Download Chapter Fourteen ANSWERS TO REVIEW QUESTIONS All the

Chapter Fourteen
ANSWERS TO REVIEW QUESTIONS
1. All the alleles of all the genes in a population.
2. There are many more phenotypes and genotypes.
3. Microscopically at the protein level and macroscopically at the population level.
4. An allele frequency describes the proportion or percentage of a given allele in a population;
phenotypic frequency, that of a particular phenotype; genotypic frequency the proportion or
percentage of a particular genotype at a given locus.
5. Nonrandom mating, migration, genetic drift, mutation, and natural selection
6. Knowing the incidence of the homozygous recessive class makes it possible to derive the
"q" part of the Hardy-Weinberg equation.
7. For females use the standard formula. For males, gene frequency equals phenotypic
frequency.
8. VNTR sequences are larger and a wider range of alleles results.
9. A heterozygote has 2 peaks for a particular locus whereas a homozygote has one.
10. Population databases are necessary to interpret DNA fingerprints because alleles are
present in different frequencies in different populations.
11. Statistics on how common a particular CNV is in a particular population.
12. When chromosomal DNA is insufficient or too damaged for analysis.
13. Innocent people may share CODIS markers with convicted felons. Secondary transfer of
DNA at the crime scene could implicate the wrong person.
ANSWERS TO APPLIED QUESTIONS
1. A gene pool refers to a population.
2. The possibility of two unrelated Caucasians without a family history of CF having an affected
child is 1/4 x 1/23 x 1/23 = 1/2116. If one person knows he or she is a carrier, the risk is 1/4
x 1/23 = 1/92.
3. For females, allele frequencies are calculated using the standard Hardy-Weinberg formula.
For males, the allele frequency equals the phenotypic frequency.
4. Related males share Y chromosome markers.
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5. q 2 =.001 q = .0316 p = .968 2pq = carrier frequency = (2)(.0316)(.968) = .061
6. q 2 = 1/8000 = 0.000125
0.011 x 0.989 = 0.022
q = 0.011
p = 1 - 0.011 = 0.989 carrier frequency = 2pq = 2 x
7. F1 (students) genotypes: 6 TT 4 Tt 10 tt
P1 (parents) genotypes:
of 6 TT students = 9 TT 2 Tt 1 tt
of 4 Tt students = 2 TT 4 Tt 2 tt
of 10 tt students = 8 Tt 12 tt
Students’ allele frequencies: T = p = 1/2Tt + TT = 1/2 (4) + 6 = 8/20 students = .4
t = q = .6. Parents allele frequencies: T = p = 1/2Tt + TT = 1/2 (14) + 11 = 18/40 parents =
.45 t = q = .55
The gene is evolving because the allele frequencies change between generations.
8. 50% of the child’s STR markers must also be in the father.
ANSWERS TO WEB ACTIVITIES
1. Answers vary with case chosen
2. Opinion. Many people would be outraged at being compelled to provide a DNA sample
when accused of a crime.
3. The dog has dozens of STR alleles. They can be used to figure out which of a pack of dogs
attacked a person, using DNA from bite wounds.
ANSWERS TO FORENSICS FOCUS
1. The three chosen STRs should be the most variable in the most populations.
2. The Global DNA Response Team should set standards to use the same STRs, specify
conditions under which to use mitochondrial DNA markers, include sample collection
protocols, and consider the cultural practices of different population so as not to offend
anyone but maintain public health.
3. a. DNA profiles of the blood, hair, and skin cells taken from under the cat’s claw.
b. The police could collect and compare DNA profiles of relatives as potential suspects, but
this would include some innocent individuals. Possibly law enforcement could triage
whom they investigate based on previous crimes committed or other information to
supplement the DNA profile.
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without the prior written consent of McGraw-Hill Education.
4. More convictions and exonerations would be obtained at the risk of invading the privacy of
many people.
5.
Hair samples from Rufus, Rufus’s claws, a squirrel, and Killer. DNA profiles used to
distinguish among the individuals.
6. a. STRs 1 and 3 generate single peaks and 2 and 4 generate two.
b. (.2)2 x (2)(.3)(.7) x (.10)2 x (2)(.4)(.2) = .04 x .42 x .01 x .16 = .0000268
c. The man might have close male relatives who cannot be ruled out as suspects.
ANSWERS TO CASE STUDIES AND RESEARCH RESULTS
1. Autosomal recessive class = 900/10,000 = .09 = q 2 q = .3 p = .7
normal lashes = p 2 + 2pq = .49 + .42
homozygote with normal lashes = .49/.49 + .42 = .538
2. a. 50%
b. increase
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without the prior written consent of McGraw-Hill Education.