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- Neurospora of genotype a + c are crossed withNeurospora of genotype + b +. (Here, + is shorthandfor the wild-type allele.) The following tetrads areobtained (note that the genotype of the four sporepairs in an ascus are listed, rather than listing alleight spores):a + c a b c + + c + b c a b + a + ca + c a b c a + c a b c a b + a b c+ b + + + + + b + + + + + + c + + ++ b + + + + a b + a + + + + c + b +137 141 26 25 2 3a. In how many cells has meiosis occurred to yieldthese data?b. Give the best genetic map to explain these results.Indicate all relevant genetic distances, both betweengenes and between each gene and the centromere.c. Diagram a meiosis that could give rise to oneof the three tetrads in the class at the far right inthe listFrom 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.Please label the tetrad type in the table as PD (parental ditype), NPD (non parental ditype) or T (tetratype) and answer the following questions a) Are the genes linked? Please explain SPECIFICALLY how you can distinguish between linked and unlinked genes in this instance. b) If the two genes are linked, calculate the % recombination between ser and thr. Show the formula used, as well as all of your calculations. c) Draw a single map illustrating the arrangement of the two genes on the chromosome with respect to each other and to the centromere of the chromosome. Make sure to map ALL three distances
- In the fungus Neurospora, a strain that is auxotrophic for thiamine (mutant allele t) was crossed with a strain that isauxotrophic for methionine (mutant allele m). Linear asci were isolated and classified into the following groups: a. Determine the linkage relations of these two genes to their centromere(s) and to each other. Specify distances in map units. b. Draw a diagram to show the origin of the ascus type with only one single representative (second from right).Alleles of genes A and B were analyzed in Neurospora according to the cross shown below. Ordered tetrads are summarized in each horizontal row with the number of tetrads in each category listed alongside. A) Analyze the data to determine the recombination frequency (RF) between A and B, along with any additional information that is available from this data. B) Draw a map of the chromosome or chromosomes with appropriate map distances. C) Use the Perkins formula to reanalyze any relationship between A and B.In tomato the mutant genes o (oblate=flattened fruit), p (peach=hairy fruit) and s (compound inflorescence) were found to be in chromosome 2. The test cross results are: 73 110 + 348 2 2 + 306 p 96 63 a) Construct the linkage map b) What are the genotypes of the homozygous parents used in making the F1 heterozygote? c) Compute for the coefficient of coincidence. O o O O +
- The expected ratio of phenotypes among the progeny of a test cross is 1:1:1:1. Out of 200 total resulting progeny, 48 occur in one of the four phenotypic classes. Given this information, which of the following must also be true? a)At least one additional cell must also contain a count of 48. b)The progeny of this cross do not conform to a 1:1:1:1ratio. c)The value of observed - expected for this cell = -2. d)Since 48 is so close to the expected value, there is no need to calculate chi square before drawing a conclusion about the ratio.A yeast geneticist irradiates haploid cells of a strain thatis an adenine-requiring auxotrophic mutant, caused bymutation of the gene ade1. Millions of the irradiatedcells are plated on minimal medium, and a small number of cells divide and produce prototrophic colonies.These colonies are crossed individually with a wildtype strain. Two types of results are obtained:(1) prototroph × wild type : progeny all prototrophic(2) prototroph × wild type : progeny 75% prototrophic,25% adenine-requiring auxotrophsa. Explain the difference between these two types ofresults.b. Write the genotypes of the prototrophs in each case.c. What progeny phenotypes and ratios do you predictfrom crossing a prototroph of type 2 by the original ade1auxotroph?Ensure answers are clearly labelled a) & b). a) Consider the following pedigree. The solid symbols represent affected individuals. Which of the following is / are possible genotypes for II-2 with respect to this disease? Please type 1 - 4, and indicate yes or no only for each. II III 1. xAxa 2. xaxa 3. Aa 4. AA2 b) Genes A, B and C are on the same chromosome linked in cis (coupling) conformation. A'is 16 cM from B, and B is 22 cM from C. The distance between A and C is 38 cM. The coefficient of coincidence is 0.55 for a trihybrid test cross. How many individuals with the genotype AabbCc do you expect to see among the offspring of the cross if 1000 offspring are obtained? Please show your calculations and round your answer off to the nearest whole number.
- You cross two yeast strains one is an ade auxotroph the other is a pro auxotroph and allow the diploid to sporulate. When you score each spore in the ascus you find the following proportions: 518 PD, 8 NPD, and 225 T. a.) What are the genotypes of each spore in all three types of the tetrads. b) Are these genes linked why or why not? c.) If these genes are unlinked what would you expect the progeny numbers and ratios to be? d.) What is the formula to determine the most accurate distance between these genes? If linked what is the map distance?Time mapping is performed in a cross involving the genes his,leu, mal, and xyl. The recipient cells are auxotrophic for all fourgenes. After 25 minutes, mating is interrupted, with the resultsin recipient cells shown below. Diagram the positions of thesegenes relative to the origin (O) of the F factor and to one another.(a) 90% are xyl+(b) 80% are mal+(c) 20% are his+(d) None are leu+Homozygous wild-type male mice (AA BB CC) were crossed with triplemutant female mice (aa bb cc), forming an F1 generation with the followinggenotype (Aa Bb Cc). The F1 males were crossed with triple mutantfemales, forming the following F2 phenotypes”“a B c” 3“A b C” 3“a b c” 8“A B c” 5“a b C” 5“A B C” 8“a B C” 6“A b c” 6 44 Determine the sequence of the genes