Chapter Five ANSWERS TO REVIEW QUESTIONS 1. a. Lethal

Chapter Five
ANSWERS TO REVIEW QUESTIONS
1. a.
b.
c.
d.
e.
f.
g.
h.
i.
j.
Lethal alleles eliminate a progeny class that Mendel's laws predict should exist.
Multiple alleles create the possibility of more than two phenotypic classes.
Incomplete dominance introduces a third phenotype for a gene with two alleles.
Codominance introduces a third phenotype for a gene with two alleles.
Epistasis eliminates a progeny class when a gene masks another's expression.
Incomplete penetrance produces a phenotype that does not reveal the genotype.
Variable expressivity can make the same genotype appear to different degrees.
Pleiotropy can make the same genotype appear as more than one phenotype because
subsets of effects are expressed.
A phenocopy mimics inheritance, but is an environmental effect.
Conditions with the same symptoms but caused by different genes will not recur with the
frequency that they would if there was only one causative gene.
2. Dominant and recessive alleles are of the same gene. Epistasis is an interaction of alleles of
different genes.
3. It can skip generations in terms of phenotype.
4. The I A allele is codominant with the I B allele; both are completely dominant to i.
5. Genetic heterogeneity (mutations that parents carry are in different genes)
6. The porphyrias exhibit variable expressivity because symptoms vary in type and severity;
pleiotropy because one genotype causes more than one symptom; genetic heterogeneity
because different mutations in the porphyrin-heme pathway may lead to porphyria.
7. The epistatic gene modifies another gene’s phenotypic expression
8. Smoking
9. No introns, no crossing over, no repair, and not associated with histones.
10. Maternal inheritance describes transmission of mitochondrial genes, which sperm do not
usually contribute to oocytes and, therefore, these traits are always passed from mothers.
Linked genes are transmitted on the same chromosome. Mendel's second law applies to
genes transmitted on different chromosomes.
11. 3 parents: sperm donor, oocyte donor, mitochondria donor
12. Only females transmit maternally inherited traits. All of a woman's children inherit a
mitochondrial trait, but a male does not pass the trait to his children.
13. 24 linkage groups, 22 pairs of autosomes, and the X and the Y
14. Markers may be variable restriction sites, variable short repeated sequences, or single
nucleotide polymorphisms (SNPs). Linkage means genes that are inherited together as on
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without the prior written consent of McGraw-Hill Education.
the same chromosome. In popular usage, both “linkage” and “marker” are used vaguely to
refer to something about DNA that occurs with a trait or illness.
ANSWERS TO APPLIED QUESTIONS
1. a. D
b. C.
c. A
d. H
e. G
f. C
g. H
h. F
i. G
2. Haplotypes can reveal if people without symptoms have the haplotype associated with the
condition. These people are non-penetrant. If the genotype is lethal, individuals with the
haplotype containing the lethal allele should not exist.
3. a. 1/2
b. 1/2
c. 0
d. 0
4. Phenotype
5. It would be Mendelian because the gene that causes the condition when mutated is in the
nucleus.
6. 0.45 x 0.05 = .0225
7. The sperm are all of genotype hh se se. The oocytes are of genotype Hh Se se, with
the alleles in coupling:
H
Se
h
se
All sperm are h se
oocytes: H Se 49.5% h se 49.5h % H se 0.5% h Se 0.5%
Chance offspring like father = 1 (h se) x 49.5 (h se) = 49.5
8. b and d
9. Calculate the percentage of individuals with the allele who have hemochromatosis
10. a. Male: RrHhTt
Female: rrhhtt
b. Parental progeny classes: Round eyeballs, hairy tail, 9 toes; square eyeballs, smooth
tail, 11 toes. Recombinant progeny classes: Round eyeballs, hairy tail, 11 toes; square
eyeballs, smooth tail, 9 toes; round eyeballs, smooth tail, 11 toes; square eyeballs, hairy
tail, 9 toes
c. Crossover frequency between eyeball shape (R) and toe number (T): Round eyeballs,
11 toes = 6+4; square eyeballs, 9 toes = 6+4; crossover frequency = 20/100 = 20%
ANSWERS TO WEB ACTIVITIES
1. Answers vary with retinal diseases selected.
2. Answers vary based on disease selected. Examples are Ehlers-Danlos syndrome, FG
syndrome, Kabuki syndrome, and VATER association.
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without the prior written consent of McGraw-Hill Education.
3. Answers vary based on disease selected. An example is Kearn-Sayre syndrome, which
causes heart failure, anemia, hearing loss, pancreatic failure, and visual loss.
1. Answers vary based on genes selected. An example is chromosome 3: genes for sodium
channel, voltage-gated, type V, alpha and contactin 3 and chondroitin sulfate proteoglycan
5.
5. Answers vary. An example is Paget disease of bone (PDB).
6. a. drugs, alcohol, infections, dietary changes, hormonal changes
b. alcohol, estrogen
c. drugs
d. drugs
e. light
ANSWERS TO FORENSICS FOCUS
1. Mitochondrial DNA. MtDNA samples from family members.
ANSWERS TO CASE STUDIES AND RESEARCH RESULTS
1. a. i, ii, iii, iv
b. People with one copy have abnormalities so mild that they are not noticeable
without a test. People with two copies can be severely affected.
c. The parents are heterozygotes, so each of their offspring has a 1 in 4 probability
of inheriting two wild type alleles, like Tina.
d. Phenotype
2. 0
3. Genetic heterogeneity
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without the prior written consent of McGraw-Hill Education.