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 30P
Now consider a mating between consanguineous people involving a recessive genetic disease. The figures that follow show the genotypes of these two people at four SNP loci (1–4).
| a. | For which of these loci is it possible to obtain information about the linkage of the SNP to the disease gene? Explain your answer for each locus and describe any special conditions that may apply. |
| b. | For any of the loci for which the mating is potentially informative, how would you tell whether the child is the product of a recombinant or non recombinant gamete? (That is, how could you solve the phase problem?) Be as specific as possible. |
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The 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:12
Consider two maize plants:a. Genotype C/cm ; Ac/Ac+, where cm is an unstableallele caused by a Ds insertionb. Genotype C/cm, where cm is an unstable allele causedby Ac insertionWhat phenotypes would be produced and in whatproportions when (1) each plant is crossed with a basepair-substitution mutant c/c and (2) the plant in part a iscrossed with the plant in part b? Assume that Ac and care unlinked, that the chromosome-breakage frequencyis negligible, and that mutant c /C is Ac+.
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
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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- The DNA of every individual in the pedigree shown in image B (below) has been sequenced at the causative locus, all the non- shaded individuals are wild type apart from III.1 and III.6. III.1 and III.6 have both been proven to have the causative allele for the condition but they do not exhibit any of the phenotypic signs or symptoms. Based on this pedigree, what is the level of penetrance for the condition? Please give your answer as a percentage to one decimal place, give the number only, no percentage symbol. ANSWER: Given the information above I calculate the level of penetrance seen in image B to be Blank 1 percent. A KEY Homozygous Homozygous Heterozygous Heterozygous Wild Type Male Female Male Female Male Note: Completely red symbol denotes an individual exhibiting the phenotype of interest CI || III IV V 1/4 1/2 1/2 1/2 1/2 Wild Type Female 1/4 1/2 Affected Known carrier Affected female Normal female Affected male Normal male D ●●●arrow_forwardConsider an example of two autosomal recessive gene a and b where a heterozygote (AaBb) is testcrossed to a double homozygous mutant (AaBb) predict the phenotype ratio under the following conditions 1.a and b are located on separate autosomes 2.a and b are linked on the same chromosome but are far apart that a crossover always occurs 3.a and b are linked on the same chromosome but very close together so that a crossover never occurarrow_forwardstion 6 of 18 Suppose that a geneticist discovers a new mutation in Drosophila melanogaster that causes the flies to shake and quiver. She calls this mutation quiver, qu, and determines that it is due to an autosomal recessive gene. She wants to determine whether the gene encoding quiver is linked to the recessive gene for vestigial wings, vg. She crosses a fly homozygous for quiver and vestigial traits with a fly homozygous for the wild-type traits, and then uses the resulting F, females in a testcross. She obtains the flies from this testcross. Phenotype Number of flies vg* qu+ 230 vg qu 224 vg qut vg* qu 97 99 Test the hypothesis that the genes quiver and vestigial assort independently by calculating the chi-squared, X², for this hypothesis. Provide the X2 to one decimal place. X2 = Does the X value support the hypothesis that the quiver and vestigial genes assort independently? Why or why not? the partial table of critical values for X2 calculations to test this hypothesis.arrow_forward
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