1. Genetic Problems

2. Question
The percentage of cytosine in a double-stranded DNA molecule is 40%. What is the percentage of thymine?
• Solution
In double-stranded DNA, A pairs with T, whereas G pairs with C; so the percentage of A equals the percentage of T, and the percentage of G equals the percentage of C. If C = 40%, then G
also must be 40%. The total percentage of C + G is therefore
40% + 40% = 80%.
All the remaining bases must be either A or T; so the total percentage of
A + T = 100% – 80% = 20%;
because the percentage of A equals the percentage of T, the
percentage of T is 20%/2 = 10%.

3. Question
The E. coli chromosome contains 4.6 million base pairs of DNA. If synthesis at each replication fork takes place at a rate of 1000 nucleotides per second, how long will it take to completely replicate the E. coli chromosome with theta replication?
Solution
Bacterial chromosomes contain a single origin of replication, and theta replication usually employs two replication forks, which proceed around the chromosome in opposite directions.
Thus, the overall rate of replication for the whole chromosome is 2000 nucleotides per second. With a total of 4.6 million base pairs of DNA, the entire chromosome will be replicated in:

4. Question
Consider the experiment conducted by Meselson and Stahl in which they used 14N and 15N in cultures of E. coli and equilibrium density gradient centrifugation. Draw pictures to represent the bands that would appear if replication were (a) semiconservative; (b) conservative; (c) dispersive.

5. Solution
a. With semiconservative replication, the two strands separate, and each serves as a template on which a new strand is synthesized. After one round of replication, the original template strand of each molecule will contain 15N and the new strand of each molecule will contain 14N; so a single band will appear in the density gradient halfway between the positions expected of DNA containing only 15N and of DNA containing only 14N. In the next round of replication, the two strands again separate and serve as templates for new strands.
Each of the new strands contains only 14N, thus some DNA molecules will contain one strand with the original 15N and one strand with new 14N, whereas the other molecules will contain two strands with 14N. This labeling will produce two bands, one at the intermediate position and one at a higher position in the tube. Additional rounds of replication should produce increasing amounts of DNA that contains only 14N; so the higher band will get darker.

6. b. With conservative replication, the entire molecule serves as a template. After one round of replication, some molecules will consist entirely of 15N, and others will consist entirely of 14N; so
two bands should be present. Subsequent rounds of replication will increase the fraction of DNA consisting entirely of new 14N; thus the upper band will get darker. However, the original DNA with 15N will remain, and so two bands will be present.

7. c. In dispersive replication, both nucleotide strands break down into fragments that serve as templates for the synthesis of new DNA. The fragments then reassemble into DNA molecules. After one round of replication, all DNA should contain approximately half 15N and half 14N, producing a single band that is halfway between the positions expected of DNA labeled with 15N and of DNA labeled with 14N. With further rounds of replication, the proportion of 14N in each molecule increases; so a single hybrid band remains, but its position in the density gradient will move upward. The band is also expected to get darker as the total amount of DNA increases.

8. Question
If there were five different types of bases in mRNA instead of four, what would be the minimum codon size (number of nucleotides) required to specify the following numbers of different amino acid types: (a) 4, (b) 20, (c) 30?
• Solution
To answer this question, we must determine the number of combinations (codons) possible when there are different numbers of bases and different codon lengths. In general, the number of different codons possible will be equal to:
blg = number of codons
where b equals the number of different types of bases and lg equals the number of nucleotides in each codon (codon length). If there are five different types of bases, then:
51 = 5 possible codons
52 = 25 possible codons
53 = 125 possible codons
The number of possible codons must be greater than or equal to the number of amino acids specified. Therefore, a codon length of one nucleotide could specify 4 different amino acids, a codon length of two nucleotides could specify 20 different amino acids, and a codon length of three nucleotides could specify 30 different amino acids: (a) one, (b) two, (c) three.