mcb421Homework 2010 2 Answers-1

mcb421Homework 2010 2 Answers-1 - MCB 421 HOMEWORK #2...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
MCB 421 HOMEWORK #2 ANSWERS FALL 2006 Page 1 of 6 1. The Salmonella typhimurium strain TR248 is auxotrophic for both histidine and cysteine due to a mutation in the hisC gene and a mutation in the cysA gene. His + revertants are found at a frequency of 1 per 10 7 cells. Cys + revertants are also found with a frequency of 1 per 10 7 cells. a.) How would you select for His + revertants only or Cys + revertants only? (What kind of medium would you plate the cells on?) ANSWER: Plate greater than 10 7 cells on a minimal plate without His (for His+ revertants) or without Cys (for Cys+). Any cells growing on the plate are revertants. b.) At what frequency would you expect to find revertants that are both His + and Cys + ? How could you directly select for such double revertants? ANSWER: If the reversions are the result of independent events, then the frequency will be the product of the two independent frequencies, i.e. 10 -7 x 10 -7 equals one in 10 14 cells. The revertants could be selected by plating on plates lacking both his and cys. c.) His + Cys + revertants are actually found at a frequency of 1 per 10 8 cells. Propose an explanation for this result. ANSWER: The higher than expected frequency tells us that reversion at both loci must occur through a single event. One explanation for this is that both the His and Cys pathways have a common step that was affected by the mutation. Another explanation is that both the mutations were nonsense mutations and the mutation that reverted the strain mutated a tRNA gene so that the altered tRNA can now read nonsense codons. (The latter explanation will be illustrated in a future lecture). 2. In the following table, briefly diagram or indicate the common properties of each type of mutation.
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
MCB 421 HOMEWORK #2 ANSWERS FALL 2006 Page 2 of 6 ANSWER: Mutation Missense Nonsense Frameshift Deletion Insertion effect on DNA base substitution base substitution resulting in a stop codon insertion or deletion of 1 or 2 base pairs loss of multiple base pairs addition of multiple base pairs effect on Protein substituted amino acid truncated polypeptide; inactive protein altered amino acid sequence downstream of mutation; usually truncated polypeptide; inactive protein usually absent; usually inactive protein may insert extra amino acid (if in frame) or cause premature truncation; usually inactive protein effect on Phenotype may result in loss of function or
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 6

mcb421Homework 2010 2 Answers-1 - MCB 421 HOMEWORK #2...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online