7022IDJITI atiO1l In yeast crosses, M. 2 where 1 and 2 are mutant sites of a nutritional gene and M and N are flanking genes. is possible to select prototrophs, which can grow on minimal medium, and they have the expected genotype M++n. But the genotype M++N_is also found. A. Please explain the origin of these two genotypes in terms of recombination. B. If M++N were more common than M++n, what would that mean?
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- A Neurospora cross was made between a strain that carried the mating-type allele A and the mutant allele arg-1and another strain that carried the mating-type allele aand the wild-type allele for arg-1 (+). Four hundred linear octads were isolated, and they fell into the sevenclasses given in the table below. (For simplicity, they areshown as tetrads.)a. Deduce the linkage arrangement of the mating-typelocus and the arg-1 locus. Include the centromere orcentromeres on any map that you draw. Label all intervalsin map units.b. Diagram the meiotic divisions that led to class 6. LabelclearlyThe Neurospora octad shown came from a cross between mt and m' strains. a. Is this an MI or an MIl octad or neither? Explain. b. Diagram the production of this octad. c. Is it possible to observe evidence of heteroduplex formation in a Neurospora ascus even if gene conversion did not occur during formation of the octad? Explain.An Hfr strain that is hisE + and pheA + was conjugated to a strain that is hisE − and pheA −. The conjugation was interrupted at different times, and the percentage of recombinants for each gene was determined by streaking on media that lacked either histidine or phenylalanine. The following results were obtained: A. Determine the map distance (in minutes) between these twogenes.B. In a previous experiment, it was found that hisE is 4 minutesaway from pabB and that PheA is 17 minutes from pabB. Drawa genetic map showing the locations of all three genes.
- On Neurospora chromosome 4, the leu3 gene is just to theleft of the centromere and always segregates at the firstdivision, whereas the cys2 gene is to the right of the centromere and shows a second-division segregation frequency of 16 percent. In a cross between a leu3 strain anda cys2 strain, calculate the predicted frequencies of thefollowing seven classes of linear tetrads where l = leu3 andc = cys2. (Ignore double and other multiple crossovers.)From a cross between e+ f+ g+ and e− f − g− strains ofNeurospora, recombination between these linkedgenes resulted in a few octads containing the followingordered set of spores:e+ f+ g+e+ f+ g+e+ f − g+e+ f − g+e− f − g−e− f − g−e− f − g−e− f − g−a. Where was recombination initiated?b. Did crossing-over occur between genes e and g?Explain.c. Why do you end up with 2 f+ : 6 f − but 4 e+: 4 e−and 4g+: 4g−?d. Could you characterize these unusual octads as MIor MII for any of the three genes involved?Explain.. A diploid strain of yeast was made by mating a haploidstrain with a genotype w−, x−, y−, and z− with a haploidstrain of opposite mating type that is wild type for thesefour genes. The diploid strain was phenotypically wildtype. Four different X-ray-induced diploid mutantswith the following phenotypes were produced fromthis diploid yeast strain. Assume a single new mutation is present in each strain.Strain 1 w− x+ y− z+Strain 2 w+ x− y− z−Strain 3 w− x+ y− z−Strain 4 w− x+ y+ z+When these mutant diploid strains of yeast go throughmeiosis, each ascus is found to contain only two viablehaploid spores.a. What kind of mutations were induced by X-rays tomake the listed diploid strains?b. Why did two spores in each ascus die?c. Are any of the genes w, x, y, or z located on thesame chromosome?d. Give the order of the genes that are found on thesame chromosome
- Robert Bost and Richard Cribbs studied a strain of E. coli (araB14)that possessed a nonsense mutation in the structural gene that encodes Lribulokinase,an enzyme that allows the bacteria to metabolize the sugararabinose (R. Bost and R. Cribbs. 1969. Genetics 62:1–8). From thearaB14 strain, they isolated some bacteria that possessed mutations thatcaused them to revert back to the wild type. Genetic analysis of theserevertants showed that they possessed two different suppressormutations. One suppressor mutation (R1) was linked to the originalmutation in L-ribulokinase and probably occurred at the same locus. Byitself, this mutation allowed the production of L-ribulokinase, but theenzyme produced was not as effective in metabolizing arabinose as theenzyme encoded by the wild-type allele. The second suppressormutation (SuB) was not linked to the original mutation. In conjunctionwith the R1 mutation, SuB allowed the production of L-ribulokinase, butSuB by itself was not able to suppress the…In a haploid yeast strain, eight recessive mutationswere found that resulted in a requirement for theamino acid lysine. All the mutations were found to revert at a frequency of about 1 × 10−6 except mutations5 and 6, which did not revert. Matings were madebetween a and α cells carrying these mutations. Theability of the resultant diploid strains to grow onminimal medium in the absence of lysine is shown inthe following chart (+ means growth and − means nogrowth.)1 2 3 4 5 6 7 81 − + + + + − + −2 + − + + + + + +3 + + − − − − − +4 + + − − − − − +5 + + − − − − − +6 − + − − − − − −7 + + − − − − − +8 − + + + + − + −a. How many complementation groups were revealedby these data? Which point mutations are foundwithin which complementation groups?The same diploid strains are now induced to undergosporulation. The vast majority of resultant spores areauxotrophic; that is, they cannot form colonies whenplated on minimal medium (without lysine). However,particular diploids can produce rare spores…You have identified a SNP marker that in one largefamily shows no recombination with the locus causinga rare hereditary autosomal dominant disease.Furthermore, you discover that all afflicted individuals in the family have a G base at this SNP on theirmutant chromosomes, while all wild-type chromosomes have a T base at this SNP. You would like tothink that you have discovered the disease locus andthe causative mutation but realize you need to consider other possibilities.a. What is another possible interpretation of the results?b. How would you go about obtaining additional genetic information that could support or eliminateyour hypothesis that the base-pair difference is responsible for the disease?
- A recent estimate of the rate of base substitutions atSNP loci is about 1 × 10−8 per nucleotide pair pergamete.a. Based on this estimate, about how many de novomutations (that is, mutations not found in the genomes of your parents) are present in your own genome?b. Where and when did these de novo mutations inyour genome most likely occur?c. It has been calculated that each sperm made in a25-year-old man is the result on average of about300 rounds of cell division, starting with the firstmitotic division of the male zygote. In contrast,each mature oocyte found in a 5-month-old femalehuman fetus is the result of about 25 rounds of division, starting with the first mitotic division of thefemale zygote. What bearing do these calculations have on the estimate of the rate of base substitutions in humans, and on your answer to part (b)?In an in situ hybridization experiment, a certain clonebound to only the X chromosome in a boy with no diseasesymptoms. However, in a boy with Duchenne musculardystrophy (X-linked recessive disease), it bound to theX chromosome and to an autosome. Explain. Could thisclone be useful in isolating the gene for Duchenne muscular dystrophy?When a female melanotic fly is crossed with a normal male, the progeny are produced: 123 normal females, 125 melanotic females, and 124 normal males. In subsequent crosses between melanotic females and normal males, melanotic females are frequently obtained, but never any melanotic males. Provide a possible explanation for the inhertiacne of the melanotic mutation (Hint: The cross produces twice as many female progeny as male progeny)