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- Chlamydomonas, 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_forwardIn 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_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_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_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
- 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_forwardYou perform a cross between a parent with the genotype WWiiNNttEErr and another parent that is wwllnnTTeerr. All genes are unlinked except for W and I which are 22 mu apart. You take an F1 from this cross and cross it with an individual that is wwiiNntteerr. a) What is the probability that this final cross yields an offspring that is wwiinntteerr? b) What is the probability that this final cross yields an offspring that is NNTT or Nntt. (You can ignore all of the other genes for this question.)arrow_forward
- In a cross between a white-eyed female (ww) and a red-eyed male (w+Y), nearly all the progeny were either red-eyed females (w+w) or white-eyed males (wY). However, about 1 in every 2000 F1 flies had an "exceptional phenotype" and was either a white-eyed female or red-eyed male. How did Bridges explain this unexpected result? A) Crossing over B) Incomplete cytokinesis C) Incorrect synapsis D) Nondisjunction E) Pseudoautosomal regionarrow_forwardhis-l and lys-3 are alleles found in baker's yeast that require histidine and lysine for growth, respectively. A cross was made between two haploid yeasts that are his-1 lys-3 and his lys". From the analysis of 900 individual tetrads, the following numbers of tetrads were obtained: GI: 2 spores are his-1 lys* +2 spores are his* lys-3 = 8 G2: 2 spores are his-1 lys-3+ 2 spores are his* lys* = 512 G3: 1 spore is his-1 lys-3 +1 spore is his-1 lys + 1 spore is his lys-3+ 1 spore is his" lys = 380 (i) Name the ascus type of each group as P, NP or T. (ii) Are the genes linked? Explain your answer. (iii) If the genes are linked, calculate the distance between the genes.arrow_forwardYou 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?arrow_forward
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