Chapter 4
Extensions and Modifications of Basic Principles
Multiple Choice
1. In humans, male gender is determined by:
a. the X chromosome.
b. the Y chromosome.
*c. the SRY gene on the Y chromosome.**
d. the SRY gene on the X chromosome.
e. the number of copies of the SRY gene on the Y chromosome.
2. In humans, extra copies of the sex chromosomes (either X or Y) can cause
which of the following?
a. underdeveloped sex organs
b. mental retardation
c. sterility
d. reduced facial hair
*e. any of the above
3. A guinea pig breeder is trying to select for unusual coat color variations that
might be especially appealing to clients. She has discovered that chocolate
brown guinea pigs are the most popular, so she has repeatedly crossed her
chocolate brown male with a female of the same phenotype. The litters are
always smaller than usual, and she gets a 2:1 ratio of brown to blonde offspring.
What might be happening?
a. The homozygote for the recessive allele is a lethal combination, so those
offspring do not survive.
*b. The homozygote for the dominant allele is a lethal combination, so those
offspring do not survive.
c. She is probably getting a 3:1 ratio and at least one of the three is the
homozygote for the brown allele.
d. There is incomplete dominance causing a lethal phenotype.
4. This same guinea pig breeder has female guinea pigs with the same coloration
as tortoiseshell cats. Assuming the same genetic mechanism is causing this
phenotype in both the cats the guinea pigs, what is the reason for this color
variation?
*a. The trait is X-linked. Because of dosage compensation and the random
inactivation of one of the X-chromosomes, each cell produces the pigment
determined by the active X-chromosome.
b. The trait causing uniform hair color is Y-linked. Because the animal is female,
there is no Y chromosome to act on the other genes.
Extensions and Modifications of Basic Principles
c. The trait is not sex-linked, but is recessive and epistatic, so it occurs more
frequently than a true Mendelian recessive phenotype.
d. The trait is hemizygous.
5. Drosophila spp. are commonly used for genetic research, as well as other
scientific research. These flies are an ideal organism with which to work
because:
a. they have a short generation time.
b. they produce a lot of offspring during their lifetime.
c. they are easy to maintain in the lab.
d. they have a small genome with large chromosomes.
* e. all of the above.
6. In many—but not all—organisms, sex is determined by the chromosomes that
the offspring carries (chromosomal sex determining systems). Typically, one
gender has two copies of the same chromosome, while the other has only one
sex chromosome or one copy each of two different sex chromosomes. The one
that produces two different gametes with regard to the sex chromosomes is
called:
a. the homogametic sex.
*b. the heterogametic sex.
c. the hemizygote.
d. the genic sex.
e. None of the above are accurate.
7. How do sex chromosomes pair and segregate during meiosis?
a. Sex chromosomes can pair because of the SRY gene.
b. Sex chromosomes don’t pair during meiosis; it is random separation.
*c. Sex chromosomes pair at pseudoautosomal regions.**
d. Sex chromosomes pair at autosomal regions.
8. What do genic sex determination and chromosomal sex determination have in
common?
a. They rely on environmental factors, as seen in reptiles.
*b. Sex is determined by genes.
c. The male and females have different phenotypes.
d. Sex is determined by the SRY gene.
9. Dosage compensation is an important biological phenomenon accomplished a
number of ways. The purpose of each of these mechanisms is to:
Chapter 4
a. lengthen the time that X-linked genes are producing protein.
b. reduce the amount of protein that X-linked genes produce.
c. make sure that Y-linked genes produce enough protein to balance the X-linked
genes.
*d. equalize the amount of protein produced by X-linked genes in the two sexes.
10. Which of these statements is true?
a. Any genetic characteristic with different phenotypes in males and females
must be sex linked.
b. All different phenotypes for a characteristic between males and females are
due to the interaction of sex hormones on autosomal genes.
c. Any characteristic observed more frequently in one gender over the other must
be sex linked.
*d. Traits observed more commonly in one sex than the other are likely to be sex
influenced.
11. John was diagnosed with a medical condition, and his doctor told him it was
hereditary. When John talked to his mother about it, she said her father had the
same condition, as did other males in her family. She didn’t know of anyone in
her husband’s family who had ever been diagnosed with this condition. Based on
this information, John reasoned that this condition was:
a. X-linked, dominant, and inherited from his paternal ancestors.
b. X-linked, dominant, and inherited from his maternal ancestors.
* c. X-linked, recessive, and inherited from his maternal ancestors.
d. X-linked, recessive, and inherited from his paternal ancestors.
e. Y-linked, recessive, and inherited from his maternal ancestors.
12. Which of these are not characteristics of dominance?
a. It is a result of interactions between genes and the same locus.
b. It is an allelic interaction.
c. It influences only how genes are expressed, not how they are inherited.
d. It varies depending on the scale on which the phenotype is being examined.
*e. It is the same no matter how we consider the phenotype.
13. Human polydactyly (having extra fingers and toes) is caused by a dominant
allele. But not every person with the genotype has the corresponding phenotype.
The percentage of people who do have the phenotype corresponding to the
genotype is referred to as:
*a. pentrance.
b. expressivity.
c. incomplete dominance.
d. hemizygote ratio.
e. none of the above.
Extensions and Modifications of Basic Principles
14. Why aren’t there more conditions that have dominant lethal alleles circulating
in the population?
a. There is no such thing as a dominant lethal allele; only recessive genes can
have lethal phenotypes.
b. Only for traits with multiple alleles can there be dominant lethal alleles.
*c. In order for dominant lethal conditions to be passed on, the afflicted person
has to reproduce before they die. Thus, only those conditions with onset later in
life can be passed to the next generation.
d. The time of onset for lethal dominant gene expression is unrelated to it being
passed to the next generation.
15. Genes can be known to interact with one another. When the effect of one
gene hides the effect of another, this is known as:
*a. epistasis.
b. codominance.
c. incomplete dominance.
d. gene interacton.
e.penetrance.
f. expressivity.
16. Red-green color blindness is X-linked recessive. A woman with normal color
vision has a father who is color blind. The woman has a child with a man with
normal color vision. Which phenotype is NOT expected?
*a. a color-blind female
b. a color-blind male
c. a noncolor-blind female
d. a noncolor-blind male
17. A Barr body is a(n):
a. active X chromosome.
*b. inactive X chromosome.
c. active Y chromosome.
d. inactive Y chromosome.
18. In Turner syndrome (45, X), how many Barr bodies would one expect to see
and what would be the sex of the specific individuals?
a. Zero and male, respectively
b. One and male, respectively
*c. None and female, respectively
d. One and female, respectively
e. Two and female, respectively
19. Klinefelter syndrome in humans, which leads to underdeveloped testes and
reduced fertility to sterility, is caused by which chromosomal condition? How
many Barr bodies will be present?
Chapter 4
a. 47, XYY; none
*b. 47, XXY; one
c. 48, XXXY; two
d. 47, XXX ; two
e. 45, X; none
20. In the experiment in which transgenic mice were used to identify the male
determining region of the Y chromosome:
a. an egg was fertilized by a sperm carrying only the Y chromosome.
b. restriction enzymes were used to detect the Sry region of the chromosome.
*c. DNA containing only the mouse Sry region was injected into normal XX
mouse eggs and most of the offspring develop into males.
d. X chromosomes were injected into an egg.
e. X and Y chromosomes were injected into the egg of a female mouse.
21. In a germ-line cell from a human male that is dividing, when do the X and Y
chromosomes segregate?
a. during mitosis
*b. during meiosis I, anaphase
c. during meiosis II, anaphase
d. They do not segregate; gametes contain a copy of X and a copy of Y.
22-23. Interactions among the human ABO blood group alleles involve _______
and ________.
*a. co-dominance
b. incomplete dominance
*c. complete dominance
d. epistasis
e. continuous variation
24. The individual from which this cell came is a:
*a. male.
b. female.
c. hermaphrodite.
d. monoecious.
25. A cell from this individual begins to go through meiosis. When the cell
reaches meiosis II, it becomes two cells. Which of the following is a possible
combination of chromosomes in one of the two cells when it goes through
metaphase of meiosis II?
a. one chromosome with A allele, one with B allele, and two X chromosomes
b. one chromosome with A allele, one with a allele, one with B allele, one with b
allele, and two X chromosomes
c. a pair of sister chromatids with A allele, a pair of sister chromatids with B
allele
d. a pair of sister chromatids with a allele, a pair of sister chromatids with B
allele, a pair of sister chromatids X
Extensions and Modifications of Basic Principles
*e. C and D are both possible.
26. What is the probability of a gamete from this individual that has the following
genotype: alleles A and b, chromosome X?
a. 1/2
b. 1/4
c. 1/6
*d. 1/8
27. Which of the following human genotypes is associated with Klinefelter
syndrome?
a. XXY
b. XXYY
c. XXXY
*d. all of the above
e. none of the above
28. A Barr body is a(n):
a. gene on the X chromosome that is responsible for female development.
b. patch of cells that has a phenotype different from surrounding cells because
of variable X inactivation.
*c. inactivated X chromosome, visible in the nucleus of a cell that is from a
female mammal.
d. extra X chromosome in a cell that is the result of nondisjunction.
e. extra Y chromosome in a cell that is the result of nondisjunction.
29. If a male bird that is heterozygous for a recessive Z-linked mutation is
crossed to a wild type female, what proportion of the progeny will be mutant
males?
*a. 0%
b. 100%
c. 75%
d. 50%
e. 25%
True/False
1. In humans, SRY is the male determining gene. (ANS: T)
2. Cytoplasmic inheritance is identical to the genetic maternal effect. (ANS: F)
3. Inactivation of the X-chromosomes is not performed on a random basis.
(ANS: F)
4. The condition XXXY is always lethal in humans. (ANS: F)
Chapter 4
5. The condition of having no X chromosomes, for example YO, is lethal in
humans. (ANS: T)
6. All human ABO blood group alleles are co-dominant. (ANS: F)
7. Epistatic genes must be dominant. (ANS: F)
8. A female with androgen insensitivity may have XY sex chromosomes rather
than XX. (ANS: T)
9. Genomic imprinting refers to the differential expression of genetic material
depending on the sex of the parent that it (the genetic material) originated from.
(ANS: T)
10. The X and Y chromosomes contain a similar number of genes and genetic
material. (ANS: F)
Fill in the Blank
1. The sex determination system used by Drosophila is called the ___________
system. ANS: genic balance
2. Male mammals inherit an X chromosome from ___________. ANS: the
maternal parent; the mother
3. An allele causing death at an early stage of development is called a
___________. ANS: lethal allele
4. Human females with XY chromosomes and a mutation in their
___________receptor gene have ___________syndrome. ANS: androgen;
androgen-insensitivity
5. Inheritance related to genes located in the cytoplasm is called ___________
inheritance. ANS: cytoplasmic
6. Human males with XY chromosomes are ___________ and produce two
different kinds of gametes, while females with XX chromosomes are
___________ and produce only one kind. ANS: heterogametic; homogametic
Short Answer Discussion
Extensions and Modifications of Basic Principles
1. List three dosage compensation strategies for equalizing the amount of sex
chromosome gene products.
ANS:
(1) Inactivation of one sex chromosome in the homogametic sex
(2) Halving the activity of genes on both sex chromosomes in the homogametic
sex
(3) Increasing the activity of genes on the sex chromosome in the heterogametic
sex
2. List six different sex determination systems and a representative organism for
each.
ANS:
(1) XX-XO
(2) XX-XY
(3) ZZ-ZW
(4) Haplodiploidy
(5) Genic balance
(6) Environmental
grasshoppers
humans
birds
bees
Drosophila
mollusks; many turtles, crocodiles, and alligators
3. Explain the differences between incomplete dominance and continuous
variation.
ANS:
(1) In incomplete dominance, the phenotype of a heterozygote is
intermediate in appearance between the phenotypes of the two
homozygotes. Incomplete dominance involves a single gene locus.
(2) Continuous variation refers to phenotypic variation exhibited by
quantitative traits that are overlapping and distributed from one extreme
to another. The continuous variation of quantitative traits is usually
controlled by several genes whose alleles have an additive effect on the
phenotype.
Use the following information for questions 4-5.
In humans, the presence of the SRY gene, normally on Y, determines maleness.
In Drosophila, an X: A (autosome) ratio of 0.5 determines maleness.
4. Explain the genders of human and Drosophila XXY individuals.
ANS: An XXY human is male. The SRY on the Y chromosome determines
maleness, in spite of the two X chromosomes. An XXY Drosophila is female.
Drosophila is diploid, so there are two of each autosome. If there are two X
chromosomes, the X:A ratio is 1.0, which is a female.
5. Explain the genders of human and Drosophila XO individuals.
Chapter 4
ANS: An XO human is female. In the absence of SRY from a Y chromosome,
the person will develop as a female, in spite of having only one X chromosome.
An XO Drosophila is male. Drosophila is diploid, so there are two of each
autosome. If there is a single X chromosome, the X:A ratio is ½, or 0.5, which is
a male.
6. Discuss the difference between “cytoplasmic inheritance” and “genetic
maternal effect.”
ANS:
(1) In cytoplasmic inheritance, the genes controlling a given trait are inherited
exclusively from the mother (through cytoplasmic organelles such as
mitochondria) and can be expressed in both male and female progeny.
(2) In the genetic maternal effect, the factors controlling a given trait are also
inherited exclusively from the mother and likewise can be expressed in both
male and female progeny, but these factors are not genes. Rather, the
offspring’s phenotype is determined by mRNA or protein factors in the oocyte.
So while genes related to the trait are inherited from both parents (not so for
the cytoplasmic inheritance), in a given generation, phenotype is determined
exclusively by the mother’s, not the offspring’s, genotype.
7. How does epistasis differ from Mendel’s principle of dominance?
ANS:
Phenotypic expression is often the result of products produced by multi-step
metabolic pathways involving several different genes; each gene encodes an
enzyme that regulates a specific biochemical step or event. Epistasis refers to
the interaction among two or more genes that control a common pathway. For
example, a mutation in any single gene contributing to a metabolic pathway can
affect the expression of other genes in the pathway, and, of course, the final
phenotype, depending on which biochemical step that gene controls.
Epistasis thus involves interaction among alleles located at different gene loci
(i.e., inter-locus, allele interaction). This is in contrast to dominance, which
involves interaction between alleles located at the same gene locus (i.e., intralocus, allele interaction).
Use the following information for questions 8–9.
The following diagram shows two pairs of autosomes and a pair of sex
chromosomes from a dioecious mammal.
Extensions and Modifications of Basic Principles
-a
B --
-b
A
8. Draw the chromosomes as they would appear in separate cells at the end of
meiosis I, after a crossover between locus A and the centromere. Indicate alleles
A, a, B, and b.
ANS:
A
After meiosis I
(Note: Independent
assortment could yield
other combination of
parental chromosomes.)
B
a
B
A
ab
b
9. Which of Mendel’s principles is demonstrated by alleles B and b in your
drawing? (Be more specific than “first principle” or “second principle.”)
ANS: Principle of segregation. (Alleles of a gene separate in meiosis. Or,
homologous chromosomes segregate in meiosis.)
10. Some organisms have more multiple X and Y chromosome and even
different numbers of X and Y chromosomes. You have discovered such a
species. Females have 8 X chromosomes while males have 4 X and 2 Y.
Describe the X and Y constitution of the gametes produced by this species—both
male and female—that allows these chromosome numbers to be stably
maintained.
ANS: Females must produce one type of gamete, with 4 X chromosomes, while
males produce two types of gametes, with 4 X chromosomes or with 2 Y
chromosomes. Fertilization using a 4 X male gamete produces an 8 X female,
while fertilization with a 2 Y gamete produces a male.
Use the following information for questions 11–12.
A man and a woman are trying to have children but are unsuccessful. As part of
a series of tests, the man is karyotyped. His autosomes appear normal, but his
sex chromosomes, shown in the following diagram, are not. The diagram also
shows a normal male’s sex chromosomes for reference.
Chapter 4
man's sex chromosomes
normal X and Y
11. In two to three sentences, explain the man’s situation, including the type of
chromosome mutation he carries; the specific regions of specific chromosomes
involved; and why he is male.
ANS: He has an X with a translocation, meaning part of one chromosome has
been moved to another. The translocated part is from Y and carries SRY, which
determines maleness. He is missing the rest of Y, including the genes required
for male fertility.
12. Can you tell if the mutation came from the man's mother or the man's father?
Explain how you can tell.
ANS: He inherited the translocation chromosome from his father because his
mother could not have carried the SRY-containing chromosome.
Use the following information for questions 13–16.
Red-green color blindness is X-linked recessive. A woman with normal color
vision has a father who is red-green color blind. The woman has four sons, none
of whom are color blind. In this family there are no instances of chromosome
loss or gain such as occurs due to nondisjunction in meiosis. Each of the next
three questions has an explanation for why none of the sons is color blind. For
each, state if color blindness is possible or not possible, then give the reason for
your choice.
13. Explanation 1: None of the sons is color blind because…the mother does not
carry the color-blindness allele.
ANS: Not possible. If there are no aneuploidies, the mother must have inherited
the X with the recessive color-blindness allele from her father.
14. Explanation 2: None of the sons is color blind because…none of them
inherited the color-blindness allele from the mother.
Extensions and Modifications of Basic Principles
ANS: Possible. She is heterozygous, so each son has a 50% chance of
inheriting the X chromosome with the dominant allele for normal color vision.
15. Explanation 3: None of the sons is color blind because…the mother
inactivated the X chromosome with the recessive color-blindness allele, and that
is the one each son inherited.
ANS: Not possible. X inactivation is reversed when an X chromosome is passed
to offspring.
16. You are trying to develop a new species of newt as an experimental model
system. You know that in other species of newt, green (G) is dominant to brown
(g) skin color and is determined by a sex-linked gene. You cross brown males to
green females and see that in the F1 all the males are green and all the females
are brown. Which is the heterogametic sex in your species of newt?
ANS: Because the F1 females have the recessive brown phenotype they must be
hemizygous (i.e., they inherited a brown allele from their father and no allele from
their mother). Therefore, the females of this species are the heterogametic sex.
We use the ZZ-ZW nomenclature for species with heterogametic females.
Therefore, your F1 females and males are ZgW and ZGZg, respectively.
17. While doing summer field work on a remote Indonesian island, you discover
a new genus of lizard closely related to Komodo dragons. You attempt to
discover what sex determination system it uses by performing a series of
controlled crosses on the island, using an isolated pair of lizards. Initially, all your
crosses yield only males (in significant numbers). As fall begins and you prepare
to leave the island, you find that your last cross yielded only females (in
significant numbers). Suggest a mode of sex determination that explains this
data.
ANS: The crosses yielded all males or all females from the same parents. Male
and female progeny were correlated with climatic conditions (summer versus
fall). Environmental sex determination that is dependent on temperature is a
likely explanation.
Use the following information for questions 18–19.
Red-green colorblindness is an X-linked recessive condition. Juliet has a bit of
difficulty passing the red-green color distinction test when she tries to get her
driver's license. She has two children with a man who is not color blind. Neither
Juliet's son (Henry) nor daughter (Roxanne) is color blind. Henry and Roxanne
have normal karyotypes. Roxanne has a son who is color blind.
18. What is Juliet's genotype for the color-blindness allele? How would you
explain her partial color blindness?
Chapter 4
ANS: Juliet is heterozygous, so every diploid cell in her eyes contains both X
chromosomes, one with the defective red-green color vision allele, and one with
the normal color vision allele. She is not completely color blind because some of
her cells are inactivating the X chromosome with the recessive, defective redgreen color vision allele and expressing the normal color vision allele, allowing
her to see color. She has some difficulty with color vision because some of her
cells are inactivating the X with the normal red-green color vision allele and
expressing the red-green color-blind phenotype. Her eyes are actually a mosaic
of cells with two different phenotypes.
19. Why aren't Henry and Roxanne color blind?
ANS: Henry is not color blind because he inherited the X with the normal redgreen color vision allele from Juliet. Roxanne is not color blind because she is
heterozygous like her mother, and does not have appreciable mosaicism for Xinactivation in her eye tissues.
20. List some examples of environmental factors affecting gene expression.
ANS:
(1) Temperature: Temperature-sensitive alleles like the Himalayan allele which is
functional only at certain temperatures.
(2) Specific foods or drugs: Glucose-6-phosphate dehydrogenase enzyme’s
function.
(3) Chemicals: The autosomal recessive mutation eyeless which will produce the
eyeless phenotype after exposing the larvae of normal flies to sodium
metaborate.
21. You cross a female rat with pink toe pads (T) and pointy ears (Xe) to a male
rat with black toe pads (t) and round ears (XE). The t and e alleles are both
recessive, and the ear-shaped gene is X-linked while the toe pad color gene is
autosomal. The F1 progeny all have pink toe pads. What is the genotype of
parental generation? What is the genotype of the F1 progeny? If the F1 are
crossed to produce F2 progeny, what proportion of the F2 will be black-padded,
pointy-eared males?
ANS: The parental generation was T/T; Xe/Xe and t/t; XE/Y. The F1 were T/t;
XEXe and T/t;XeY. Black-padded, pointy-eared males are tt and Xe/Xe. ¼ of the F2
progeny will be t/t, ¼ will be XeY. Therefore, 1/16 of the progeny will be blackpadded, pointy-eared males.
Extended Answer Discussion
Extensions and Modifications of Basic Principles
1. Listed below are blood types for several children and their mothers. Give all
possible genotypes and blood types for the father of each child.
Child’s blood type
a. A
b. O
c. AB
d. B
Mother’s blood type
A
B
A
AB
ANS:
a. Child is either iAiA or iOiA. If iAiA, then the mother contributed an iA allele
and the father contributed an iA allele; the father could be type A (iAiA or
iOiA) or type AB (iAiB). If the child is iAiO, then the mother contributed an iA
allele or an iO allele. In the former case the father would have contributed
an iO allele, and thus can be either type A (iAiO) or type B (iBiO) or type O
(iOiO); in the latter case, the father would have contributed an iA allele, and
thus can be either type A (iAiA or iOiA) or type AB (iAiB).
b. Child must be iOiO (because the iO allele is recessive to both iA and iB
alleles) and receive the iO allele from each parent. Therefore, the mother
must be iO iB. The father, who contributed an iO allele, could be either type
A (iOiA), type B (iOiB), or type O (iOiO).
c. Child must be iAiB. The mother contributed the iA allele. The father, who
contributed an iB allele, may be either type AB (iAiB) or type B (iOiB or iBiB).
d. Child is either iBiB or iOiB. The mother did not contribute the iA allele so she
must have contributed an iB allele. The father could have contributed
either an iB or an iO allele. In the former case he may be either type B (iBiB
or iOiB) or type AB (iAiB); in the latter case he may be either type A (iAiO),
type B (iBiO), or type O (iOiO).
Matching
Match numbers with the most appropriate letter choice
1. Pleiotropism (ANS: g)
2. Complementation test (ANS:
h)
3. Phenocopy (ANS: a)
4. Incomplete dominance (ANS:
e)
5. Cytoplasmic inheritance
(ANS: d)
6. Conditional mutation (ANS: b)
7. Sex-influenced trait (ANS: i)
8. Genetic maternal effect (ANS:
j)
9. Co-dominant (ANS: f)
a. Environmentally produced phenotype
b. Temperature-sensitive
c. Polygenes and environment
d. Non-nuclear
e. Intermediate phenotype
f. Discrete and equal
g. Unrelated effects
h. Homozygous mutations
i. Variable expressivity
j. Maternal genotype
Chapter 4
10. Multi-factorial trait (ANS: c)
Problems and Calculations
1. You are studying a coat color gene (B, brown) in Mexican bats. You have
isolated a recessive allele (b) that causes yellow coat color, but you suspect that
the phenotype may be sensitive to environmental conditions. To test the norm of
reaction for temperature, you examine the segregation ratio of phenotypes in F1
progeny from a cross between two heterozygotes. You do this once at normal
laboratory temperatures (28C) and once at temperatures closer to their native
habitat (34C) and record the following data:
Brown
28C 153
34C 170
Yellow
47
30
a. What ratio do you expect in each experiment?
ANS: 3:1
b. What test can you use to determine if the ratio you observed is significantly
different from the expected ratio?
ANS: The chi-square test
c. Using that statistical test, is either ratio more different that the expected ratio
than one would expect from chance alone? If so, suggest a biological
explanation.
ANS: The chi-square test for the treatment at 34C yields a value of 2 = 10.67,
indicating a significant difference from the expected ratio of 3:1. This suggests
that elevated temperatures affect the penetrance of the yellow phenotype.