Biology, 7e (Campbell) Chapter 14: Mendel and the Gene Idea Chapter Questions 1) 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 completely 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. Answer: D Topic: Concept 14.1 Skill: Comprehension 2) A plant
with purple flowers is allowed to self-pollinate. Generation after generation, it produces purple
flowers. This is an example of A) hybridization. B) incomplete dominance. C) true-breeding. D)
the law of segregation. E) polygenetics. Answer: C Topic: Concept 14.1 Skill: Knowledge 3)
Which of the following statements about Mendel's breeding experiments is correct? A) None of
the parental (P) plants were true-breeding. B) All of the F2 progeny showed a phenotype that was
intermediate between the two parental (P) phenotypes. C) Half of the F1 progeny had the same
phenotype as one of the parental (P) plants, and the other half had the same phenotype as the
other parent. D) All of the F1 progeny resembled one of the parental (P) plants, but only some of
the F2 progeny did. E) none of the above Answer: D Topic: Concept 14.1 Skill: Comprehension
4) What is the difference between a monohybrid cross and a dihybrid cross? A) A monohybrid
cross involves a single parent, whereas a dihybrid cross involves two parents. B) A monohybrid
cross produces a single progeny, whereas a dihybrid cross produces two progeny. C) A
monohybrid cross involves organisms that are heterozygous for a single character, whereas a
dihybrid cross involves organisms that are heterozygous for two characters. D) A monohybrid
cross is performed only once, whereas a dihybrid cross is performed twice. E) A monohybrid
cross results in a 9:3:3:1 ratio whereas a dihybrid cross gives a 3:1 ratio. Answer: C Topic:
Concept 14.1 Skill: Knowledge 1 5) A cross between homozygous purple-flowered and
homozygous white-flowered pea plants results in offspring with purple flowers. This
demonstrates A) the blending model of genetics. B) true-breeding. C) dominance. D) a dihybrid
cross. E) the mistakes made by Mendel. Answer: C Topic: Concept 14.1 Skill: Comprehension 6)
The F1 offspring of Mendel's classic pea cross always looked like one of the two parental
varieties because A) one allele was completely dominant over another. B) each allele affected
phenotypic expression. C) the traits blended together during fertilization. D) no genes interacted
to produce the parental phenotype. E) different genes interacted to produce the parental
phenotype. Answer: A Topic: Concept 14.1 Skill: Knowledge 7) What was the most significant
conclusion that Gregor Mendel drew from his experiments with pea plants? A) There is
considerable genetic variation in garden peas. B) Traits are inherited in discrete units, and are not
the results of "blending." C) Recessive genes occur more frequently in the F1 than do dominant
ones. D) Genes are composed of DNA. E) An organism that is homozygous for many recessive
traits is at a disadvantage. Answer: B Topic: Concept 14.1 Skill: Comprehension 8) Which of the
following is (are) true for alleles? A) They can be identical or different for any given gene in a
somatic cell. B) They can be dominant or recessive. C) They can represent alternative forms of a
gene. D) Only A and B are correct. E) A, B, and C are correct. Answer: E Topic: Concept 14.1
Skill: Knowledge 9) What is genetic cross between an individual showing a dominant phenotype
(but of unknown genotype) and a homozygous recessive individual called? A) a self-cross B) a
testcross C) a hybrid cross D) an F1 cross E) a dihybrid cross Answer: B Topic: Concept 14.1
Skill: Knowledge 2 10) How many unique gametes could be produced through independent
assortment by an individual with the genotype AaBbCCDdEE? A) 4 B) 8 C) 16 D) 32 E) 64
Answer: B Topic: Concept 14.1 Skill: Application 11) Two plants are crossed, resulting in
offspring with a 3:1 ratio for a particular trait. This suggests A) that the parents were truebreeding for contrasting traits. B) incomplete dominance. C) that a blending of traits has
occurred. D) that the parents were both heterozygous. E) that each offspring has the same alleles.
Answer: D Topic: Concept 14.1 Skill: Knowledge 12) Two characters that appear in a 9:3:3:1
ratio in the F2 generation should have which of the following properties? A) Each of the
characters is controlled by a single gene. B) The genes controlling the characters obey the law of
independent assortment. C) Each of the genes controlling the characters has two alleles. D) Only
A and C are correct. E) A, B, and C are correct. Answer: E Topic: Concept 14.1 Skill:
Comprehension 13) A 9:3:3:1 phenotypic ratio is characteristic of which of the following? A) a
monohybrid cross B) a dihybrid cross C) a trihybrid cross D) linked genes E) both A and D
Answer: B Topic: Concept 14.1 Skill: Knowledge 14) A sexually reproducing animal has two
unlinked genes, one for head shape (H) and one for tail length (T). Its genotype is HhTt. Which
of the following genotypes is possible in a gamete from this organism? A) HT B) Hh C) HhTt D)
T E) tt Answer: A Topic: Concept 14.1 Skill: Comprehension 3 15) It was important that Mendel
examined not just the F1 generation in his breeding experiments, but the F2 generation as well,
because A) he obtained very few F1 progeny, making statistical analysis difficult. B) parental
traits that were not observed in the F1 reappeared in the F2, suggesting that the traits did not
truly disappear in the F1. C) analysis of the F1 progeny would have allowed him to discover the
law of segregation, but not the law of independent assortment. D) the dominant phenotypes were
visible in the F2 generation, but not in the F1. E) all of the above Answer: B Topic: Concept 14.1
Skill: Comprehension 16) When crossing a homozygous recessive with a heterozygote, what is
the chance of getting an offspring with the homozygous recessive phenotype? A) 0% B) 25% C)
50% D) 75% E) 100% Answer: C Topic: Concept 14.1 Skill: Application 4 Use the following
diagram and description to answer the questions below. In a particular plant, leaf color is
controlled by gene D. Plants with the dominant allele D have dark green leaves, and plants with
the homozygous recessive dd genotype have light green leaves. A true-breeding dark-leaved
plant is crossed with a light-leaved one, and the F1 offspring is allowed to self-pollinate. The
predicted outcome of this cross is diagrammed in the Punnett square shown below, where 1, 2, 3,
and 4 represent the genotypes corresponding to each box within the square. 17) Which of the
boxes marked 1-4 correspond to plants with dark leaves? A) 1 only B) 1 and 2 C) 2 and 3 D) 4
only E) 1, 2, and 3 Answer: E Topic: Concept 14.1 Skill: Application 18) Which of the boxes
correspond to plants with a heterozygous genotype? A) 1 B) 1 and 2 C) 1, 2, and 3 D) 2 and 3 E)
2, 3, and 4 Answer: D Topic: Concept 14.1 Skill: Application 19) Which of the plants will be
true-breeding? A) 1 and 4 B) 2 and 3 C) 1-4 D) 1 only E) none Answer: A Topic: Concept 14.1
Skill: Application 5 20) P = purple, pp = white. The offspring of a cross between two
heterozygous purple-flowering plants (Pp Pp) results in A) all purple-flowered plants. B) purpleflowered plants and white-flowered plants. C) two types of white-flowered plants: PP and Pp. D)
all white-flowered plants. E) all pink-flowered plants. Answer: B Topic: Concept 14.1 Skill:
Comprehension 21) Mendel accounted for the observation that traits which had disappeared in
the F1 generation reappeared in the F2 generation by proposing that A) new mutations were
frequently generated in the F2 progeny, "reinventing" traits that had been lost in the F1. B) the
mechanism controlling the appearance of traits was different between the F1 and the F2 plants.
C) traits can be dominant or recessive, and the recessive traits were obscured by the dominant
ones in the F1. D) the traits were lost in the F1 due to blending of the parental traits. E) members
of the F1 generation had only one allele for each character, but members of the F2 had two
alleles for each character. Answer: C Topic: Concept 14.1 Skill: Comprehension 22) What are
Punnett squares used for? A) predicting the result of genetic crosses between organisms of
known genotypes B) determining the DNA sequence of a given gene C) identifying the gene
locus where allelic variations are possible D) testing for the presence of the recessive allele E)
more than one of the above Answer: A Topic: Concept 14.1 Skill: Knowledge 23) Which of the
following is false, regarding the law of segregation? A) It states that each of two alleles for a
given trait segregate into different gametes. B) It can be explained by the segregation of
homologous chromosomes during meiosis. C) It can account for the 3:1 ratio seen in the F2
generation of Mendel's crosses. D) It can be used to predict the likelihood of transmission of
certain genetic diseases within families. E) It is a method that can be used to determine the
number of chromosomes in a plant. Answer: E Topic: Concept 14.1 Skill: Knowledge 6 24) The
fact that all seven of the pea plant traits studied by Mendel obeyed the principle of independent
assortment means that A) none of the traits obeyed the law of segregation. B) the diploid number
of chromosomes in the pea plants was 7. C) all of the genes controlling the traits were located on
the same chromosome. D) all of the genes controlling the traits behaved as if they were on
different chromosomes. E) the formation of gametes in plants occurs by mitosis only. Answer: D
Topic: Concept 14.1 Skill: Comprehension 25) Black fur in mice (B) is dominant to brown fur
(b). Short tails (T) are dominant to long tails (t). What fraction of the progeny of the cross BbTt
BBtt will have black fur and long tails? A) 1/16 B) 3/16 C) 3/8 D) 1/2 E) 9/16 Answer: D Topic:
Concept 14.2 Skill: Application 26) In certain plants, tall is dominant to short. If a heterozygous
plant is crossed with a homozygous tall plant, what is the probability that the offspring will be
short? A) 1/2 B) 1/4 C) 0 D) 1 E) 1/6 Answer: C Topic: Concept 14.2 Skill: Comprehension 27)
A couple has three children, all of whom have brown eyes and blond hair. Both parents are
homozygous for brown eyes (BB), but one is a blond (rr) and the other is a redhead (Rr). What is
the probability that their next child will be a brown-eyed redhead? A) 1/16 B) 1/8 C) 1/4 D) 1/2
E) 1 Answer: D Topic: Concept 14.2 Skill: Application 7 28) Two true-breeding stocks of pea
plants are crossed. One parent has red, axial flowers and the other has white, terminal flowers; all
F1 individuals have red, axial flowers. If 1,000 F2 offspring resulted from the cross,
approximately how many of them would you expect to have red, terminal flowers? (Assume
independent assortment.) A) 65 B) 190 C) 250 D) 565 E) 750 Answer: B Topic: Concept 14.2
Skill: Application 29) In a cross AaBbCc AaBbCc, what is the probability of producing the
genotype AABBCC? A) 1/4 B) 1/8 C) 1/16 D) 1/32 E) 1/64 Answer: E Topic: Concept 14.2
Skill: Application 30) Given the parents AABBCc AabbCc, assume simple dominance and
independent assortment. What proportion of the progeny will be expected to phenotypically
resemble the first parent? A) 1/4 B) 1/8 C) 3/4 D) 3/8 E) 1 Answer: C Topic: Concept 14.2 Skill:
Application 31) A 1:2:1 phenotypic ratio in the F2 generation of a monohybrid cross is a sign of
A) complete dominance. B) multiple alleles. C) incomplete dominance. D) polygenic
inheritance. E) pleiotropy. Answer: C Topic: Concept 14.3 Skill: Knowledge 8 Refer to the
following result to answer the next several questions. A tall plant is crossed with a short plant,
and the progeny are all intermediate in size between the two parental plants. 32) This could be an
example of A) incomplete dominance. B) polygenic inheritance. C) complete dominance. D) A
and B E) B and C Answer: D Topic: Concept 14.3 Skill: Comprehension 33) If the intermediate
F1 progeny were allowed to self-pollinate, and the F2 progeny were also intermediate in size, but
following a normal distribution, this would suggest A) incomplete dominance. B) polygenic
inheritance. C) complete dominance. D) a strong environmental influence. E) codominance.
Answer: B Topic: Concept 14.3 Skill: Application 34) If the intermediate F1 progeny were
allowed to self-pollinate, and 25% of the F2 progeny were tall, 50% were intermediate in size,
and 25% were short, this would suggest A) incomplete dominance. B) polygenic inheritance. C)
complete dominance. D) pleiotropy. E) multifactorial inheritance. Answer: A Topic: Concept
14.3 Skill: Application 35) In snapdragons, heterozygotes have pink flowers, whereas
homozygotes have red or white flowers. When plants with red flowers are crossed with plants
with white flowers, what proportion of the offspring will have pink flowers? A) 0% B) 25% C)
50% D) 75% E) 100% Answer: E Topic: Concept 14.3 Skill: Application 9 36) Tallness (T) is
dominant to dwarfness (t), while red (R) flower color is dominant to white (r). The heterozygous
condition results in pink (Rr) flower color. A dwarf, red snapdragon is crossed with a plant
homozygous for tallness and white flowers. What are the genotype and phenotype of the F1
individuals? A) ttRr-dwarf and pink B) ttrr-dwarf and white C) TtRr-tall and red D) TtRr-tall and
pink E) TTRR-tall and red Answer: D Topic: Concept 14.3 Skill: Application 37) Skin color in a
fish is inherited via a single gene with four different alleles. How many different types of
gametes would be possible in this system? A) 1 B) 2 C) 4 D) 8 E) 16 Answer: C Topic: Concept
14.3 Skill: Comprehension 38) In cattle, roan coat color (mixed red and white hairs) occurs in the
heterozygous (Rr) offspring of red (RR) and white (rr) homozygotes. Which of the following
crosses would produce offspring in the ratio of 1 red:2 roan:1 white? A) red white B) roan roan
C) white roan D) red roan E) The answer cannot be determined from the information provided.
Answer: B Topic: Concept 14.3 Skill: Application Refer to the following to answer the questions
below. Gene S controls the sharpness of spines in a type of cactus. Cactuses with the dominant
allele, S, have sharp spines, whereas homozygous recessive ss cactuses have dull spines. At the
same time, a second gene, N, determines whether cactuses have spines. Homozygous recessive
nn cactuses have no spines at all. 39) The relationship between genes S and N is an example of
A) incomplete dominance. B) epistasis. C) complete dominance. D) pleiotropy. E) codominance.
Answer: B Topic: Concept 14.3 Skill: Comprehension 10 40) A cross between a true-breeding
sharp-spined cactus and a spineless cactus would produce A) all sharp-spined progeny. B) 50%
sharp-spined, 50% dull-spined progeny. C) 25% sharp-spined, 50% dull-spined, 25% spineless
progeny D) all spineless progeny E) It is impossible to determine the phenotypes of the progeny.
Answer: A Topic: Concept 14.3 Skill: Application 41) If doubly heterozygous SsNn cactuses
were allowed to self-pollinate, the F2 would segregate in which the of following ratios? A) 3
sharp-spined : 1 spineless B) 1 sharp-spined : 2 dull-spined : 1 spineless C) 1 sharp spined : 1
dull-spined : 1 spineless D) 1 sharp-spined 1 : dull-spined E) 9 sharp-spined : 3 dull-spined : 4
spineless Answer: E Topic: Concept 14.3 Skill: Application Use the information given here to
answer the following questions. Feather color in budgies is determined by two different genes Y
and B. YYBB, YyBB, or YYBb is green; yyBB or yyBb is blue; YYbb or Yybb is yellow; and
yybb is white. 42) A blue budgie is crossed with a white budgie. Which of the following results is
not possible? A) green offspring B) yellow offspring C) blue offspring D) A and B E) A, B, and
C Answer: D Topic: Concept 14.3 Skill: Application 43) Two blue budgies were crossed. Over
the years, they produced 22 offspring, 5 of which were white. What are the most likely genotypes
for the two blue budgies? A) yyBB and yyBB B) yyBB and yyBb C) yyBb and yyBb D) yyBB
and yybb E) yyBb and yybb Answer: C Topic: Concept 14.3 Skill: Application 11 44) Three
babies were mixed up in a hospital. After consideration of the data below, which of the following
represent the correct baby and parent combinations? A) I-3, II-1, III-2 B) I-1, II-3, III-2 C) I-2,
II-3, III-1 D) I-2, II-1, III-3 E) I-3, II-2, III-1 Answer: C Topic: Concept 14.3 Skill: Application
Use the following information to answer the questions below. A woman who has blood type A,
has a daughter who is type O positive and a son who is type B negative. Rh positive is a simple
dominant trait over Rh negative. 45) Which of the following is a possible genotype for the son?
A) IBIB B) IBIA C) ii D) IBi E) IAIA Answer: D Topic: Concept 14.3 Skill: Application 46)
Which of the following is a possible genotype for the mother? A) IAIA B) IBIB C) ii D) IAi E)
IAIB Answer: D Topic: Concept 14.3 Skill: Application 47) Which of the following is a possible
phenotype for the father? A) A B) O C) B D) AB E) impossible to determine Answer: C Topic:
Concept 14.3 Skill: Application 12 48) Which of the following is the probable genotype for the
mother? A) IAIARR B) IAIARr C) IAirr D) IAiRr E) IAiRR Answer: D Topic: Concept 14.3,
Concept 14.4 Skill: Application 49) Which of the following is a possible phenotype of the father?
A) A negative B) O negative C) B positive D) A positive E) O positive Answer: C Topic:
Concept 14.3, Concept 14.4 Skill: Application The questions below refer to the following terms.
Each term may be used once, more than once, or not at all. A. B. C. D. incomplete dominance
multiple alleles pleiotropy epistasis 50) the ability of a single gene to have multiple phenotypic
effects Answer: C Topic: Concept 14.3 Skill: Knowledge 51) the ABO blood group system
Answer: B Topic: Concept 14.3 Skill: Knowledge 52) the phenotype of the heterozygote differs
from the phenotypes of both homozygotes Answer: A Topic: Concept 14.3 Skill: Knowledge 53)
cystic fibrosis affects the lungs, the pancreas, the digestive system, and other organs, resulting in
symptoms ranging from breathing difficulties to recurrent infections Answer: C Topic: Concept
14.3 Skill: Knowledge 13 54) Which of the following is an example of polygenic inheritance? A)
pink flowers in snapdragons B) the ABO blood groups in humans C) Huntington's disease in
humans D) white and purple flower color in peas E) skin pigmentation in humans Answer: E
Topic: Concept 14.3 Skill: Knowledge 55) Hydrangea plants of the same genotype are planted in
a large flower garden. Some of the plants produce blue flowers and others pink flowers. This can
be best explained by A) environmental factors such as soil pH. B) the allele for blue hydrangea
being completely dominant. C) the alleles being codominant. D) the fact that a mutation has
occurred. E) acknowledging that multiple alleles are involved. Answer: A Topic: Concept 14.3
Skill: Comprehension Use the following information to answer the questions below. A woman
and her spouse both show the normal phenotype for pigmentation, but both had one parent who
was an albino. Albinism is an autosomal recessive trait. 56) What is the probability that their first
child will be an albino? A) 0 B) 1/4 C) 1/2 D) 3/4 E) 1 Answer: B Topic: Concept 14.4 Skill:
Application 57) If their first two children have normal pigmentation, what is the probability that
their third child will be an albino? A) 0 B) 1/4 C) 1/2 D) 3/4 E) 1 Answer: B Topic: Concept 14.4
Skill: Application 14 58) Huntington's disease is caused by a dominant allele. If one of your
parents has the disease, what is the probability that you, too, will have the disease? A) 1 B) 3/4
C) 1/2 D) 1/4 E) 0 Answer: C Topic: Concept 14.4 Skill: Knowledge 59) A woman has six sons.
The chance that her next child will be a daughter is A) 1. B) 0. C) 1/2. D) 1/6. E) 5/6. Answer: C
Topic: Concept 14.4 Skill: Knowledge The following questions refer to the pedigree chart in
Figure 14.1 for a family, some of whose members exhibit the recessive trait, wooly hair. Affected
individuals are indicated by an open square or circle. Figure 14.1 60) What is the genotype of
individual B-5? A) WW B) Ww C) ww D) WW or ww E) ww or Ww Answer: C Topic: Concept
14.4 Skill: Application 15 61) What is the likelihood that the progeny of D-3 and D-4 will have
wooly hair? A) 0% B) 25% C) 50% D) 75% E) 100% Answer: C Topic: Concept 14.4 Skill:
Application 62) What is the probability that individual C-1 is Ww? A) 3/4 B) 1/4 C) 2/4 D) 2/3
E) 1 Answer: E Topic: Concept 14.4 Skill: Application 63) People with sickle-cell trait A) are
heterozygous for the sickle-cell allele. B) are usually healthy. C) have increased resistance to
malaria. D) produce normal and abnormal hemoglobin. E) all of the above Answer: E Topic:
Concept 14.4 Skill: Knowledge 64) When a disease is said to have a multifactorial basis, it
means that A) many factors, both genetic and environmental, 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. Answer: A Topic: Concept 14.4 Skill:
Knowledge 65) Which of the following terms is least related to the others? A) pedigree B)
karyotype C) amniocentesis D) chorionic villus sampling E) epistasis Answer: E Topic: Concept
14.4 Skill: Knowledge 16 The questions below will use the following answers. Each answer may
be used once, more than once, or not at all. A. B. C. D. E. Huntington's disease Tay-Sachs
disease phenylketonuria cystic fibrosis sickle-cell disease 66) Substitution of the "wrong" amino
acid in the hemoglobin protein results in this disorder. Answer: E Topic: Concept 14.4 Skill:
Knowledge 67) Individuals with this disorder are unable to metabolize certain lipids, affecting
proper brain development. Affected individuals die in early childhood. Answer: B Topic:
Concept 14.3 Skill: Knowledge 68) This is caused by a dominant single gene defect and
generally does not appear until the individual is 35-45 years of age. Answer: A Topic: Concept
14.4 Skill: Knowledge 69) Effects of this recessive disorder can be completely overcome by
regulating the diet of the affected individual. Answer: C Topic: Concept 14.4 Skill: Knowledge
70) This results from a defect in membrane proteins that normally function in chloride ion
transport. Answer: D Topic: Concept 14.4 Skill: Knowledge 71) Which of the following
techniques involves the preparation of a karyotype? A) amniocentesis B) chorionic villus
sampling C) fetoscopy D) A and B only E) A, B, and C Answer: D Topic: Concept 14.4 Skill:
Knowledge 17 Media Activity Questions 72) All the offspring of a cross between a black-eyed
MendAlien and an orange-eyed MendAlien have black eyes. This means that the allele for black
eyes is ________ the allele for orange eyes. A) codominant to B) recessive to C) more aggressive
than D) dominant to E) better than Answer: D Topic: Web/CD Activity: Monohybrid Crosses 73)
What is the expected phenotypic ratio of a cross between two orange-eyed MendAliens? A) 3
black-eyed:1 orange-eyed B) 0 black-eyed:1 orange-eyed C) 1 black-eyed:3 orange-eyed D) 1
black-eyed:0 orange-eyed E) 1 black-eyed:1 orange-eyed Answer: B Topic: Web/CD Activity:
Monohybrid Crosses 74) Andalusian chickens with the genotype CBCB are black, those with the
genotype CBCW are gray. What is the relationship between the CB and the CW alleles? A) CB is
dominant to CW. B) CB is recessive to CW. C) CW is dominant to CB. D) The relationship is
one of incomplete dominance. E) CB and CW are codominant. Answer: A Topic: Web/CD
Activity: Incomplete Dominance 75) Black eyes are dominant to orange eyes, and green skin is
dominant to white skin. Sam, a MendAlien with black eyes and green skin, has a parent with
orange eyes and white skin. Carole is MendAlien with orange eyes and white skin. If Sam and
Carole were to mate, the predicted ratio of their offspring would be: A) 1 black eyes, green skin :
1 black eyes, white skin : 1 orange eyes, green skin : 1 orange eyes, white skin B) 3 black eyes,
green skin : 3 black eyes, white skin : 9 orange eyes, green skin : 1 orange eyes, white skin C) 1
black eyes, green skin : 3 black eyes, white skin : 3 orange eyes, green skin : 9 orange eyes,
white skin D) 9 black eyes, green skin : 3 black eyes, white skin : 3 orange eyes, green skin : 1
orange eyes, white skin E) There is insufficient information to determine Sam's genotype.
Answer: A Topic: Web/CD Activity: Dihybrid Crosses 76) All the offspring of a cross between a
red-flowered plant and a white-flowered plant have pink flowers. This means that the allele for
red flowers is ________ to the allele for white flowers. A) dominant B) codominant C)
pleiotropic D) incompletely dominant E) recessive Answer: D Topic: Web/CD Activity:
Incomplete Dominance 18 Self-Quiz Questions 77) In some plants, a true-breeding, red-flowered
strain gives all pink flowers when crossed with a whiteflowered strain: RR (red) rr (white) Rr
(pink). If flower position (axial or terminal) is inherited as it is in peas (see Table 14.1 in your
text), what will be the ratios of genotypes and phenotypes of the F1 generation resulting from the
following cross: axial-red (true-breeding) terminal-white? What will be the ratios in the F2
generation? Answer: Parental cross is AARR aarr. Genotype of F1 is AaRr, phenotype is all
axial-pink. Genotypes of F2 are 4 AaRr : 2 AaRR : 1 AARr : 2 aaRr : 2 Aarr : 1AARR : 1 aaRR :
1 AArr : 1 aarr. Phenotypes of F2 are 6 axial-pink : 3 axial-white : 2 terminal-pink : 1 terminalwhite : 1 terminal red. 78) Flower position, stem length, and seed shape were three characters
that Mendel studied. Each is controlled by an independently assorting gene and has dominant
and recessive expression as follows: If a plant that is heterozygous for all three characters is
allowed to self-fertilize, what proportion of the offspring would you expect to be as follows?
(Note: Use the rules of probability instead of a huge Punnett square.) A) homozygous for the
three dominant traits B) homozygous for the three recessive traits C) heterozygous for all three
characters D) homozygous for axial and tall, heterozygous for seed shape Answer: A) B) C) D)
1/64 1/64 1/8 1/32 79) A black guinea pig crossed with an albino guinea pig produces 12 black
offspring. When the albino is crossed with a second black one, 7 blacks and 5 albinos are
obtained. What is the best explanation for this genetic situation? Write genotypes for the parents,
gametes, and offspring. Answer: Albino (b) is a recessive trait; black (B) is dominant. First cross:
parents BB bb; gametes B and b; offspring all Bb (black coat). Second cross: parents bb Bb;
gametes 1/2 B and 1/2 b (heterozygous parent) and b; offspring 1/2 Bb and 1/2 bb. 19 80) In
sesame plants, the one-pod condition (P) is dominant to the three-pod condition (p), and normal
leaf (L) is dominant to wrinkled leaf (l). Pod type and leaf type are inherited independently.
Determine the genotypes for the two parents for all possible matings producing the following
offspring: A) 318 one-pod, normal leaf : 98 one-pod, wrinkled leaf B) 323 three-pod, normal leaf
: 106 three-pod, wrinkled leaf C) 401 one-pod, normal leaf D) 150 one-pod, normal leaf : 147
one-pod, wrinkled leaf : 51 three-pod, normal leaf : 48 three-pod, wrinkled leaf E) 223 one-pod,
normal leaf : 72 one-pod, wrinkled leaf : 76 three-pod, normal leaf; 27 three-pod, wrinkled leaf
Answer: A) PPLl PPLl, PpLl, or ppLl B) ppLl ppLl C) PPLL any of the 9 possible genotypes or
PPll ppLL D) PpLl Ppll E) PpLl PpLl 81) A man with type A blood marries a woman with type B
blood. Their child has type O blood. What are the genotypes of these individuals? What other
genotypes, and in what frequencies, would you expect in offspring from this marriage? Answer:
Man IAi; woman IBi; child ii. Other genotypes for children are 1/4 IAIB ,1/4 IAi, 1/4 IBi 82)
Phenylketonuria (PKU) is an inherited disease caused by a recessive allele. If a woman and her
husband, who are both carriers, have three children, what is the probability of each of the
following? A) All three children are of normal phenotype. B) One or more of the three children
have the disease. C) All three children have the disease. D) At least one child is phenotypically
normal. (Note: Remember that the probabilities of all possible outcomes always add up to 1.)
Answer: A) 3/4 3/4 3/4 = 27/64 B) 1 - 27/64 = 37/64 C) 1/4 1/4 1/4 = 1/64 D) 1 - 1/64 = 63/64
83) The genotype of F1 individuals in a tetrahybrid cross is AaBbCcDd. Assuming independent
assortment of these four genes, what are the probabilities that F2 offspring will have the
following genotypes? A) aabbccdd B) AaBbCcDd C) AABBCCDD D) AaBBccDd E)
AaBBCCdd Answer: A) 1/256 B) 1/16 C) 1/256 D) 1/64 E) 1/128 20 84) What is the probability
that each of the following pairs of parents will produce the indicated offspring? (Assume
independent assortment of all gene pairs.) A) AABBCC aabbcc AaBbCc B) AABbCc AaBbCc
AAbbCC C) AaBbCc AaBbCc AaBbCc D) aaBbCC AABbcc AaBbCc Answer: A) 1 B) 1/32 C)
1/8 D) 1/2 85) Karen and Steve each have a sibling with sickle-cell disease. Neither Karen nor
Steve nor any of their parents have the disease, and none of them have been tested to reveal
sickle-cell trait. Based on this incomplete information, calculate the probability that if this couple
has a child, the child will have sicklecell disease. Answer: 1/9 86) In 1981, a stray black cat with
unusual rounded, curled-back ears was adopted by a family in California. Hundreds of
descendants of the cat have since been born, and cat fanciers hope to develop the curl cat into a
show breed. Suppose you owned the first curl cat and wanted to develop a true-breeding variety.
How would you determine whether the curl allele is dominant or recessive? How would you
obtain true-breeding curl cats? How could you be sure they are true-breeding? Answer: Matings
of the original mutant cat with true-breeding noncurl cats will produce both curl and noncurl F1
offspring if the curl allele is dominant, but only noncurl offspring if the curl allele is recessive.
You would obtain some true-breeding offspring homozygous for the curl allele from matings
between the F1 cats resulting from the original curl noncurl crosses whether the curl trait is
dominant or recessive. You know that cats are true-breeding when curl curl matings produce only
curl offspring. As it turns out, the allele that causes curled ears is dominant. 21