BSCI410/Project#1/SP12BSCI 410-Liu /Chang Project #1 (Homework) 50 points Fall 2012 Due: Thu (Sept. 20) at 11:00 AM in class1. (5 points) Multiple choices (pick one)(1) Double stranded breaks are caused by B(a) UV; (b) gamma ray; (c) DNA replication error(2). Which type of mutation is least likely to revert? E(a) Deletion; (b) Translocation; (c) transposition; (d) transition; (e) a and b(3). The hydrolysis of an -NH2 group from a base is called C(a) hydroxylation; (b) alkylation; (c) deamination; (d) intercalation(4). In an Ames test for mutagenicity, rat liver enzymes are accidentally excluded from the test. What can you say about the result? Ca) the mutation rate will probably rise b) the mutation rate will probably decline c) the results are less relevant to mammals d) the results are less relevant to bacteria e) no mutations will form(5). To isolated mutants in essential genes (genes absolutely needed for survival), we need to isolate Ea) mutations that only partially disrupt the function of the geneb) recessive mutation that can be kept alive as heterozygotec) mutations that disrupt the gene function only under certain conditionsd) mutations that can be revertede) a, b, and c2. (6 points) Balancer chromosome works to keep a lethal mutation as a heterozygote.In following Drosophila example, b is a newly isolated recessive lethal mutation. C* marks the balancer chromosome (in red) and is a dominant mutation causing cyan color body as a heterozygote but lethality as a homozygote. A, B, C in capital letters are all wild type genes. 1A C* BA b CXA C* BA b CBSCI410/Project#1/SP12(a) Draw the progeny genotypes of above cross. Label phenotype of each genotype (note: death is one of the phenotype)A b C - CYAN COLOR and aliveA B C* A b C - LETHALA b CA B C* - CYAN COLOR and alive (same as the top one)A b CA B C* - LETHALA B C*(b) Use drawings in (a) to explain why one could maintain a heterozygote stock of theb mutation.Unless there is recombination, the C* will always be on the strain with a wild type B while the C will always be on the strain with the recessive mutant b. In this way, whenever there are heterozygotes, there will be cyan color along with heterzygosity at the B locus to keep b mutation in the stock. A shorter version: All other combinations will be dead except ABC*/AbC, where b is always kept alive as a herozygote. Since there is no recombination between ABC* and AbC, this heterozygote will be always there generations after generation through self-crossing.3. (5 points) Describe a genetic selection scheme to isolate yeast mutants defective in ADH (Alcohol Dehydrogenase). (Note: start with mutagenesis). Mutagenize the yeast with a mutagen such as EMS, plate the mutagenized yeast population on a media (either complete or minimal media) containing allyl alcohol. Only mutants that CAN NOT metabolize allyl alcohol (such as a mutant in ADH gene) can grow and form colony. All wild type yeast will be killed by metabolizing allyl alcohol into toxic acrolein aldehyde. Therefore, the colony on the plate represent ADH mutants.2BSCI410/Project#1/SP12 4. (5 points) What effect on DNA could each of following mutagen cause?(1) UV irradiationPyrimidine dimers – Thymine Thymine dimers(2) X-rayDOUBLE STRANDED BREAKS(3) HydroxylamineMutates G to A or C to T to form Transition mutation(4) Nitrous acidCauses C to T, and A to G change. Both are transitions(5) Transposon insertion to the exon of a geneDisrupting protein coding region and destroing gene’s activity. 5. (5 points) You have isolated five recessive pea mutants (s1-s5) which are abnormal in seed shape. While wild type seeds are Round (R); these s1-s5 mutant seeds are Moon shape (M). To test if any of these mutations are allelic to each other, you did pairwise crosses among all five mutations and observed F1 phenotypes, which are summarized in the table below. How many genes are represented by these five mutations? How are thesefive mutations grouped into different genes? S1 and s2 affect the same gene (gene1), S3 is different from all and affects gene 2, S4 and S5 affect same gene (gene3). The five mutations define three different genes. s1 s2 s3 s4 s5 s1 M M R R Rs2 M R R Rs3 M R Rs4 M Ms5 M (R: Round; M: Moon-shaped)6. (3 points) Translate following mRNA sequence:5’UUGCUCCAUGGCCUGGAUUUACUUACCCCCAGGCCCCAAAUGAGGGUA3’Use the underlined as START and STOP codon respectively.3BSCI410/Project#1/SP12Met A W I Y L P P G P K StopMet Ala Trp Ile Tyr Leu Pro Pro Gly Pro Lys Stop7. (6 points) Mr. Hawkins isolated five rose flower mutants, which are named r1, r2, r3, r4, and r5. While the wild type flowers are red, all these r1-r5 mutant flowers are white.(a) Mr. Hawkins first wants to determine if r1 mutations is recessive or dominant. How should he proceed with the cross? What is the phenotype of the cross progeny if r1 is recessive? What is the phenotype of the cross progeny if r1 is semi-dominant? He should cross r1 with WT to see F1 progeny color. If r1 is recessive, F1 (r1/+) will be WT red color. If r1 is dominant, F1 (r1/+) will be white. If r1 is semi-dominant, F1 (r1/+) will be pink.(b) Mr. Hawkins discovered that r2 and r4 are both recessive. He then proceeded to perform the complementation tests between r2 and r4. Briefly describe the cross and the likely outcome (ie. phenotype) if r2 and r4 reside in the same gene.Cross homozygous r2 with homozygous r4. If r2 and r4 reside in the same gene, The F1 progeny will be white.8. (6 points) TA David Lin isolated a new strain of worms from China. He crossed the China worm with the worm he studies in the lab. First, he crossed female worm from China with male worm from the lab, he noticed that lots of F1 progeny are sterile. Second, when he crossed male worm from China with female worm in the lab, all F1 progeny are fertile and healthy. (a) What phenomenon (a specific terminology) did David observe?Hybrid dysgenesis(b) Briefly explain what is happening in the first cross that yielded sterile progeny?The lab strain has a transposon that is kept immobile by the presence of repressors in the cytoplasm. The China strain has no transposon and no repressors. In the first cross, the oocyte of the China strain was crossed with the lab strain sperm, which brings the transposon into the F1 progeny but did not bring repressors with it as sperm has little cytoplasm. The transposon