Concept explainers
A family consisting of a mother (
a. Following the illustration style of Figure
b. Identify all the possible genotypes of children of this couple by specifying PCR fragment lengths in each genotype.
c. What genetic term best describes the pattern of inheritance of this DNA marker? Explain your choice.
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Chapter 7 Solutions
Genetic Analysis: An Integrated Approach (3rd Edition)
- A blood stain from a crime scene and blood samples from four suspects were analyzed by PCR using fluorescent primers associated with three STR loci: D3S1358, vWA, and FGA. The resulting electrophoretograms are shown below. The numbers beneath each peak identify the allele (upper box) and the height of the peak in relative fluorescence units (lower box). Solve, (a) Since everyone has two copies of each chromosome and therefore, two alleles of each gene, what accounts for the appearance ofonly one allele at some loci? (b) Which suspect is a possible source of the blood? (c) Could the suspect be identifi ed using just one of the three STR loci? (d) What can you conclude about the amount of DNA obtained from Suspect 1 compared to Suspect 4?arrow_forwardIf you are comparing the two telomeres in each entryin the following list, in which cases would you expectthe two telomeres always to have exactly the samenumber of TTAGGG repeats?a. One telomere at one end of a chromosome, one telomere at one end of a nonhomologous chromosome.b. One telomere at one end of a chromosome, onetelomere at the corresponding end of the homologous chromosome.arrow_forwardIf the bandicoot genome is 3.62 x 109 base pairs, and the "highly repetitive DNA" fraction is composed entirely of copies of sequence 5'TGCGTGTGTGC3' and its complement, how many copies of this sequence are present in the bandicoot genome?arrow_forward
- You have identified a SNP marker that in one largefamily shows no recombination with the locus causinga rare hereditary autosomal dominant disease.Furthermore, you discover that all afflicted individuals in the family have a G base at this SNP on theirmutant chromosomes, while all wild-type chromosomes have a T base at this SNP. You would like tothink that you have discovered the disease locus andthe causative mutation but realize you need to consider other possibilities.a. What is another possible interpretation of the results?b. How would you go about obtaining additional genetic information that could support or eliminateyour hypothesis that the base-pair difference is responsible for the disease?arrow_forwardThe region of the normal hemoglobin gene used for genetic testing for sickle cell anemia contains a restriction site such that homozygous normal individuals show two DNA fragments. If a single nucleotide change in hemoglobin destroys that restriction site, then how many DNA fragments will be visible on a gel from individuals that are homozygous mutant? What about heterozygotes?arrow_forwardWith only one exception, the most commonly used STR loci are spread across the human genome,each on a separate chromosome.a. Why is it important to analyze STRs that are all on their own chromosomes?b. What is the exception to this? (i.e., which two loci are on the same human chromosomeand which chromosome are they on?)c. Using the UCSC Genome browser, look up the two loci that you answered in 2.b. one ata time. Include a rough hand-drawing or screenshot of the ideogram (pictograph,example below) of the chromosome and mark the approximate location of each locus.Example:d. Using your answer in 2.c. and your knowledge of the frequency of crossover events onhuman chromosomes (see chapters 5 and 6), in your opinion is the fact that these twoloci are on the same chromosome a problem for forensic genetic analysis? Explain youranswerarrow_forward
- A pair of paralogous repeats, A and B, have 96% sequence similarity and therefore can promote non-allelic homologous recombination (NAHR). They exist in four possible arrangements in a genome, illustrated below as arrangements 1 – 4. What is the result of NAHR between repeats A and B in arrangement 1? A.Translocation between chromosomes 1 and 2 resulting in monocentric chromosomes B.Deletion or duplication of the region between A and B C.Translocation between chromosomes 1 and 2 resulting in acentric and dicentric chromosomes D.Inversion of the region between A and Barrow_forwardIn Figure , what do the red and blue parts of the DNA labeled by balloon 6 represent?arrow_forwardWhat is the specific base sequence found in human telomeres, and how does the base sequence contained in the telomeric regions of chromosomes differ from that found elsewhere in the chromosome? (Please indicate both the base sequence and the 5’/3’ polarity.)arrow_forward
- Given the Percentage Composition of One Nucleotide ina Genome, Can We Predict the Percentages of the OtherThree Nucleotides?arrow_forwardBased on the attached image, if we are using the Holliday junction model of recombination, where exactly would be the positions where DNA is cut? Would it be to the right because of branch migration?arrow_forwardUsing the figure and the following background information answer the following questions about the figure Background: Identification of the genetic cause of hornlessness in cattle has been the subject of intensive genetic and genomic research, culminating in the nomination of two different candidate neomutations on cattle chromosome 1 that are predicted to have arisen 500-1,000 years ago: a complex allele of Friesian origin (PF), an 80,128 base pair (bp) duplication (1909352-1989480 bp), and a second, simple allele of Celtic origin (PC) corresponding to a duplication of 212 bp (chromosome 1 positions 1705834-1706045) in place of a 10-bp deletion (1706051-1706060)We report the use of genome editing using transcription activator-like effector nucleases (TALENs) to introgress the putative PC POLLED allele into the genome of bovine embryo fibroblasts to try and produce a genotype identical to what is achievable using natural mating, but without the attendant genetic drag and admixture. In…arrow_forward