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Concept explainers
a.
To determine:
The appearance of the mosaic ommatidia that have a wild-type
Introduction:
The eyes of arthropods are made up of units called ommatidia. An ommatidium contains a cluster of photoreceptor cells that are surrounded by support cells and pigment cells.
b.
To draw:
The chromosome including the positions of the FRTs, the P[w+], boss- and boss+ alleles, and the mitotic crossover and the segregation of the chromosomes into daughter cells to generate the mosaics.
Introduction:
In fruit flies, mosaics are usually generated by site-specific recombination between nonsister chromatids during mitosis using the FLP/FRT recombination system.
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Chapter 18 Solutions
Genetics: From Genes to Genomes, 5th edition
- Time mapping is performed in a cross involving the genes his, leu, mal, and xyl. The recipient cells were auxotrophic for all four genes. After 25 minutes, mating was interrupted with the following results in recipient cells. Diagram the positions of these genes relative to the origin (O) of the F factor and to one another. (a) 90% were xyl+ (b) 80% were mal+ (c) 20% were his+ (d) none were leu+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_forwardComplementation tests of recessive mutants a through f produced the data shown in the table below. "+" means wild-type phenotype; "-" means mutant phenotype. PSCO a b с 0 0 0 0 0 a e a f b + C + d e + + + + + - d e f A new mutant, g, is isolated and it fails to complement c. Which other mutant(s) would g also fail to complement? (Select all correct answers.) ++ f ++ + + -arrow_forward
- Contrast the role of the MCM1 protein in different yeastcell types shown in Figure 12-10. How are the a-specificgenes controlled differently in different cell types?arrow_forwardMap distances were determined for four differentgenes (MAT, HIS4, THR4, and LEU2) on chromosome III of the yeast Saccharomyces cerevisiae:HIS4 ↔ MAT 37 cMTHR4 ↔ LEU2 35 cMLEU2 ↔ HIS4 23 cMMAT ↔ LEU2 16 cMMAT ↔ THR4 20 cMWhat is the order of genes on the chromosome?arrow_forwardThe following results are derived from crosses between Neurospora strain xy and strain ++: Tetrad Class 3 4 ху x+ x+ xy ++ ++ ++ +y +y ху +y 25 ++ 3 124 4 (i) Name the ascus type of each class from 1 to 4 as P, NP or T. (ii) Are genes x and y linked? Explain your answer. (iii) If they are linked, determine the map distance between the two genes. If they are unlinked, provide all the information you can about why you draw this conclusion.arrow_forward
- One issue with interrupted-mating experimentssuch as that in Problem 19 is that gene order may beambiguous if the genes are close together. Anothershortcoming is that such experiments do not provide accurate map distances. The reason is that researchers select for the first Hfr marker transferred intothe recipient, but the recovery of F− exconjugantswith a later Hfr marker is complex, depending bothon transfer of the marker into the cell and on crossovers that transfer the marker into the recipientchromosome.To make more accurate maps, bacterial geneticistsoften do Hfr × F− crosses in a different way: Theyselect for exconjugants that contain a late Hfr marker,and then screen for the presence of the earlier markers.This method ensures that all of the markers haveentered the F− cell, so relative gene distances arenow accounted for solely by crossover frequencies.Furthermore, gene order is clarified by considering thecrossovers responsible for each class of exconjugants.As an example,…arrow_forwardThe oncogenic protein BETA promotes entry into the S phase of the cell cycle. Phosphorylation of BETA at the amino acid Tyr98 causes BETA to be degraded by the proteasome, thus limiting its abundance. A mutation in the codon encoding Tyr98 changes this residue to Cys, which cannot be phosphorylated. What is the best description of this mutant allele?a) antimorphb) hypermorphc) hypomorphd) amorphe) neomorpharrow_forwardConsidering Figure 2-13, if you had a homozygous double mutant m3/m3 m5/m5, would you expect it to be mutant in phenotype? (Note: This line would have two mutant sites in the same coding sequence.)arrow_forward
- In Arabidopsis, it is well-known that a pulse of full-spectrum light during the night (in an otherwise long night) will induce flowering. This suggests that plants measure the length of night, and not the length of day. If the pulse of light during the night was blue light instead of full spectrum light, what would be the flowering time response of a plant with a knockout in cry2 (relative to wild type in the same conditions)? Explain.arrow_forwardIn 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…arrow_forwardIn considering the formation of the A–P and D–V axes inDrosophila, we noted that, for mutations such as bcd,homozygous mutant mothers uniformly produce mutant offspring with segmentation defects. This outcomeis always true regardless of whether the offspring themselves are bcd+/bcd or bcd/bcd. Some other maternaleffect lethal mutations are different, in that the mutantphenotype can be “rescued” by introducing a wild-typeallele of the gene from the father. In other words, forsuch rescuable maternal-effect lethals, mut+/mut animals are normal, whereas mut/mut animals have the mutant defect. Explain the difference between rescuableand nonrescuable maternal-effect lethal mutations.arrow_forward
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