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- 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.arrow_forwardPlease answer all parts along with the reason. I'll definitely give a like. Thank you in advance! 1A) From the cross Ab/aB x ab/ab, what is the recombination frequency if the progeny numbers are 17 AB/ab, 72 Ab/ab, 68 aB/ab, and 21 ab/ab? 1B)In human gene mapping, a LOD score is calculated to see if a gene causing a rare disease is linked to a known SNP. The LOD score is -4. This means that 1C) A three-point testcross is used to determine the order of three linked genes. The following crossover progeny result: single crossovers, double crossovers, and no crossovers. To determine the order, the no-crossover progeny must be compared to what other class of progeny?arrow_forwardChlamydomonas, a eukaryotic green alga, may be sensitive to the antibiotic erythromycin, which inhibits protein synthesis in bacteria. There are two mating types in this alga, mt+ and mt-. If an mt+ cell sensitive to the antibiotic is crossed with an mt- cell that is resistant, all progeny cells are sensitive. The reciprocal cross (mt+ resistant and mt- sensitive) yields all resistant progeny cells. Assuming that the mutation for resistance is in the chloroplast DNA, what can you conclude from the results of these crosses?arrow_forward
- In rice, male sterility is controlled by maternal cytoplasmic elements. This phenotype renders the male part of rice plants (i.e. the stamen) unable to produce fertile pollen; the female parts, however, remain receptive to pollination by pollen from male fertile rice plants. However, the presence of a nuclear fertility restorer gene F restores fertility to male-sterile lines. Give the result(s) of the cross and explain the phenotype of the offspring.arrow_forwardDraw all possible linkage maps for these three genes.arrow_forwardIn com, male sterility is controlled by maternal cytoplasmic elements. However, the presence of a nuclear fertility restorer gene (F_) restores fertility to male sterile lines. a. What are the crosses male sterile female x FF male? Give the genotypes and phenotypes of the offspring in each cross. Explain.arrow_forward
- 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.arrow_forwardThe wild-type (W) Abraxas moth has large spots on its wings, but the lacticolor (L) form of this species has very small spots. Crosses were made between strains differing in this character, with the following results:Provide a clear genetic explanation of the results in these two crosses, showing the genotypes of all individual moths.arrow_forwardThree autosomal recessive mutations in yeast, all producing the same phenotype (m1, m2, and m3), are subjected to complementation analysis. Of the results shown below, which, if any, are alleles of one another? Predict the results of the cross that is not shown—that is, m2 * m3. Cross 1: m1 * m24 F1: all wild-type progeny Cross 2: m1 * m34 F1: all mutant progenyarrow_forward
- One yeast strain carries the alleles lys+ and arg+, whereas another strain has lys-3 and arg-2. The two strains were crossed toeach other, and an ascus obtained from this cross has four spores with the following genotypes: lys+ arg+, lys+ arg-2, lys-3arg+, and lys-3 arg 2. This ascus has a. a parental ditype.b. a tetratype.c. a nonparental ditype.d. either a tetratype or a nonparental ditype.arrow_forwardIn tomato, the following genes are located on chromosome 2: + tall plant d dwarf plant + uniformly green leaves m mottled green leaves + smooth fruit p pubescent (hairy) fruit Results of the cross +++ / dmp and dmp / dmp were: + + + = 470 + m p = 1 + + p = 14 d + p = 25 d + + = 0 d m p = 441 + m + = 19 d m + = 30 Identify the single and double crossovers among the progeny.arrow_forwardA cross was performed between a yeast strain that requires methionine and lysine for growth (met− lys−)and another yeast strain, which is met+ lys+. One hundred asci were dissected, and colonies were grownfrom the four spores in each ascus. Cells from thesecolonies were tested for their ability to grow on petriplates containing either minimal medium (min), min+ lysine (lys), min + methionine (met), or min + lys+ met. The asci could be divided into two groupsbased on this analysis:Group 1: In 89 asci, cells from two of the four spore colonies couldgrow on all four kinds of media, while the other two spore coloniescould grow only on min + lys + met.Group 2: In 11 asci, cells from one of the four spore colonies couldgrow on all four kinds of petri plates. Cells from a second one ofthe four spore colonies could grow only on min + lys plates andon min + lys + met plates. Cells from a third of the four sporecolonies could only grow on min + met plates and on min + lys+ met. Cells from the…arrow_forward
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