BCOR101 Midterm II
Wednesday, October 26, 2005 Name__Key_______________________
Please show all of your work.
1. A donor strain is trp+, pro+, met+ and a recipient strain is trp-, pro-, met-. The donor strain is infected with phage an the lysate is used to inoculate the recipient (-) strain. After a few minutes, the inoculated recipients are plated on plates without tryptophan. 100 colonies were chosen at random and those colonies were tested for their ability to also grow on plates without proline or on plates without methionine or both. Here are the results:
Type of plate Number of colonies that can grow
All nutrients except tryptophan
100 All except proline and
tryptophan
40 All except methionine and tryptophan
10 All except proline
methioinine and tryptophan 0
a) What is the order of the three genes trp, pro, and met?
Trp is in the middle.
b) Which gene (pro or met) is closer to trp?
Pro is closer because the cotransduction frequency is higher
2. White Leghorn chickens are homozygous for a dominant allele C that produces colored feathers, but also homozygous for the dominant inhibitor allele at another locus (I) that inhibits color formation and prevents expression of C. Another breed (White Wyandottes) are homozygous for both recessive alleles (ccii). They make neither the pigment nor the inhibitor.
a) If White leghorns and White Wyandottes are crossed, how many of the F1 chickens will have white feathers?
CCII x ccii -> CcIi
All of the chickens will be white because they have one copy of the dominant inhibitor, I.
b) If those F1s are randomly crossed among themselves, what proportions of offspring are expected to be white in the F2?
13/16
CcIi x CcIc ->
9 C-I- white 3 C-ii dark 3 ccIi white 1 ccii white
3. The substrates A, B, C, D, and E are all involved in the same biosynthetic pathway for an essential nutrient (X). Four mutants (1-4) were identified which were unable to grow without an external source of nutrient X. They were then tested for their ability to grow on each of the intermediate substrates. “+”
means the cells grew; “0” means they could not grow.
A B C D E 1 + 0 + + + 2 + 0 0 + + 3 0 0 0 + 0 4 0 0 0 + +
a) What is the order of substrates in that pathway?
B - C- A- E- D
b) For mutant #1, which substrate is likely to build up to high concentrations in the cell?
B.
Mutation 1 blocks the B->C step, so the precursor (B) will build up in the cell.
4. The petals of the plant “blue-eyed Mary” are usually blue, but occasionally you find pink or white flowered plants. Pure true-breeding lines of blue, pink, and white flowered plants were crossed, to produce an F1. Then the F1 plants were selfed to make an F2.
Here are the results:
True-breeding line cross F1 F2 phenotypes
Blue x Blue Blue 90 blue
Genotypes: PPBB x PPBB PPBB PPBB
Blue x Pink Blue 301 blue; 102 pink
Genotypes: PPBB x PPbb PPBb PPB-; PPbb
Blue x White Blue 99 blue; 33 white
Genotypes: PPBB x ppBB PpBB P-BB; ppBB
White x Pink Blue 91 blue; 40 white; 32 pink
Genotypes: ppBB x PPbb PpBb P-B-; pp--; P-bb
a) Provide a genetic explanation for these results. Define some allele symbols of your choice. List the genotypes for each color class in the table above.
Must be two loci, because white x pink gives blue F1 and 9:4:3 ratio.
Assume the first locus P makes a pink pigment with P dominant to p (white).
The second locus (B) converts pink to blue, with P dominant to p.
b) A certain white plant and a certain pink plant (from the F2 in the last line of the table) were crossed and gave 50 pink and 50 blue. What must have been the genotypes of those two plants?
Must be ppBb (white) x PPbb (pink)
Offspring would then be PpBb (pink) and (PpBb (blue)
5. For the replication bubble illustrated here
a) indicate the leading strand and the lagging strand at each replication fork b) identify the ends of each of the fragments below as 3’ or 5’.
Fragment
(Label 5’ and 3’ ends) Does this fragment represent a leading or lagging strand?
3’-a---b-5’ leading strand 3’- c---d-5’ lagging strand 5’-e---f-3’ lagging strand 5’-g---h-3’ leading strand
6. You have isolated a strain of E. coli with a mutation in DNA ligase. The enzyme functions when cells are grown at 22 C but is inactive when cells are grown at 37 C. Cells were grown at 22 C in media containing 15N until all of their DNA contained 15N. The cells were then shifted to 37 C and grown in media containing 14N for one generation. Using solid lines for 15N DNA and dashed lines for 14N DNA, show what the products of replication would look like and compare these to what they would look like if the cells were grown at 22 C.
At 22 C, both the leading and lagging strand would be synthesized normally. At 37 C, the lagging strand would contain Okazaki fragments that would not be joined together due to the mutation in DNA ligase. Therefore, there will be gaps between the Okazaki fragments bound to the template strand.
Lagging strand Leading strand 22 C:
- - - - - - 37 C:
- - - - - - - - - - - - - - -
7. For the RNA sequence 5’ – CAUCAUGACAGACCCUUGCUAACGC-3’
a) Show the sequence of both strands of the DNA from which this RNA was transcribed.
5’ – CATCATGACAGACCCTTGCTAACGC – 3’
3’ - GTAGTACTGTCTGGGAACGATTGCG – 5’
b) indicate the 5’ and 3’ ends of each DNA strand and label the template strand.
See above
c) what is the sequence of the peptide encoded by this RNA?
Met – Thr – Asp – Pro – Cys
8. The anticodon of a tRNA molecule is 5’-AUG-3’. What amino acid is attached to the 3’ end of this tRNA? Explain your answer.
The codon that interacts with 5’-CAU-‘ which codes for histidine.
9. A friend brings you three samples of nucleic acid and asks you to determine each sample’s chemical identity (whether DNA or RNA) and whether the molecules are double-stranded or single-stranded.
You use powerful nucleases to degrade each sample to its constituent nucleoside monophosphates and then determine the approximate relative proportions of the nucleosides. The results of your assay are shown below. What can you tell your friend about the nature of these samples?
Sample 1: dGMP: 13% dCMP: 14% dAMP: 36% dTMP: 37% - double stranded DNA Sample 2: dGMP: 12% dCMP: 36% dAMP: 47% dTMP: 5% - single stranded DNA Sample 3: GMP: 22% CMP: 47% AMP 17% UMP: 14% - single stranded RNA
10. Two E. coli genes, A and B, are known from mapping experiments to be very close to each other.
A deletion mutation is isolated that eliminates the activity of both A and B. Neither the A nor the B protein can be found in the mutant, but a novel protein is isolated in which the amino-terminal 30 amino acids are identical to those of the B gene product and the carboxyl-terminal 30 amino acids are identical to those of the A gene product.
With regard the 5’ – 3’ orientation of the nontranscribed DNA strand, is the order of the genes AB or BA? Explain your answer.
The order of the genes is BA. If you are looking at the nontranscribed DNA strand, the orientation of the A and B genes is the same as found in the mRNA. The novel protein found in the mutant has the same amino terminus as the B protein and the same carboxyterminus as the A protein. Thus, the mutation must have resulted in a deletion between A and B, and fused the 5’ end of the B gene with the 3’ end of the A gene.
11. Please complete the table below, indicating all of the nucleotides in the DNA, mRNA and tRNAs and the amino acid sequence of the encoded peptide.
DNA: ATGTATGAAAAATGG DNA: TACATACTTTTTACC RNA: AUGUAUGAAAAAUGG tRNA: UACAUACUUUUUACC protein: met-tyr-glu-lys-trp
DNA DNA
mRNA tRNA protein
