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Concept explainers
Five
a. Use the data in Table A to place each point mutation as precisely as you can along the chromosome.
b. Use the complementation data in Table B to determine where the division between rIIA and rIIB is located on the rII region.
c. Based on the data and on your analysis, draw a complementation table for the five point mutants
d. Add mutant
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Chapter 6 Solutions
GENETIC ANALYSIS: AN INTEG. APP. W/MAS
- 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?arrow_forwardIn the Holliday model for homologous recombination shown, the resolution steps can produce recombinant or nonrecombinantchromosomes. Explain how this can occur.arrow_forwardA yeast geneticist irradiates haploid cells of a strain that is an adenine-requiring auxotrophic mutant, caused by mutation of the gene ade1. Millions of the irradiated cells 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 of results.b. Write the genotypes of the prototrophs in each case.c. What progeny phenotypes and ratios do you predict from crossing a prototroph of type 2 by the original ade1auxotroph?arrow_forward
- Compared to the normal A allele, the disease-causing allele in sickle cell anemia (S allele) is missing an MstII restriction site. On a Southern blot of genomic DNA cut with MstII and hybridized with the probe shown on the diagram below, a person with sickle anemia, carrying two S alleles, will show Choose an answer below: a single band at 1.1 kb. a single band at 1.3 kb. a single band at 0.2 kb. one band at 0.2 and one at 1.3 kb. one band at 1.1 and one at 1.3 kb.arrow_forwardConsider a maize plant: Genotype C/cm ; Ac/Ac+ where cm is an unstable colorless allele caused by Ds insertion. What phenotypic ratios would be produced and in what proportions when this plant is crossed with a mutant c/c Ac+/Ac+? Assume that the Ac and c loci are unlinked, that the chromosome-breakage frequency is negligible, and the C allele encodes pigment production.arrow_forwardWhat is a recombinant vector? How is a recombinant vector constructed? Explain how X-Gal is used in a method of identifying recombinant vectors that contain segments of chromosomal DNA.arrow_forward
- Consider two maize plants:a. Genotype C/c m ; Ac/Ac+, where cm is an unstable allele caused by a Ds insertionb. Genotype C/c m, where cm is an unstable allele caused by Ac insertionWhat phenotypes would be produced and in what proportions when (1) each plant is crossed with a basepair-substitution mutant c/c and (2) the plant in part a is crossed with the plant in part b? Assume that Ac and c are unlinked, that the chromosome-breakage frequency is negligible, and that mutant c /C is Ac+.arrow_forwardYou have two mutants of pQE. 1-CRYGD. One is 66 ng/uL and the other is 337 ng/uL. You need to add 1 ng to your competent NEB Express Iq cells. A. Can you accurately pipet the plasmid for each? Why or why not? B. How could you dilute the plasmid to obtain 50 uL of 0.5 ng/uL? C.How much of the new 0.5 ng/uL sample would you need to pipette for 1 ngarrow_forwardThe data below are from a DNAse-Seq experiment of chromosome 22. DNAse-seq is another method for measuring chromatin organization. In the experiment, regions of chromatin are digested with DNAse I and DNA is sequenced to determined to identify cleavage sites. See this short article for complete explanation. Which of the following statements are true based on the data below (select all that apply)? A. Region C appears to be an actively transcribed region in all tissue types except trophoblasts B. Region A could be an origin of replication for the chromosome C. Region B is highly associated with nucleosomes in BE2C and Natural KIller Cells D. The genes in the fetal kidney tissue are the least actively transcribed on chromosome 22. E. The data for each tissue must be from separate people to have such a variation in DNAse sensitivityarrow_forward
- Consider the first category of test-cross offspring shown in figure 8.2 (+b, LS). Consider also that the parents of the heterozygous female flies in the test cross had the following genotypes: bb, SS, and +, LL. A. What would be the physical phenotype of these flies? B. If PC was conducted with the DNA of one of these flies using the primers for the molecular marker, what would be the appearance of the bands on an electrophoresis gel with the PC products? C. If the gene for black body and the locus for the molecular marker (L long or S short) were unlinked, what proportion of the test-cross progeny would be black flies that are heterozygous for the molecular marker? What proportion would be flies with normal body color, which are homozygous for one form of the molecular marker? D. If the gene for black body and the locus for the molecular marker were linked, how would the proportion of flies be different?arrow_forwardAn aberrant corn plant gives the following RF values when testcrossed: (The locus order is centromere-d–f–b–x–y–p.) The aberrant plant is a healthy plant, but it produces far fewer normal ovules and pollen than does the control plant. a. Propose a hypothesis to account for the abnormal recombination values and the reduced fertility in the aberrant plant. b. Use diagrams to explain the origin of the recombinants according to your hypothesis.arrow_forwardA small-scale pedigree study involving 15 families with a total of 50 children was performed to determine whether a SNP in a candidate gene is linked to a disease. The result showed that among 22 children who developed the disease, 19 belonged to non-recombinants while 3 recombinants. For the remaining 28 children, 21 were non-recombinants and 7 belonged to recombinants. A Lod (Z) score can be calculated and used to determine whether this SNP is linked to the disease. By using a very stringent threshold, which of the following conclusions is correct? O A.Z is approximately 4.03, indicating a linkage O B. Z=0.25, indicating no linkage OCZ=0.2, indicating no linkage O D.Z is approximately 4.19, indicating a linkagearrow_forward
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