Transcript Ch 11 Clicker Questions

CAMPBELL BIOLOGY IN FOCUS
URRY • CAIN • WASSERMAN • MINORSKY • REECE
11
Mendel and
the Gene Idea
Questions prepared by
Douglas Darnowski, Indiana University Southeast
James Langeland, Kalamazoo College
Murty S. Kambhampati, Southern University at New Orleans
Roberta Batorsky, Temple University
© 2016 Pearson Education, Inc.
SECOND EDITION
Imagine crossing a pea heterozygous at the loci for flower
color (Pp) and seed color (Yy) with a second pea
homozygous for flower color (pp) and seed color (yy).
What genotypes of gametes will the first pea produce?
A.
B.
C.
D.
two gamete types: pp and Pp
two gamete types: pY and Py
four gamete types: pY, py, PY, and Py
four gamete types: Pp, Yy, pp, PP
© 2016 Pearson Education, Inc.
Imagine crossing a pea heterozygous at the loci for flower
color (Pp) and seed color (Yy) with a second pea
homozygous for flower color (pp) and seed color (yy).
What genotypes of gametes will the first pea produce?
A.
B.
C.
D.
two gamete types: pp and Pp
two gamete types: pY and Py
four gamete types: pY, py, PY, and Py
four gamete types: Pp, Yy, pp, PP
© 2016 Pearson Education, Inc.
Pea plants were particularly well suited for use in
Mendel’s breeding experiments for all of the following
reasons except that
A. peas show easily observed variations in a number of
characters, such as pea shape and flower color.
B. it is possible to control matings between different pea
plants.
C. it is possible to obtain large numbers of progeny from
any given cross.
D. peas have an unusually long generation time.
E. many of the observable characters that vary in pea
plants are controlled by single genes.
© 2016 Pearson Education, Inc.
Pea plants were particularly well suited for use in
Mendel’s breeding experiments for all of the following
reasons except that
A. peas show easily observed variations in a number of
characters, such as pea shape and flower color.
B. it is possible to control matings between different pea
plants.
C. it is possible to obtain large numbers of progeny from
any given cross.
D. peas have an unusually long generation time.
E. many of the observable characters that vary in pea
plants are controlled by single genes.
© 2016 Pearson Education, Inc.
A cross between homozygous purple-flowered and
homozygous white-flowered pea plants results in
offspring with purple flowers. This demonstrates
A.
B.
C.
D.
E.
the blending model of genetics.
true breeding.
dominance.
a dihybrid cross.
the mistakes made by Mendel.
© 2016 Pearson Education, Inc.
A cross between homozygous purple-flowered and
homozygous white-flowered pea plants results in
offspring with purple flowers. This demonstrates
A.
B.
C.
D.
E.
the blending model of genetics.
true breeding.
dominance.
a dihybrid cross.
the mistakes made by Mendel.
© 2016 Pearson Education, Inc.
Imagine a genetic counselor working with a couple who
have just had a child who is suffering from Tay-Sachs
disease. Neither parent has Tay-Sachs, nor does anyone
in their families. Which of the following statements
should this counselor make to this couple?
A.
B.
C.
D.
E.
“Because no one in either of your families has Tay-Sachs, you are not
likely to have another baby with Tay-Sachs. You can safely have another
child.”
“Because you have had one child with Tay-Sachs, you must each carry the
allele. Any child you have has a 50% chance of having the disease.”
“Because you have had one child with Tay-Sachs, you must each carry the
allele. Any child you have has a 25% chance of having the disease.”
“Because you have had one child with Tay-Sachs, you must both carry
the allele. However, since the chance of having an affected child is 25%,
you may safely have three more children without worrying about having
another child with Tay-Sachs.”
“You must both be tested to see who is a carrier of the Tay-Sachs allele.”
© 2016 Pearson Education, Inc.
Imagine a genetic counselor working with a couple who
have just had a child who is suffering from Tay-Sachs
disease. Neither parent has Tay-Sachs, nor does anyone
in their families. Which of the following statements
should this counselor make to this couple?
A.
B.
C.
D.
E.
“Because no one in either of your families has Tay-Sachs, you are not
likely to have another baby with Tay-Sachs. You can safely have another
child.”
“Because you have had one child with Tay-Sachs, you must each carry the
allele. Any child you have has a 50% chance of having the disease.”
“Because you have had one child with Tay-Sachs, you must each
carry the allele. Any child you have has a 25% chance of having the
disease.”
“Because you have had one child with Tay-Sachs, you must both carry
the allele. However, since the chance of having an affected child is 25%,
you may safely have three more children without worrying about having
another child with Tay-Sachs.”
“You must both be tested to see who is a carrier of the Tay-Sachs allele.”
© 2016 Pearson Education, Inc.
Albinism in humans occurs when both alleles at a locus
produce defective enzymes in the biochemical pathway
leading to melanin. Given that heterozygotes are
normally pigmented, which of the following statements
is correct?
A. One normal allele produces as much melanin as two
normal alleles.
B. Each defective allele produces a little bit of melanin.
C. Two normal alleles are needed for normal melanin
production.
D. The two alleles are codominant.
E. The amount of sunlight will not affect skin color of
heterozygotes.
© 2016 Pearson Education, Inc.
Albinism in humans occurs when both alleles at a locus
produce defective enzymes in the biochemical pathway
leading to melanin. Given that heterozygotes are
normally pigmented, which of the following statements
is correct?
A. One normal allele produces as much melanin as
two normal alleles.
B. Each defective allele produces a little bit of melanin.
C. Two normal alleles are needed for normal melanin
production.
D. The two alleles are codominant.
E. The amount of sunlight will not affect skin color of
heterozygotes.
© 2016 Pearson Education, Inc.
In humans, alleles for dark hair are genetically dominant,
while alleles for light hair are recessive. Dark hair is also
more prevalent in southern Europe than in northern
Europe. Which of the following statements about hair
color alleles is most likely to be correct?
A. Dark hair alleles are more common than light hair alleles in
all areas of Europe.
B. Dark hair alleles are more common than light hair alleles
in southern Europe but not in northern Europe.
C. Dark hair alleles are equally common in all parts of Europe.
D. Dark hair is dominant to light hair in southern Europe but
recessive to light hair in northern Europe.
E. Dark hair is dominant to light hair in northern Europe but
recessive to light hair in southern Europe.
© 2016 Pearson Education, Inc.
In humans, alleles for dark hair are genetically dominant,
while alleles for light hair are recessive. Dark hair is also
more prevalent in southern Europe than in northern
Europe. Which of the following statements about hair
color alleles is most likely to be correct?
A. Dark hair alleles are more common than light hair alleles in
all areas of Europe.
B. Dark hair alleles are more common than light hair
alleles in southern Europe but not in northern Europe.
C. Dark hair alleles are equally common in all parts of Europe.
D. Dark hair is dominant to light hair in southern Europe but
recessive to light hair in northern Europe.
E. Dark hair is dominant to light hair in northern Europe but
recessive to light hair in southern Europe.
© 2016 Pearson Education, Inc.
Imagine a locus with four different alleles for fur color in
an animal, Da, Db, Dc, and Dd. If you crossed two
heterozygotes, DaDb and DcDd, what genotype proportions
would you expect in the offspring?
A.
B.
C.
D.
E.
25% DaDc, 25% DaDd, 25% DbDc, 25% DbDd
50% DaDb, 50% DcDd
25% DaDa, 25% DbDb, 25% DcDc, 25% DdDdDcDd
50% DaDc, 50% DbDd
25% DaDb, 25% DcDd, 25% DcDc, 25% DdDd
© 2016 Pearson Education, Inc.
Imagine a locus with four different alleles for fur color in
an animal, Da, Db, Dc, and Dd. If you crossed two
heterozygotes, DaDb and DcDd, what genotype proportions
would you expect in the offspring?
A.
B.
C.
D.
E.
25% DaDc, 25% DaDd, 25% DbDc, 25% DbDd
50% DaDb, 50% DcDd
25% DaDa, 25% DbDb, 25% DcDc, 25% DdDdDcDd
50% DaDc, 50% DbDd
25% DaDb, 25% DcDd, 25% DcDc, 25% DdDd
© 2016 Pearson Education, Inc.
Envision a family in which a man, age 47, has just been
diagnosed with Huntington’s disease, which is caused by
a dominant allele (and the man is a heterozygote). His
daughter, age 25, has a 2-year-old son. No one else in the
family has the disease. What is the probability that the
daughter will contract the disease?
A.
B.
C.
D.
E.
0%
25%
50%
75%
100%
© 2016 Pearson Education, Inc.
Envision a family in which a man, age 47, has just been
diagnosed with Huntington’s disease, which is caused by
a dominant allele (and the man is a heterozygote). His
daughter, age 25, has a 2-year-old son. No one else in the
family has the disease. What is the probability that the
daughter will contract the disease?
A.
B.
C.
D.
E.
0%
25%
50%
75%
100%
© 2016 Pearson Education, Inc.
Review the family described in the previous question.
What is the probability that the 2-year-old son will
contract the disease?
A.
B.
C.
D.
E.
0%
25%
50%
75%
100%
© 2016 Pearson Education, Inc.
Review the family described in the previous question.
What is the probability that the 2-year-old son will
contract the disease?
A.
B.
C.
D.
E.
0%
25%
50%
75%
100%
© 2016 Pearson Education, Inc.
Imagine that you are the daughter in the family described
in the previous questions. You had been planning on
having a second child. What kind of choices would you
make about genetic testing, for yourself and for your
child?
© 2016 Pearson Education, Inc.
When a disease is said to have a multifactorial basis, it
means that
A. both genetic and environmental factors contribute to
the disease.
B. it is caused by a gene with a large number of alleles.
C. it affects a large number of people.
D. it has many different symptoms.
E. it tends to skip a generation.
© 2016 Pearson Education, Inc.
When a disease is said to have a multifactorial basis, it
means that
A. both genetic and environmental factors contribute
to the disease.
B. it is caused by a gene with a large number of alleles.
C. it affects a large number of people.
D. it has many different symptoms.
E. it tends to skip a generation.
© 2016 Pearson Education, Inc.
Which of the following observations supported Mendel’s
hypothesis that inheritance is “particulate” rather than
due to blending?
A. There are two distinct flower colors in pea plants.
B. White-flowered plants are true-breeding.
C. Crossing true-breeding purple-flowered and whiteflowered plants produced all purple-flowered plants.
D. Crossing two purple-flowered heterozygotes produced
purple-flowered and white-flowered plants.
© 2016 Pearson Education, Inc.
Which of the following observations supported Mendel’s
hypothesis that inheritance is “particulate” rather than
due to blending?
A. There are two distinct flower colors in pea plants.
B. White-flowered plants are true-breeding.
C. Crossing true-breeding purple-flowered and whiteflowered plants produced all purple-flowered plants.
D. Crossing two purple-flowered heterozygotes
produced purple-flowered and white-flowered
plants.
© 2016 Pearson Education, Inc.
Manx cats have characteristic stubby tails due to being
heterozygous at a single locus. Homozygotes for the Manx
allele die before birth with severe spinal deformities.
What is the expected phenotypic ratio of live offspring of
two Manx cats?
A.
B.
C.
D.
3 normal:1 Manx
3 Manx:1 normal
2 normal:1 Manx
2 Manx:1 normal
© 2016 Pearson Education, Inc.
Manx cats have characteristic stubby tails due to being
heterozygous at a single locus. Homozygotes for the Manx
allele die before birth with severe spinal deformities.
What is the expected phenotypic ratio of live offspring of
two Manx cats?
A.
B.
C.
D.
3 normal:1 Manx
3 Manx:1 normal
2 normal:1 Manx
2 Manx:1 normal
© 2016 Pearson Education, Inc.
The allele for inflated pods (I) is dominant to that for
constricted pods (i), and the allele for green pods (G) is
dominant to that for yellow pods (g).
A plant of unknown genotype with inflated green pods is
crossed with a plant with constricted yellow pods. Among
the offspring are 49 plants with inflated green pods, and
53 plants with constricted green pods. What was the
previously unknown genotype?
A.
B.
C.
D.
IiGg
IiGG
IIGg
IIGG
© 2016 Pearson Education, Inc.
The allele for inflated pods (I) is dominant to that for
constricted pods (i), and the allele for green pods (G) is
dominant to that for yellow pods (g).
A plant of unknown genotype with inflated green pods is
crossed with a plant with constricted yellow pods. Among
the offspring are 49 plants with inflated green pods, and
53 plants with constricted green pods. What was the
previously unknown genotype?
A.
B.
C.
D.
IiGg
IiGG
IIGg
IIGG
© 2016 Pearson Education, Inc.
Imagine a cross of two triple heterozygous pea plants
with purple flowers and yellow round seeds (genotype
PpYyRr). If you were to create a Punnett square for this
cross (not recommended!) what would be its dimensions?
Recall that all three loci assort independently.
A.
B.
C.
D.
3×3
4×4
6×6
8×8
© 2016 Pearson Education, Inc.
Imagine a cross of two triple heterozygous pea plants
with purple flowers and yellow round seeds (genotype
PpYyRr). If you were to create a Punnett square for this
cross (not recommended!) what would be its dimensions?
Recall that all three loci assort independently.
A.
B.
C.
D.
3×3
4×4
6×6
8×8
© 2016 Pearson Education, Inc.
Imagine the same cross of two triple heterozygous pea
plants with purple flowers and yellow round seeds
(genotype PpYyRr). Using the rules of probability (and
not a Punnett square!), determine what proportion of
offspring will have purple flowers and green wrinkled
seeds.
A.
B.
C.
D.
¼ × ¼ × ¼ = 1/64
¾ + ¼ + ¼ = 5/4
¾ × ¼ × ¼ = 3/64
½ × ½ × ½ = 1/8
© 2016 Pearson Education, Inc.
Imagine the same cross of two triple heterozygous pea
plants with purple flowers and yellow round seeds
(genotype PpYyRr). Using the rules of probability (and
not a Punnett square!), determine what proportion of
offspring will have purple flowers and green wrinkled
seeds.
A.
B.
C.
D.
¼ × ¼ × ¼ = 1/64
¾ + ¼ + ¼ = 5/4
¾ × ¼ × ¼ = 3/64
½ × ½ × ½ = 1/8
© 2016 Pearson Education, Inc.
The Rh blood factor in humans is perhaps the most
important after the ABO system. The + phenotype is
dominant to the – phenotype, and it is encoded on a
separate autosome from the ABO locus (i.e., the two loci
assort independently).
A child of blood type A+ is born to a mother of blood type
O. What can be concluded about the blood type of the
biological father?
A.
B.
C.
D.
A+ or AB+
A+ only
AB+ only
any possible blood type
© 2016 Pearson Education, Inc.
The Rh blood factor in humans is perhaps the most
important after the ABO system. The + phenotype is
dominant to the – phenotype, and it is encoded on a
separate autosome from the ABO locus (i.e., the two loci
assort independently).
A child of blood type A+ is born to a mother of blood type
O. What can be concluded about the blood type of the
biological father?
A.
B.
C.
D.
A+ or AB+
A+ only
AB+ only
any possible blood type
© 2016 Pearson Education, Inc.
Assume there are 50 people in the classroom.
Theoretically speaking, what is the maximum number of
different alleles there could be at a hypothetical
autosomal locus?
A.
B.
C.
D.
2
3
50
100
© 2016 Pearson Education, Inc.
Assume there are 50 people in the classroom.
Theoretically speaking, what is the maximum number of
different alleles there could be at a hypothetical
autosomal locus?
A.
B.
C.
D.
2
3
50
100
© 2016 Pearson Education, Inc.
Consider the case of dominant epistasis in squash, where
YY and Yy = yellow and yy = green. At a second,
independently assorting locus, WW and ww = white, and
ww allows for yellow or green.
What is the expected phenotypic ratio among the progeny
of the cross YyWw × YyWw? (Note these are white
squash.)
A.
B.
C.
D.
1:1:1:1
9:3:3:1
9:4:3
12:3:1
© 2016 Pearson Education, Inc.
Consider the case of dominant epistasis in squash, where
YY and Yy = yellow and yy = green. At a second,
independently assorting locus, WW and ww = white, and
ww allows for yellow or green.
What is the expected phenotypic ratio among the progeny
of the cross YyWw × YyWw? (Note these are white
squash.)
A.
B.
C.
D.
1:1:1:1
9:3:3:1
9:4:3
12:3:1
© 2016 Pearson Education, Inc.
Regarding the simplified model for polygenic inheritance
of human skin color presented in the text, since there are
more than seven skin color phenotypes in humans, what
modification(s) could make the model more complete?
More than one may apply.
A.
B.
C.
D.
E.
more loci
more alleles per locus
more individuals in the population
differing levels of effect per allele
differing levels of environmental
exposure
© 2016 Pearson Education, Inc.
Regarding the simplified model for polygenic inheritance
of human skin color presented in the text, since there are
more than seven skin color phenotypes in humans, what
modification(s) could make the model more complete?
More than one may apply.
A.
B.
C.
D.
E.
more loci
more alleles per locus
more individuals in the population
differing levels of effect per allele
differing levels of environmental
exposure
© 2016 Pearson Education, Inc.