Answers for the second set of problems

1. The strand has both ribo- and deoxyribonucleotides. Therefore, untwisting and separation of DNA strands, RNA primer synthesis, and synthesis of replicated DNA has been initiated.  Removal of Okazaki fragments by DNA Polymerase I and ligation of DNA fragments by Ligase have not occurred.

2a. DNA Polymerase III initiates replication from an RNA primer; DNA Polymerase I (by its unique 5' to 3' exonuclease activity) removes the RNA primer and finishes DNA synthesis

b. Helicase opens double-stranded DNA; gyrase prevents supercoiling.

c. Deoxyribonucleotides are the building blocks of DNA; dideoxyribonucleotides, because they lack the 3' OH, prevent chain elongation and terminate replication (their use is the basis of the Sanger method of DNA sequencing).

3a. Ligase joins DNA fragments together; DNA replication would be complete and there would be no RNA primers left, but DNA fragments would not be fully joined together.

b. DNA polymerase III initiates and carries out DNA replication; the strands would be separated and the RNA primers would be bound to the single strand DNA

c. Single Stranded Binding protein binds to single stranded DNA and prevent its digestion; DNA would be cut in all the places opened by helicase

d. Gyrase uncoils supercoiled DNA; the replication fork would stop because helicase could no longer open up the strands

e. Primase makes the RNA primer; the strands would separate and bind SSB, but no replication would occur

4a. Primase is defective because no RNA primers are made for initiation of replication.

b. Ligase is defective because as the fragmented strands haven't been ligated together.

c. Gyrase is defective because supercoiling hasn't been relieved.

d. DNA Polymerase I is defective because RNA primers have not been eliminated.

5a. False, A-T base pairs melt first, because their binding is less stable (they have two rather than three hydrogen bonds).

b. False, DNA polymerase III synthesizes DNA from an RNA primer during replication.

c. False, it takes more energy (and therefore more stringent conditions) to pull apart GC-rich DNA

6. Both of these methods label DNA.

In nick translation, a small amount of DNase is used to create nicks in the DNA.  Labeled oligonucleotides are then added with DNA Polymerase I in order to fill in at the nicks.

Random priming uses the Klenow fragment and short random oligonucleotides to prime synthesis through the added DNA.  Labelled oligonucleotides are used in the synthesis of DNA at the primers.

7a. The template DNA strand, from which the mRNA is synthesized, is 5' CAAACTACCCTGGGTTGCCAT 3'

(RNA synthesis proceeds in a 5' -> 3' direction, so the template strand and the mRNA will be complementary to each other)

b. The coding DNA strand, which is complementary to the template strand, is 5' ATGGCAACCCAGGGTAGTTTG 3'

c. The sequence of the mRNA is 5' AUGGCAACCCAGGGUAGUUUG 3'

(the sequence of the mRNA is complementary to the template strand and identical to the coding strand with U substituted for T)

d. The third codon is 5' ACC 3'.  Therefore, the corresponding anti-codon is 5' GGU 3'

8. Below is a table for the genetic code:

a. The following codons can be mutated by one base to produce an amber codon:

CAG    Gln

AAG    Lys

GAG    Glu

TCG    Ser

TTG    Leu

TGG    Trp

TAA    Stop

TAT    Tyr

TAC    Tyr

b. From part a, CAG (Gln) and TGG (Trp) can become amber stop codons through EMS.

c. From part b, both of the resulting amber codons could be suppressed by amber nonsense suppressors generated by EMS.

9a. The codon is the three nucleotide sequence in the mRNA that indicates which amino acid should be incorporated in the growing polypeptide chain.  The anticodon is the complementary three nucleotide sequence in the appropriate tRNA.

b. Template strand is the DNA strand off which the mRNA is synthesized.  The coding, or non-template, strand is the DNA strand complementary to the template strand; it has the same sequence (except for T for U substitutions) as the mRNA.

10a. False, a wobble allows the anticodon in the tRNA to hybridize with different codons in mRNA.

b. False, a frameshift mutation affects all the subsequent amino acids.

c. False, only one codon (AUG) encodes for the start of protein synthesis; three codons signal the end of protein synthesis.

d. False, the wobble is first base (5 to 3) in the anticodon.

e. True, RNA can be used as a template for DNA synthesis in a process known as reverse transcription.

f. True.  For example, a single base substitution causing CAT to change to AAT would signal a termination.

g. False, the Wobble Hypothesis explains how alternate base pairing can occur with the first nucleotide (going from 5' to 3') in the anticodon.

11a. Digestion of RNA with alkali will cleave the strand after each 3 phosphate.  Therefore, the products remaining will consist of pppNp, Np, and N-OH

b. If RNA was synthesized in the 3' to 5' direction (i.e. by adding ribonucleotides to the 5' end), then the pppNp and Np fragments should be labeled with tritium.

c. If RNA was synthesized in the 5' to 3' direction (i.e. by adding ribonucleotides to the 3' end), then the Np and N-OH fragments should be labeled with tritium.

d. Since the N-OH fragments were labeled with tritium, RNA synthesis must occur in a 5' to 3' direction.

12. In a missense mutation, the new nucleotide alters the codon so as to produce an altered amino acid in the protein product.  With a nonsense mutation, the new nucleotide changes a codon that specified an amino acid to one of the stop codons (TAA, TAG, or TGA). Therefore, translation of the messenger RNA transcribed from this mutant gene will stop prematurely.