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- Below is a partially filled in complementation table. Please answer the following questions. 1 2 3 4 LO 5 6 1 2 I 3 4 5 4 + + 5 6 a. If you were to perform a complementation test with mutants 1 and 2, what would be the result? [Select] b. If you were to perform a complementation test with mutants 5 and 6, what would be the result? [Select] c. If you were to perform a complementation test with mutants 3 and 4, what would be the result? [Select] d. There is one complementation group that you can determine from the information above. If you performed a complementation test with mutant 1 and 6 and there was growth, would you say that the mutants were part of that complementation group? [Select] e. What is the minimum number of complementation groups that could exist based on the information above? [Select] f. What is the maximum number of complementation groups that could exist based on the information above? [Select]Now, figure out the complementation groups. This shows how many genes are involved in this pathway. Some of the mutant classes in the previous problem must contain more than one complementation group (gene) For example class 2 is defined by mutants 1,3,4,6, and 7. These mutants may, or may not, all be in the same gene. Complementation testing will help you know if they are, or are not, in the same gene. So there are actually additional steps between some of the letters. For example, maybe instead of XàY, it is really XàFàGàY.b d f a + + + + d + e The normal flower color in garden peas is purple. A researcher interested in the biochemical pathway leading to purple flower color has isolated six white mutants, each of which is recessive to purple. The six white mutants have been designated a, b, c, d, e and f. Complementation tests have been carried out for each pair of mutants, and the results are indicated in the table above, where a “+" indicates complementation (purple flower color in the offspring of the complementation test) and a “-" indicates no complementation (white flower color). What is the number of genes (complementation groups) identified by these mutants? +| +|+|+|+
- For each genotype in the table below, determine whether or not functional B-gal will be produced in the presence or absence of the inducer. Write a plus (+) if B-gal is produced or a minus (-) if it is not. ma Chromosome F' lac (plasmid) - Inducer + Inducer ---- --- --- I*O*Z I*O°Z+ I*O*Z* I* = wildtype repressor |- - no functional repressor produced O* - wildtype operator OC = operator mutation prevents repressor from binding %3DCystic fibrosis (CF) is a genetic disorder affecting a number of organs, including the lung airways, pancreas, and sweat glands. Mutations in both copies of the CFTR gene causes cystic fibrosis. Imagine that you have sweat gland samples from several Cystic Fibrosis patients (A-C) with unknown mutations in CFTR. You also have normal (+) sweat gland sample to use as a positive control. А В С А В С Choose which mutation would explain the RNA and protein results in A, B, & C: 1. Promoter/Regulatory mutation 2. Silent mutation 3. Missense mutation 4. Deletion mutation 5. Splice site mutation 6. Nonsense mutation RNA gel Protein gelKnock out mice that are mutant for a gene product X die. Variants of the X gene that were reintroduced through homologous recombination indicate that variant proteins that lack the C-terminal region of the protein cannot rescue the lethality (did not allow them to live), and always localized to the cytoplasm. Those that retained the C-terminal region rescued the lethality (conferred viability!) and were consistently localized within the nucleus. RNA-seq analysis of the mutant cells vs the wild-type cells indicated that the expression of many genes that are essential for neural function was reduced in the knock-out mutant cells. The N-terminal region of Protein X is 100% conserved between mouse and humans at both the amino acid and the nucleotide level. The predicted mouse mRNA sequence is shown below where the AUG corresponds to the translational start site (AUG). 5’-AUGUUUACAGAGGGGAAU... -3’ d) What motif could be present to direct this protein to its correct destination? e)…
- Knock out mice that are mutant for a gene product X die. Variants of the X gene that were reintroduced through homologous recombination indicate that variant proteins that lack the C-terminal region of the protein cannot rescue the lethality (did not allow them to live), and always localized to the cytoplasm. Those that retained the C-terminal region rescued the lethality (conferred viability!) and were consistently localized within the nucleus. RNA-seq analysis of the mutant cells vs the wild-type cells indicated that the expression of many genes that are essential for neural function was reduced in the knock-out mutant cells. The N-terminal region of Protein X is 100% conserved between mouse and humans at both the amino acid and the nucleotide level. The predicted mouse mRNA sequence is shown below where the AUG corresponds to the translational start site (AUG). 5’-AUGUUUACAGAGGGGAAU... -3’ Q3a) Using a total RNA sample obtained from human cells, what is the first primer you would…E. coli cells are simultaneously infected with two strains of phage λ. One strain has a mutant host range, is temperature sensitive, and produces clear plaques (genotype h st c); another strain carries the wildtype alleles (genotype h+ st+ c+). Progeny phages are collected from the lysed cells and are plated on bacteria. The following numbers of different progeny phages are obtained: Progeny phage genotype Number of plaques h+ c+ st+ 321 h c st 338 h+ c st 26 h c+ st+ 30 h+ c st+ 106 h c+ st 110 h+ c+ st 5 h c st+ 6 a. Determine the order of the three genes on the phage chromosome. b. Determine the map distances between the genes. c. Determine the coefficient of coincidence and the interferenceT. Miyake and M. Demerec examined proline-requiring mutations in the bacterium Salmonella typhimurium (). On the basis of complementation testing, they found four proline auxotrophs: proA, proB, proC, and proD. To determine whether proA, proB, proC, and proD loci were located close together on the bacterial chromosome, they conducted a transduction experiment. Bacterial strains that were proC+ and had mutations at proA, proB, or proD were used as donors. The donors were infected with bacteriophages, and progeny phages were allowed to infect recipient bacteria with genotype proC− proA+ proB+ proD+. The recipient bacteria werethen plated on a selective medium that allowed only proC+ bacteria to grow. After this, the proC+ transductants were plated on selective media to reveal their genotypes at the other three pro loci. The following results were obtained: Q.Why are there no proC− genotypes among the transductants?
- T. Miyake and M. Demerec examined proline-requiring mutations in the bacterium Salmonella typhimurium (). On the basis of complementation testing, they found four proline auxotrophs: proA, proB, proC, and proD. To determine whether proA, proB, proC, and proD loci were located close together on the bacterial chromosome, they conducted a transduction experiment. Bacterial strains that were proC+ and had mutations at proA, proB, or proD were used as donors. The donors were infected with bacteriophages, and progeny phages were allowed to infect recipient bacteria with genotype proC− proA+ proB+ proD+. The recipient bacteria werethen plated on a selective medium that allowed only proC+ bacteria to grow. After this, the proC+ transductants were plated on selective media to reveal their genotypes at the other three pro loci. The following results were obtained: Q.Is there evidence that proA, proB, and proD are located close to proC? Explain your answer.T. Miyake and M. Demerec examined proline-requiring mutations in the bacterium Salmonella typhimurium (). On the basis of complementation testing, they found four proline auxotrophs: proA, proB, proC, and proD. To determine whether proA, proB, proC, and proD loci were located close together on the bacterial chromosome, they conducted a transduction experiment. Bacterial strains that were proC+ and had mutations at proA, proB, or proD were used as donors. The donors were infected with bacteriophages, and progeny phages were allowed to infect recipient bacteria with genotype proC− proA+ proB+ proD+. The recipient bacteria werethen plated on a selective medium that allowed only proC+ bacteria to grow. After this, the proC+ transductants were plated on selective media to reveal their genotypes at the other three pro loci. The following results were obtained: Q.Which genotypes represent single transductants and which represent cotransductants?Drosophila (fruit-flies) mutants A and B have the same phenotype: the absence of red pigment in the eyes. Homozygous mutants were crossed, and all offspring possessed red eyes. Does the appearance of red eyes in the progeny indicate complementation or a failure to complement? [Select] Do the mutants likely have mutations in the same or different genes? [ Select]
