ND STONY BROOK UNIVERSITY LOOSELEAF GENETICS: FROM GENES TO GENOMES
6th Edition
ISBN: 9781260406092
Author: HARTWELL, Leland, HOOD, Leroy, Goldberg, Michael
Publisher: Mcgraw-hill Education/stony Brook University
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Chapter 11, Problem 29P
One of the difficulties faced by human geneticists is that matings are not performed with a scientific goal in mind, so pedigrees may not always provide desired information. As an example, consider the following matings (W, X, Y, and Z):
| a. | Which of these matings are informative and which noninformative for testing the linkage between anonymous loci A and B? (A1 and A2 are different alleles of locus A, B1 and B2 are different alleles of locus B, etc.) Explain your answer for each mating. |
| b. | Is locus A more l b. ikely to be a SNP or an SSR? What about locus B? Explain. |
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A series of three-point testcrosses is made to determine the genetic map order of seven linked allele pairs: A/a, B/b, G/g, H/h, Q/q, R/r, and Y/y.From each cross between a triply heterozygous parent listed below, two recombinant classes were noticed as the least frequent among all 8 progeny classes, and are listed at the right in the table.
A. For each testcross write the genotype of the F1 heterozygous parent.
F1 Parental Phenotype Least frequent F2 Phenotype
1.AHB&ahb AHb & ahB
2.RYh&ryH RYH & ryh
3.BhY&bHy Bhy & bHY
4.qYB&Qyb qYb & QyB
5.AbQ&aBq Abq & aBQ
6.ghR&GHr ghr & GHR
B. Write the unified map order of these genes, showing your reasoning.
You are studying four linked genes located on chromosome 2 in the fruit fly Drosophila
melanogaster: adp (which contributes to obesity), b (which contributes to body color), pr (which
contributes to metabolism), and vg (which contributes to wing formation). A series of crosses
between pair-wise combinations of these mutations yielded the following recombination frequencies
between the indicated loci:
pr and adp 9%
adp and b 6%
pr and b 3%
vg and b 10%
vg and pr 7%
What is the genetic distance in map units (cM) between the adp and vg loci?
16
9
6
4
Following the analysis of a pedigree, a genetic link at 4cM is considered between a mutation leading to a pathology and the molecular marker HUMTH01. The
study counts 14 "parental" and 3 "recombinant" individuals. We call p(theta=0.04) is the probability of obtaining such a pedigree in case of a 4cM genetic
linkage, p(theta=0.5) is the probability of obtaining such a pedigree in case of independence between the mutation and the marker, Z(theta=0.04) the value of
the Lod-score under the assumption of 4cM genetic linkage. Tick all the correct answers:
p(theta=0,04)=1,79.10E-9 and Z(theta=0,04)=0,47
p(theta=0,5)=7,18.10E-12
and Z(theta=0,04)=0,77
and Z(theta=0,04)=0,67
p(theta=0,04)=2,75.10E-10
p(theta=0,5)=6,04.10E-10 and Z(theta=0,04)=0,47
p(theta=0,5)=9,36.10E-12 and Z(theta=0,04)=1,33
p(theta =0,5)=5,82.10E-11 and Z(theta=0,04)=0,67
no correct answer
p(theta=0,04)=4,31.10E-11
p(theta=0,04)=2,01.10E-10
and Z(theta=0,04)=0,77
and Z(theta=0,04)=1,33
Chapter 11 Solutions
ND STONY BROOK UNIVERSITY LOOSELEAF GENETICS: FROM GENES TO GENOMES
Ch. 11 - Choose the phrase from the right column that best...Ch. 11 - Would you characterize the pattern of inheritance...Ch. 11 - Would you be more likely to find single nucleotide...Ch. 11 - A recent estimate of the rate of base...Ch. 11 - If you examine Fig. 11.5 closely, you will note...Ch. 11 - Approximately 50 million SNPs have thus far been...Ch. 11 - Mutations at simple sequence repeat SSR loci occur...Ch. 11 - Humans and gorillas last shared a common ancestor...Ch. 11 - In 2015, an international team of scientists...Ch. 11 - Using PCR, you want to amplify an approximately 1...
Ch. 11 - Prob. 11PCh. 11 - The previous problem raises several interesting...Ch. 11 - You want to make a recombinant DNA in which a PCR...Ch. 11 - You sequence a PCR product amplified from a...Ch. 11 - Prob. 15PCh. 11 - The trinucleotide repeat region of the Huntington...Ch. 11 - Sperm samples were taken from two men just...Ch. 11 - Prob. 18PCh. 11 - a. It is possible to perform DNA fingerprinting...Ch. 11 - On July 17, 1918, Tsar Nicholas II; his wife the...Ch. 11 - The figure that follows shows DNA fingerprint...Ch. 11 - Microarrays were used to determine the genotypes...Ch. 11 - A partial sequence of the wild-type HbA allele is...Ch. 11 - a. In Fig. 11.17b, PCR is performed to amplify...Ch. 11 - The following figure shows a partial microarray...Ch. 11 - Scientists were surprised to discover recently...Ch. 11 - The microarray shown in Problem 25 analyzes...Ch. 11 - The figure that follows shows the pedigree of a...Ch. 11 - One of the difficulties faced by human geneticists...Ch. 11 - Now consider a mating between consanguineous...Ch. 11 - The pedigree shown in Fig. 11.22 was crucial to...Ch. 11 - You have identified a SNP marker that in one large...Ch. 11 - The pedigrees indicated here were obtained with...Ch. 11 - Approximately 3 of the population carries a mutant...Ch. 11 - The drug ivacaftor has recently been developed to...Ch. 11 - In the high-throughput DNA sequencing protocol...Ch. 11 - A researcher sequences the whole exome of a...Ch. 11 - As explained in the text, the cause of many...Ch. 11 - Figure 11.26 portrayed the analysis of Miller...Ch. 11 - A research paper published in the summer of 2012...Ch. 11 - Table 11.2 and Fig. 11.27 together portray the...Ch. 11 - The human RefSeq of the entire first exon of a...Ch. 11 - Mutations in the HPRT1 gene in humans result in at...Ch. 11 - Prob. 44P
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- A researcher crosses mice with brown eyes and long tails, and the F1 progeny were recovered in the following numbers and phenotypic classes: F1: 6 apricot, short : 30 brown, long : 15 brown, short : 9 apricot, long You know the genes encoding these traits are autosomal, completely dominant and assort independently. You want to use a chi-square test to analyse these results. a) Making use of the appropriate genetic convention for naming alleles, give the genotype of the male parent in this cross. b) What is your null hypothesis for the chi-square test? c) Give the expected number of individuals in the "brown, long" class. d) You obtain a value of 3.47 for the chi-square test. What conclusion can you make from the results of the chi-square test? P df 0.995 0.975 0.9 0.5 0.1 0.05* 0.025 0.01 0.005 1 0.000 0.000 0.016 0.455 2.706 3.841 5.024 6.635 7.879 2 0.010 0.051 0.211 1.386 4.605 5.991 7.378 9.210 10.597 0.072 0.216 0.584 2.366 6.251 7.815 9.348 11.345 12.838 4 0.207 0.484 1.064 3.357…arrow_forwardConsider 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 PCR 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 PCR 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_forwardWhile studying the genetics of the "Cute Faced Bat" face structure gene, three alleles ) are identified that produce 6 genotypes and 4 phenotypes “Cute”, “Super Cute Nose”, “Super Cute Face”, and “Super, Duper Cute” (equally expressing both Super Cute Face and Super Cute Nose). Given these observations, what can you conclude about the allele interactions?arrow_forward
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- You are mapping three linked loci in Drosophila melanogaster (the common laboratory fruit fly). You cross flies that are triply mutant for apricot (pale eyes), bristle (extra bristles) and clipped (notched wings) to wild-type flies. The F+ flies are wild-type in appearance. You then backcross the F+ females to pure-breeding (apricot, bristle, clipped) males and score the phenotypes of 1000 F progeny for all three loci. Here are the results: 359 wild-type 361 apricot, bristle, clipped 89 bristle, clipped 91 apricot 42 apricot, bristle 38 clipped 9 apricot, clipped 11 bristle Using these data, first determine what gametes from the F; trihybrid produced each of the eight F2 categories. Note that apricot = aa (recessive to wild-type A); bristle = bb (recessive to wild-type B); and clipped = cc (recessive to wild-type C). Then determine if each gamete is recombinant (R) or nonrecombinant (R) for each pair of alleles (that is, for each genetic interval). Complete the table by dragging the…arrow_forwardIn a three point cross problem for genes Q, L, and T, you have identified the parentals and double recombinants. Using that data (shown below), what is the order of the genes? Type: Alleles parental q L t parental Q l T double cross over q L T double cross over Q l t Group of answer choices L Q T L T Q T L Q the correct answer is not availablearrow_forwardA trihybrid individual with the genotype QqRrTt is testcrossed with a qgrtt individual. The resulting offspring are as follows: QrT 11 50 343 76 grT qRT grt QRt 76 357 78 9 QRt Qrt QRT Total - 1000 a) Determine the allele arrangement and gene order in the trihybrid parent. Give your answer using the AbC/aBc notation. b) Draw a chromosome map for these genes. Give your map distances to one decimal place. (Use underline or underscore () to draw the line for the map)arrow_forward
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