Genetic Analysis: An Integrated Approach (2nd Edition)
2nd Edition
ISBN: 9780321948908
Author: Mark F. Sanders, John L. Bowman
Publisher: PEARSON
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Chapter 10, Problem 16P
Suppose the mating couple (
Is it possible that their fetus could have sickle cell disease? If so, what is the probability? If not, explain whynot.
Fetal DNA is collected and analyzed by Southern blotting. The fetus has a single DNA band that is
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Susan’s grandfather was deaf, and passed down a hereditary form of deafness within Susan’s family as shown in Figure Q19–12.A. Is this mutation most likely to be dominant or recessive?B. Is it carried on an autosome or a sex chromosome? Why?C. A complete SNP analysis has been done for all of the 11 grandchildren (4 affected, and 7 unaffected). In comparing these 11 SNP results, how long a haplotype block would you expect to find around the critical gene? How might you detect it?
Using figure 1 and the following background information answer the following questions.
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 our previous studies, we…
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Chapter 10 Solutions
Genetic Analysis: An Integrated Approach (2nd Edition)
Ch. 10 - Define the following terms as described in this...Ch. 10 - 2. Using sickle cell disease as an example,...Ch. 10 -
3. Compare and contrast the contributions of...Ch. 10 - Why do differences in protein electrophoretic...Ch. 10 - Prob. 5PCh. 10 - Prob. 6PCh. 10 - Prob. 7PCh. 10 - 8. Wildtype βglobin protein is composed of amino...Ch. 10 - Prob. 9PCh. 10 - Prob. 10P
Ch. 10 - 11. How is an autoradiograph produced from a...Ch. 10 - Prob. 12PCh. 10 - Prob. 13PCh. 10 - Prob. 14PCh. 10 - The family represented in the pedigree and...Ch. 10 - Suppose the mating couple (I-1 and I-2) shown in...Ch. 10 - What are restriction endonucleases, and why are...Ch. 10 - 18. Following restriction digestion, DNA fragments...Ch. 10 - 19. The doublestranded DNA sequence below is part...Ch. 10 - 20. Restriction enzymes recognize specific...Ch. 10 - Prob. 21PCh. 10 - Prob. 22PCh. 10 - Prob. 23PCh. 10 - Prob. 24PCh. 10 - 25. A second strain of dwarf plants has a...Ch. 10 - During gel electrophoresis of linear DNA...Ch. 10 - Prob. 27PCh. 10 - 28. In molecular biology, restriction...Ch. 10 - A complete plant gene containing four introns and...Ch. 10 - Prob. 30PCh. 10 - The map below illustrates three alleles in a...Ch. 10 - Prob. 32PCh. 10 - 33. Northern blot analysis is performed on mRNA...
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- Question 7 Which of the following is inconsistent (not consistent) with inheritance of a maternal effect trait? A) Progeny from the same parents have the same maternal effect phenotype as each other but not their mother. B) Progeny can show a recessive phenotype even though the progeny have dominant genotypes. C) Both parents contribute recessive alleles to an offspring yet it has a dominant phenotype for the maternal effect trait. D) Progeny from the same parents have different traits related to the maternal effect gene in question. CS Scanned with CamScannerarrow_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_forwardFamilial retinoblastoma, a rare autosomal dominant defect, arose in a large family that had no prior history of the disease. Consider the following pedigree (the darkly colored symbols represent affected individuals): a. Circle the individual(s) in which the mutation most likely occurred. b. Is the person who is the source of the mutation affected by retinoblastoma? Justify your answer. c. Assuming that the mutant allele is fully penetrant, what is the chance that an affected individual will have an affected child?arrow_forward
- The Klinefelter syndrome (disomy of the X chromosome in males) is a genetic disease caused by aneuploidy of the sex chromosomes in humans. Patients with the 47,XXY karyotype are male. Describe two genetic scenarios leading to patients with the Klinefelter syndrome.arrow_forwardThe DNA of every individual in the pedigree shown below has been sequenced at the causative locus. All the non-shaded individuals are wild type apart from III.1. III.1 has been proven to have the causative mutation for this autosomal dominant condition, but they exhibit no symptoms. Based on this small pedigree, what is the level of penetrance for the condition? Please give your answer as a WHOLE percentage, give the number only, no percentage symbol. Answer: The level of penetrance for the condition shown in the pedigree below is Blank 1 percent. 1:1 1:2 Il:1 I1:2 I1:3 Il:4 I1:5 I1:6 II:1 I:2 III:3 III:4 III:3 III:6 III:7 III:8 III:9 III:10 III:11 III12 II:13 III:14 IV:1 | IV:2 IV:3 IV:4 IV:5 IV:6 IV:7 IV:8 IV:9 IV:10 IV:11 IV:12 IV:13 IV:14 IV:15 IV:16 IV:17 IV:18 IV:19 V:1 V:2 V:3 V:4 V:5 V:6 V:7 V:8 V:9 V:10 V:11 V:12arrow_forwardCampomelic dysplasia (CMD1) is a congenital humansyndrome featuring malformation of bone and cartilage.It is caused by an autosomal dominant mutation of agene located on chromosome 17. Consider the followingobservations in sequence, and in each case, draw whateverappropriate conclusions are warranted.(a) Of those with the syndrome who are karyotypically46,XY, approximately 75 percent are sex reversed,exhibiting a wide range of female characteristics.(b) The nonmutant form of the gene, called SOX9, isexpressed in the developing gonad of the XY male,but not the XX female.(c) The SOX9 gene shares 71 percent amino acid codingsequence homology with the Y-linked SRY gene.(d) CMD1 patients who exhibit a 46,XX karyotypedevelop as females, with no gonadal abnormalities.arrow_forward
- A diploid (ie, contains TWO sets of chromosomes) organisms with a 45,000-kb haploid (counts only one set of its chromosomes) genome contains 21% G residues. Calculate the number of A, C, G< and T residues in the DNA of each cell in this organism. Can you help explain why this is the answer, thank you! Answer: Since the haploid genome contains 21% G, it must contain 21% C (Because G=C) and 58% A + T, or 29% A and 29% T. Each cell is a diploid, containing 90,000 kb or 9x10^7 bases. Therefore, A=T = (0.29)(9x10^7) = 2.61 x 10^7 bases and G=C=(0.21)(9x10^7) = 1.89x10^7 bases.arrow_forwardQuestion is attachedarrow_forwardHuman adult hemoglobin is a tetramer containing two alpha (a) and two beta (b) polypeptide chains. The a gene cluster on chromosome 16 and the b gene cluster on chromosome 11 share amino acid similarities such that 61 of the amino acids of the a@globin polypeptide (141 amino acids long) are shared in identical sequence with the b@globin polypeptide (146 amino acids long). How might one explain the existence of two polypeptides with partially shared function and structure on two different chromosomes?arrow_forward
- The fruit fly Drosophila melanogaster has about 2 x 10^8 base pairs of DNA per haploid genome, of which about 75% in nonrepeated DNA. The DNA is distributed among four pairs of homologous chromosomes, which have a total of about 5,000 visible bands when in polytene forms in the salivary gland. The number of genes initially estimated from mutational studies was also about 5,000 but recent DNA sequencing studies suggest that the gene number may be somewhat higher. a. Why was it tempting to speculate that each band corresponds to a single gene? What does this suggest about the number of different proteins Drosophila can make? Does that seem like a reasonable number to you? b. Assuming all the nonrepeated DNA is uniformly distributed in the chromosomes, how much nonrepeated DNA (in base pairs) is there in average band?arrow_forwardSuppose a man carries a very uncommon dominant mutation for a deadly muscular disease that doesn't show up until people are usually over 50 years of age (he has the disease). He and his wife are expecting their first child--a girl. What is the probability she will also have the disease if... (1) it is autosomal(2) it is X-linked(3) it is Y-linked Explain each answer in detail. Assume the penetrance of the trait is 100%.arrow_forwardIn Drosophila, the genes ct(cut wing margin), y (yellow body), and v (vermilion eye color) are X-linked. Females heterozygous for all three markers were mated with wildtype males and the following male progeny were obtained. As is conventional in Drosophila genetics, the wild-type allele of each gene is designated by a “+” sign in the appropriate column. Use the data to (A) create a genetic map of the genes, (B) calculate interference, and (C) interpret the value of interference. Show your work.arrow_forward
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