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BS 50—Genetics and Genomics Week of Oct 31 Additional Practice Problems for Section Answer Key Question 1. Loss-of-function mutations are significantly more frequent than gain-of-function mutations. Why is this so? Mutations are more likely to reduce or eliminate gene function than to enhance it. As stated in the text, “By randomly changing or removing one of the components of a machine, it is much easier to break it (that is, loss of function) than alter the way it works (that is, gain of function) Question 2 a) Define “transition mutations” and “transversion mutations.” Give one example of each. Transitions: Change from a purine to a purine or a pyrimidine to a pyrimidine. Examples: A to G; G to A; C to T; T to C Transversions: Change from a purine to a pyrimidine or vice versa. Examples: A to C or T; G to C or T; C to A or G; T to A or G. b) The frequency of spontaneous transitions is significantly higher than the rate of spontaneous transversions. Using everything you know about transitions and transversions, propose an explanation for why transitions are more common. In a normal double-stranded piece of DNA, purines are always paired with pyrimidines and vice versa. Because purines and pyrimidines are different sized molecules (made of two or one ring, respectively), the consistent pairing results in a consistent distance between the two sugar-phosphate backbones. If two purines or two pyrimidines pair, the double helix will be distorted, and thus be easily recognized by the various repair systems. For a transversion to occur, it is necessary for two purines or two pyrimidines to pair (for at least one round of replication). This will result in a distortion and repair. Thus, transversions are not likely to occur because they are easy to detect by repair systems. Whereas transitions do require mispairing, they do not cause the severe distortions of the double helix, and are thus more likely to “escape” the repair systems. Question 3 Shown below is a list of statements (a-k) and types of mutations (1-8). On the blank line following each mutation, write the letter(s) of all statements that apply to that type of mutation. Hint: Each statement may be used more than once and each type of mutation may have more than one correct statement. a) A mutation that changes UUA to UUG 1 b) A mutation that gives methionine instead of leucine c) Created by the addition of a nucleotide to a coding region d) A stop codon is read as an amino acid e) A chemically similar amino acid is replaced by the mutation f) A mutation that changes CCU to ACU g) Deleting a nucleotide in a coding region gives this type of mutation h) Mutation does not alter the peptide i) A mutation changing UAU to UAG j) Premature termination codon is responsible for this mutation k) A chemically different amino acid is replaced by the mutation 1) Missense mutation: b, e, f, k 2) Silent mutation: a, h 3) Frameshift mutation: c, g 4) Nonsense mutation: i, j 5) Synonymous mutation: a, h 6) Suppressor mutation: d Question 4. A partial peptide sequence for the wild type and three mutant alleles of a gene, PET1, are shown below. Each mutant was caused by a single point mutation. Wild Type: met ile arg met asp Mutant 1: met ile gln Mutant 2: met ile arg met Mutant 3: met ser met asp ile asn lys trp… gly … gly lys trp… lys trp… a) Using the amino acid sequence of the wild-type and the three mutants, deduce the exact DNA sequence of the coding strand of the wild-type allele. Each mutation helps determine the correct codon possible for the amino acids. Students therefore cannot get full credit for indicating alternate codons for an amino acid (except for the lys codon, which could be AAA or AAG). Wild Type: met ile arg met asp lys ATG ATC AGA ATG GAT AA(G/A) Mutant 1: met ile gln asn gly ATG ATC CAG AAT GGA TAA trp…… TGG 2 Mutant 2: met ile arg met gly lys ATG ATC AGA ATG GGT AA(G/A) Mutant 3: met ile ser ATG ATC AG(C/T) trp…… TGG met asp lys ATG GAT AA(G/A) trp…… TGG b) What type of mutation (transition vs. transversion vs. indel; missense vs. nonsense vs. frameshift) has occurred in each mutant? Mutant 1: Insertion of either a C in the first base of the third codon or a T, C or A in the third base of the second codon; Frameshift mutation. Mutant 2: A to G transition in the second base of the fifth codon; Missense mutation Mutant 3: A to C or T transversion in the third base of the third codon; Missense mutation. Question 5. 5-bromouracil is an unstable base analog of thymine that spontaneously converts between an enol and keto form. In the enol form, 5-bromouracil pairs with G. In the keto form, 5bromouracil pairs with A. Use a diagram to show how this chemical can cause AT to GC transitions during DNA replication. Answer: T T T A A T A 5BU added during first round of replication. 5BU is incorporated in the daughter strand across from A 5BUketo nd A A 2 DNA replication. No 5BU in the cell enol 5BU 5BUenol G T G C G A 3rd DNA replication. No 5BU in the cell 3 Question 6. A) (6 pts) Yes. Some combination of: each has only one centromere; each has a telomere at its two termini; there is no aneuploidy. B) (6 pts) No. Any combination of 1-2-6-7 has two centromeres; 5-3-4-7 has three telomeres; segment 7 appears twice. C) (8 pts) Yes. In a heterozygote, during meiosis, translocated chromosomes and normal chromosomes pair. Adjacent I segregation (or independent assortment) results in aneuploidy for segments 1 and 5; in each case, one is duplicated and one is missing. (An indication of understanding of the concept of aneuploidy must be demonstrated for full credit). Question 7. a) In humans, the red photoreceptor gene is 96% identical to the green photoreceptor gene. The red and green photoreceptor genes are also next to each other on the X chromosome. Diagram how unequal crossing over can result in red/green colorblindness and gene duplication. Red Green Non-functional hybrid protein Gene duplication b) Gene duplication can lead to the formation of gene families. The globin family includes two clusters of genes, the alpha globin cluster and beta globin cluster. Why is it important for humans to have both fetal and adult forms of hemoglobin? Fetal hemoglobin has a higher affinity for oxygen than adult hemoglobin. This allows the fetus to draw oxygen from its mother’s bloodstream. 4