enzyme primase.
5). If species A has more DNA per nucleus than species B, it is not necessary that species A should have more genes.
It depends on the presence of complexity of genome, non coding genes and other proteins that constituted along with the DNA.
6). Given that G + C =60%
DNA = 1kb = 1000base pairs.
(G + C) + (A + T) = 100
Therefore A + T = 40%
For 1 kb DNA molecule G+C are 600 and A+T are 400
G and C are bonded by three hydrogen bonds. So number of bonds are 600 * 3 = 1800
A and T are bonded by double hydrogen bonds. So number of bonds are 400 * 2 = 800
Total number of hydrogen bonds that hold the two strands of the molecule together are 1800 + 800 = 2600
7). DNA molecule of composition [5’ – AAAAAAAAAAA – 3’] is replicated in a solution of adenine nucleoside triphosphate with all its phosphorus atoms in the form of the radioactive isotope 32P.
The daughter strands will not exhibit radioactivity because the replicated strand will not take adenine. All will be thymine only those are complementary to it.
5’ – AAAAAAAAAAA – 3’
5’ – TTTTTTTTTTT – 3’
But, in the DNA molecule of composition [5’ – ATATATATATAT – 3’] , the daughter strand will exhibit radioactivity because adenine molecules will be incorporated that possess radioactive isotope 32P.
5’ – ATATATATATAT – 3’
5’ – TATATATATATA – 3’
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