Genetics: From Genes to Genomes
6th Edition
ISBN: 9781259700903
Author: Leland Hartwell Dr., Michael L. Goldberg Professor Dr., Janice Fischer, Leroy Hood Dr.
Publisher: McGraw-Hill Education
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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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Students have asked these similar questions
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.
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…
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
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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