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Biology, e Campbell Chapter Mendel and the Gene Idea MultipleChoice Questions
Pea plants were particularly well suited for use in Mendels 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. Answer D
Topic Concept . Skill Knowledge/Comprehension
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 dihybrid cross involves organisms that are heterozygous for two characters and
a monohybrid only one. D A monohybrid cross is performed for one generation, whereas a
dihybrid cross is performed for two generations. E A monohybrid cross results in a ratio
whereas a dihybrid cross gives a ratio. Answer C
Topic Concept . Skill Knowledge/Comprehension
A cross between homozygous purpleflowered and homozygous whiteflowered pea plants
results in offspring with purple flowers. This demonstrates A the blending model of genetics.
B truebreeding. C dominance. D a dihybrid cross. E the mistakes made by Mendel. Answer
C
Topic Concept . Skill Knowledge/Comprehension
The F offspring of Mendels classic pea cross always looked like one of the two parental
varieties because A one phenotype 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 . Skill Knowledge/Comprehension
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 quotblending.quot C Recessive genes
occur more frequently in the F 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 . Skill Knowledge/Comprehension
How many unique gametes could be produced through independent assortment by an
individual with the genotype AaBbCCDdEE A B C D E Answer B
Topic Concept . Skill Application/Analysis
Two plants are crossed, resulting in offspring with a 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 . Skill Knowledge/Comprehension
what is the chance of producing an offspring with the homozygous recessive phenotype A B
C D E Answer C Topic Concept . one for head shape H and one for tail length T. but the F
generation as well. E many of the F progeny died. making statistical analysis difficult.
because A he obtained very few F progeny. D the dominant phenotypes were visible in the F
generation. B The genes controlling the characters obey the law of independent assortment.
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 . Skill Application/Analysis . C Each of the genes
controlling the characters has two alleles. B parental traits that were not observed in the F
reappeared in the F. Skill Knowledge/Comprehension A sexually reproducing animal has two
unlinked genes. Answer B Topic Concept . Answer B Topic Concept . but not in the F. Skill
Application/Analysis It was important that Mendel examined not just the F generation in his
breeding experiments. Its genotype is HhTt. E Sixteen different phenotypes are possible. but
not the law of independent assortment. Two characters that appear in a ratio in the F
generation should have which of the following properties A Each of the traits is controlled by
single genes. D Four genes are involved. Skill Synthesis/Evaluation When crossing an
organism that is homozygous recessive for a single trait with a heterozygote. C analysis of
the F progeny would have allowed him to discover the law of segregation.
and represent the genotypes corresponding to each box within the square. and Answer E
Topic Concept . In a particular plant. . Skill Application/Analysis Which of the plants will be
truebreeding A and B and C D only E None Answer A Topic Concept . . Plants with at least
one allele D have dark green leaves. Skill Application/Analysis . and the F offspring is
allowed to self pollinate. A true breeding darkleaved plant is crossed with a lightleaved one.
Skill Application/Analysis Which of the boxes correspond to plants with a heterozygous
genotype A B and C . . and D and E . Figure . leaf color is controlled by gene locus D. and
Answer D Topic Concept .. The predicted outcome of the F is diagrammed in the Punnett
square shown in Figure . where . Which of the boxes marked correspond to plants with dark
leaves A only B and C and D only E . . and the following description to answer the questions
below. .Use Figure . and plants with the homozygous recessive dd genotype have light green
leaves.
E It is a method that can be used to determine the number of chromosomes in a plant.
Answer D Topic Concept . quotreinventingquot traits that had been lost in the F. D the traits
were lost in the F due to blending of the parental traits. C All of the genes controlling the
traits were located on the same chromosome. C traits can be dominant or recessive. but
members of the F had two alleles for each character. B The diploid number of chromosomes
in the pea plants was . Answer C Topic Concept . B the mechanism controlling the
appearance of traits was different between the F and the F plants. Skill
Knowledge/Comprehension Which of the following about the law of segregation is false A It
states that each of two alleles for a given trait segregate into different gametes. E The
formation of gametes in plants occurs by mitosis only. B It can be explained by the
segregation of homologous chromosomes during meiosis. C It can account for the ratio seen
in the F generation of Mendels crosses. Skill Knowledge/Comprehension The fact that all
seven of the pea plant traits studied by Mendel obeyed the principle of independent
assortment most probably indicates which of the following A None of the traits obeyed the
law of segregation. Answer E Topic Concept . Skill Synthesis/Evaluation Mendels
observation of the segregation of alleles in gamete formation has its basis in which of the
following phases of cell division A Prophase I of meiosis B Prophase II of meiosis C
Metaphase I of meiosis D Anaphase I of meiosis E Anaphase of mitosis Answer D Topic
Concept . D It can be used to predict the likelihood of transmission of certain genetic
diseases within families. and the recessive traits were obscured by the dominant ones in the
F. . Skill Synthesis/Evaluation Mendel was able to draw his ideas of segregation and
independent assortment because of the influence of which of the following A His reading and
discussion of Darwins Origin of Species B The understanding of particulate inheritance he
learned from renowned scientists of his time C His discussions of heredity with his
colleagues at major universities D His reading of the scientific literature current in the field E
His experiments with the breeding of plants such as peas Answer E Topic Concept . D All of
the genes controlling the traits behaved as if they were on different chromosomes. E
members of the F generation had only one allele for each character. Mendel accounted for
the observation that traits which had disappeared in the F generation reappeared in the F
generation by proposing that A new mutations were frequently generated in the F progeny.
What fraction of the progeny of the cross BbTt BBtt will have black fur and long tails A / B / C
/ D / E / Answer D Topic Concept . If a heterozygous plant is crossed with a homozygous tall
plant. Skill Synthesis/Evaluation Black fur in mice B is dominant to brown fur b.Skill
Synthesis/Evaluation Mendels second law of independent assortment has its basis in which
of the following events of meiosis I A Synapsis of homologous chromosomes B Crossing
over C Alignment of tetrads at the equator D Separation of homologs at anaphase E
Separation of cells at telophase Answer C Topic Concept . what is the probability that the
offspring will be short A B / C / D / E Answer B Topic Concept . tall is dominant to short.
Short tails T are dominant to long tails t. Skill Application/Analysis . Skill Application/Analysis
In certain plants.
C Yellow is dominant to black. or yellow. / black to / brown puppies. Skill
Application/Analysis Given the parents AABBCc AabbCc. assume simple dominance and
independent assortment. terminal flowers A B C D E Answer B Topic Concept . These
results indicate which of the following A Brown is dominant to black. B Black is dominant to
brown and to yellow. Labrador retrievers are black. Skill Application/Analysis . terminal
flowers. What proportion of the progeny will be expected to phenotypically resemble the first
parent A / B / C / D / E Answer C Topic Concept . or / black to / yellow puppies. axial flowers
and the other has white. approximately how many of them would you expect to have red.
what is the probability of producing the genotype AABBCC A / B / C / D / E / Answer E Topic
Concept . Skill Application/Analysis In a cross AaBbCc AaBbCc. E Epistasis is involved. F
offspring resulted from the cross. all F individuals have red. One parent has red. Two
truebreeding stocks of pea plants are crossed. Answer E Topic Concept . Skill
Application/Analysis Use the following information to answer the questions below. brown. If .
In a cross of a black female with a brown male. axial flowers. results can be either all black
puppies. D There is incomplete dominance. The genes for flower color and location assort
independently.
the F will be expected to be which of the following A Red and long B Red and oval C White
and long D Purple and long E Purple and oval Answer D Topic Concept . with long being the
dominant characteristic. which of the following phenotypic ratios would be expected A B C D
E Answer E Topic Concept . How many genes must be responsible for these coat colors in
Labrador retrievers A B C D Answer B Topic Concept . purple. the results were as follows /
black / yellow / brown The genotype aabb must result in which of the following A Black B
Brown C Yellow D A lethal result Answer C Topic Concept . Skill Application/Analysis In one
type cross of black black. . If truebreeding red long radishes are crossed with true breeding
white oval radishes. The part of the radish we eat may be oval or long. If a homozygous bent
wing fly is mated with a homozygous vestigial wing fly. Skill Application/Analysis In the F
generation of the above cross. A cross between a redflowered plant and a whiteflowered
plant yields allpurple offspring. Skill Application/Analysis Drosophila fruit flies usually have
long wings but mutations in two different genes can result in bent wings bt or vestigial wings
vg. Skill Application/Analysis Use the following information to answer the questions below.
Radish flowers may be red. or white. which of the following offspring would you expect A All
bt vg heterozygotes B / bent and / vestigial flies C All homozygous flies D / bent to / vestigial
ratio E / bent and vestigial to / normal Answer A Topic Concept .
B multiple alleles. C incomplete dominance. Skill Knowledge/Comprehension In
snapdragons. E pleiotropy. Answer C Topic Concept .Skill Application/Analysis The flower
color trait in radishes is an example of which of the following A A multiple allelic system B
Sex linkage C Codominance D Incomplete dominance E Epistasis Answer D Topic Concept .
heterozygotes for one of the genes have pink flowers. D polygenic inheritance. whereas
homozygotes have red or white flowers. Skill Application/Analysis A phenotypic ratio in the F
generation of a monohybrid cross is a sign of A complete dominance. When plants with red
flowers are crossed with plants with white flowers. Skill Application/Analysis . what proportion
of the offspring will have pink flowers A B C D E Answer E Topic Concept .
How many different types of gametes would be possible in this system A B C D E Answer C
Topic Concept . Tallness T in snapdragons is dominant to dwarfness t. Which of the following
crosses would produce offspring in the ratio of red roan white A red white B roan roan C
white roan D red roan E The answer cannot be determined from the information provided. A
dwarf. The heterozygous condition results in pink Rr flower color. roan coat color mixed red
and white hairs occurs in the heterozygous Rr offspring of red RR and white rr homozygotes.
red snapdragon is crossed with a plant homozygous for tallness and white flowers. What are
the genotype and phenotype of the F individuals A ttRrdwarf and pink B ttrrdwarf and white C
TtRrtall and red D TtRrtall and pink E TTRRtall and red Answer D Topic Concept . Answer B
Topic Concept . Skill Application/Analysis Skin color in a certain species of fish is inherited
via a single gene with four different alleles. Skill Knowledge/Comprehension In cattle. while
red R flower color is dominant to white r. Skill Application/Analysis .
the F would segregate in which of the following ratios A sharpspined spineless B
sharpspined dullspined spineless C sharp spined dullspined spineless D sharpspined
dullspined E sharpspined dullspined spineless Answer E Topic Concept . E codominance. B
epistasis. N. D pleiotropy. At the same time. Skill Knowledge/Comprehension A cross
between a truebreeding sharpspined cactus and a spineless cactus would produce A all
sharpspined progeny. E It is impossible to determine the phenotypes of the progeny. Skill
Application/Analysis . Answer A Topic Concept . C complete dominance. determines whether
cactuses have spines. The relationship between genes S and N is an example of A
incomplete dominance. Gene S controls the sharpness of spines in a type of cactus.Refer to
the following to answer the questions below. S. Homozygous recessive nn cactuses have no
spines at all. Skill Application/Analysis If doubly heterozygous SsNn cactuses were allowed
to selfpollinate. C sharpspined. Cactuses with the dominant allele. have sharp spines. a
second gene. Answer B Topic Concept . spineless progeny D all spineless progeny.
dullspined. dullspined progeny. B sharpspined. whereas homozygous recessive ss cactuses
have dull spines.
they produced offspring. YYBB. yyBB or yyBb is blue. Skill Application/Analysis Use the
following information to answer the questions below. YyBB. Which of the following results is
not possible A Green offspring only B Yellow offspring only C Blue offspring only D Green
and yellow offspring E a ratio Answer D Topic Concept . Which of the following is a possible
partial genotype for the son A IBIB B IBIA C ii D IBi E IAIA Answer D Topic Concept . one for
pigment on the outside and one for the inside of the feather. or YYBb is green.Use the
information given here to answer the following questions. 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 . of which were white. Over the years.
respectively. Skill Application/Analysis Two blue budgies were crossed. A third gene for the
MN blood group has codominant alleles M and N. and yybb is white. A woman who has
blood type A positive has a daughter who is type O positive and a son who is type B
negative. A blue budgie is crossed with a white budgie. YYbb or Yybb is yellow. Feather
color in budgies is determined by two different genes Y and B. Rh positive is a trait that
shows simple dominance over Rh negative and is designated by the alleles R and r. Skill
Application/Analysis .
Skill Knowledge/Comprehension . B Each parent must be type M. Skill Application/Analysis
Which of the following is a possible phenotype for the father A A negative B O negative C B
positive D AB negative E Impossible to determine Answer C Topic Concept . C Both children
are heterozygous for this gene. 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 . which of the following is possible A
Each parent is either M or MN. E The MN blood group is recessive to the ABO blood group.
Answer A Topic Concept . Skill Application/Analysis Which describes the ability of a single
gene to have multiple phenotypic effects A Incomplete dominance B Multiple alleles C
Pleiotropy D Epistasis Answer C Topic Concept . Skill Application/Analysis 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 . D Neither parent can have the N allele. Concept . Skill
Application/Analysis If both children are of blood group MM..
white eyes can be due to an Xlinked gene or to a combination of other genes. Answer C
Topic Concept . D In Drosophila fruit flies. Some of the plants produce blue flowers and
others pink flowers. Skill Knowledge/Comprehension Cystic fibrosis affects the lungs. Which
of the following terms best describes this A Incomplete dominance B Multiple alleles C
Pleiotropy D Epistasis Answer C Topic Concept . Which describes the ABO blood group
system A Incomplete dominance B Multiple alleles C Pleiotropy D Epistasis Answer B Topic
Concept . and other organs. Skill Knowledge/Comprehension Which of the following terms
best describes when the phenotype of the heterozygote differs from the phenotypes of both
homozygotes A Incomplete dominance B Multiple alleles C Pleiotropy D Epistasis Answer A
Topic Concept . C In rabbits and many other mammals. B The allele b produces a dominant
phenotype. the pancreas. although b through b do not. one genotype cc prevents any fur
color from developing. Skill Knowledge/Comprehension Which of the following provides an
example of epistasis A Recessive genotypes for each of two genes aabb results in an albino
corn snake. Skill Knowledge/Comprehension Which of the following is an example of
polygenic inheritance A Pink flowers in snapdragons B The ABO blood groups in humans C
Huntingtons disease in humans D White and purple flower color in peas E Skin pigmentation
in humans Answer E Topic Concept . . Skill Knowledge/Comprehension Hydrangea plants of
the same genotype are planted in a large flower garden. the digestive system. resulting in
symptoms ranging from breathing difficulties to recurrent infections. This can be best
explained by which of the following 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 .
E More than two alleles in a genotype is lethal. E /. B Most of the alleles will never be found
in a liveborn organism. what is the probability that his or her child will have the disease A B /
C / D / E Answer C Topic Concept .Skill Application/Analysis Most genes have many more
than two alleles. C All of the alleles but one will produce harmful effects if homozygous. Skill
Synthesis/Evaluation Huntingtons disease is a dominant condition with late age of onset in
humans. Skill Knowledge/Comprehension . D /. Skill Knowledge/Comprehension A woman
has six sons. The chance that her next child will be a daughter is A . C /. Answer C Topic
Concept . If one parent has the disease. D There may still be only two phenotypes for the
trait. Answer D Topic Concept . B . However. which of the following is also true A At least
one allele for a gene always produces a dominant phenotype.
Skill Application/Analysis What is the probability that individual III is Ww A / B / C / D / E
Answer E Topic Concept . for a family.The following questions refer to the pedigree chart in
Figure . wooly hair. some of whose members exhibit the dominant trait. Skill
Application/Analysis . Skill Application/Analysis What is the likelihood that the progeny of IV
and IV will have wooly hair A B C D E Answer C Topic Concept . Affected individuals are
indicated by an open square or circle. What is the genotype of individual II A WW B Ww C
ww D WW or ww E ww or Ww Answer C Topic Concept . Figure .
E All of the above Answer E Topic Concept . Answer A Topic Concept . it means that A both
genetic and environmental factors contribute to the disease. D it has many different
symptoms. which is the best use of this discovery A To screen all newborns of an atrisk
population B To design a test for identifying heterozygous carriers of the allele C To
introduce a normal allele into deficient newborns D To follow the segregation of the allele
during meiosis E To test schoolage children for the disorder Answer B Topic Concept . C
have increased resistance to malaria. Skill Synthesis/Evaluation . This and only this allele.
produces an effect that results in death at or about the time of birth. Of the following. if
homozygous. Skill Synthesis/Evaluation A scientist discovers a DNAbased test for the allele
of a particular gene. E it tends to skip a generation. People with sicklecell trait A are
heterozygous for the sicklecell allele. B it is caused by a gene with a large number of alleles.
C it affects a large number of people. Skill Knowledge/Comprehension When a disease is
said to have a multifactorial basis. Skill Knowledge/Comprehension An ideal procedure for
fetal testing in humans would have which of the following features A Lowest risk procedure
that would provide the most reliable information B The procedure that can test for the
greatest number of traits at once C A procedure that provides a D image of the fetus D The
procedure that can be performed at the earliest time in the pregnancy E A procedure that
could test for the carrier status of the fetus Answer A Topic Concept . B are usually healthy.
D produce normal and abnormal hemoglobin.
C Regulate the diet of the affected persons to severely limit the uptake of the amino acid. D
Feed the patients the missing enzymes in a regular cycle. i. An obstetrician knows that one
of her patients is a pregnant woman whose fetus is at risk for a serious disorder that is
detectable biochemically in fetal cells. Skill Synthesis/Evaluation . presumably because this
reduces the frequency of malaria. B Transfuse the patients with blood from unaffected
donors. The probability of their next child being normal for this characteristic is which of the
following A B / C / D / E / Answer C Topic Concept . This amino acid is not otherwise
produced by humans. Skill Application/Analysis Phenylketonuria PKU is a recessive human
disorder in which an individual cannot appropriately metabolize a particular amino acid. twice
per week. Answer C Topic Concept . Therefore the most efficient and effective treatment is
which of the following A Feed them the substrate that can be metabolized into this amino
acid. Skill Synthesis/Evaluation Cystic fibrosis CF is a Mendelian disorder in the human
population that is inherited as a recessive.e. Such a relationship is related to which of the
following A Mendels law of independent assortment B Mendels law of segregation C Darwins
explanation of natural selection D Darwins observations of competition E The malarial
parasite changing the allele Answer C Topic Concept . Skill Synthesis/Evaluation The
frequency of heterozygosity for the sickle cell anemia allele is unusually high. The
obstetrician would most reasonably offer which of the following procedures to her patient A
CVS B Ultrasound imaging C Amniocentesis D Fetoscopy E Xray Answer C Topic Concept .
Two normal parents have two children with CF..
Which of the following represents the most likely assumption A All cases must occur in
relatives. there must be only one mutant allele. If a young child is the first in her family to be
diagnosed. Skill Synthesis/Evaluation . who look very old even as children. at an average
age of approximately . Skill Synthesis/Evaluation A pedigree analysis for a given disorders
occurrence in a family shows that. HutchinsonGilford progeria is an exceedingly rare human
genetic disorder in which there is very early senility. although both parents of an affected
child are normal. each of the parents has had affected relatives with the same condition.
Answer B Topic Concept . The disorder is then which of the following A Recessive B
Dominant C Incompletely dominant D Maternally inherited E A new mutation Answer A Topic
Concept . Skill Application/Analysis One of two major forms of a human condition called
neurofibromatosis NF is inherited as a dominant. Answer C Topic Concept . B One of the
parents has very mild expression of the gene. C The condition skipped a generation in the
family. which of the following is the best explanation A The mother carries the gene but does
not express it at all. and death. do not live to reproduce. Patients. D Each patient will have
had at least one affected family member in a previous generation. D The child has a different
allele of the gene than the parents. C The disorder may be due to mutation in a single
proteincoding gene. B Successive generations of a family will continue to have more and
more cases over time. usually of coronary artery disease. therefore. although it may be either
mildly to very severely expressed. E The disease is autosomal dominant.
Gene Allele Character Trait Dominant allele Recessive allele Genotype Phenotype
Homozygous Heterozygous Testcross Monohybrid cross H Determines phenotype in a
heterozygote I A heritable feature that varies among individuals J The genetic makeup of an
individual K A cross between individuals heterozygous for a single character L Has no effect
on phenotype in a heterozygote G J E F I C A D H B L K G A heritable unit that determines a
character and can exist in different forms.SelfQuiz Questions Match each term on the left
with a statement on the right. C Having two identical alleles for a gene D Having two different
alleles for a gene E An alternative version of a gene F An organisms appearance or
observable traits A A variant for a character B A cross between an individual with an
unknown genotype and a homozygous recessive individual .
stem length. 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 . redflowered strain gives all pink flowers when
crossed with a whiteflowered strain CRCR red CWCW white CRCW pink. D . B . and seed
shape were three characters that Mendel studied. phenotype is all axialpink. what proportion
of the offspring would you expect to be as follows Note Use the rules of probability instead of
a huge Punnett square. Phenotypes of F are axialpink axialred axialwhite terminalpink
terminalwhite terminal red. Genotype of F is AaCRCW. Answer In some plants. Two pea
plants heterozygous for the characters of pod color and pod shape are crossed. in your
textbook. what will be the ratios of genotypes and phenotypes of the F generation resulting
from the following cross axialred truebreeding terminalwhite What will be the ratios in the F
generation Answer Parental cross is AACRCR aaCWCW. Flower position. Draw a Punnett
square to determine the phenotypic ratios of the offspring. a truebreeding. C . Genotypes of
F are AaCRCW AaCRCR AACRCW aaCRCW AaCWCW AACRCR aaCRCR AACWCW
aaCWCW. If flower position axial or terminal is inherited as it is in peas see Table . Each is
controlled by an independently assorting gene and has dominant and recessive expression
as follows Character Dominant Recessive Flower position Axial A Terminal a Stem length
Tall T Dwarf t Seed shape Round R Wrinkled r If a plant that is heterozygous for all three
characters is allowed to selffertilize.
and threepod. normal leaf and threepod. Answer A . Other genotypes for children are IAIB.
wrinkled leaf Answer A PPLl PPLl. and in what frequencies. When the albino is crossed with
a second black one. onepod. Answer Albino b is a recessive trait. E / . E PpLl PpLl.
Assuming independent assortment of these four genes. C PPLL any of the possible
genotypes or PPll ppLL. C All three children have the disease. woman IBi. Determine the
genotypes for the two parents for all possible matings producing the following offspring A
onepod. and offspring. and normal leaf L is dominant to wrinkled leaf l. wrinkled leaf E
onepod. wrinkled leaf. offspring all Bb black coat. offspring Bb and / bb. PPLl PpLl. normal
leaf. onepod. wrinkled leaf. normal leaf. normal leaf. blacks and albinos are obtained. black B
is dominant. What are the genotypes of these individuals What other genotypes. C . B ppLl
ppLl. In sesame plants. gametes. A man with type A blood marries a woman with type B
blood. If a woman and her husband. First cross parents BB bb. normal leaf D onepod.
Second cross parents Bb bb. D . A black guinea pig crossed with an albino guinea pig
produces black offspring. normal leaf and onepod. have three children. Pod type and leaf
type are inherited independently. the onepod condition P is dominant to the threepod
condition p. gametes B and b. normal leaf. C . would you expect in offspring from this
marriage Answer Man IAi. What is the best explanation for this genetic situation Write
genotypes for the parents. or PPLl ppLl. threepod. wrinkled leaf B threepod. IAi. Their child
has type O blood. D At least one child is phenotypically normal. and threepod. wrinkled leaf
C onepod. D PpLl Ppll. B . who are both carriers. Phenylketonuria PKU is an inherited
disease caused by a recessive allele. child ii. IBi. what is the probability of each of the
following A All three children are of normal phenotype. what are the probabilities that F
offspring will have the following genotypes A aabbccdd B AaBbCcDd C AABBCCDD D
AaBBccDd E AaBBCCdd Answer A . B One or more of the three children have the disease.
gametes B and b heterozygous parent and b. D The genotype of F individuals in a
tetrahybrid cross is AaBbCcDd. B . Note Remember that the probabilities of all possible
outcomes always add up to . threepod.
Overall. while the homozygous recessive genotype pp causes red kernels. and thus the
genotypic ratio for the F generation will be IP colorless Ipp colorless iiP purple iipp red. all of
the crosseyed offspring will also be white. recessively inherited disease is expressed only in
individuals with type O blood. while the recessive allele i permits color when homozygous. A
AABBCC aabbcc AaBbCc B AABbCc AaBbCc AAbbCC C AaBbCc AaBbCc AaBbCc D
aaBbCC AABbcc AaBbCc Answer A . and none of them have been tested to reveal sicklecell
trait. C . a stray black cat with unusual rounded. what percentage of their offspring will be
crosseyed What percentage will be white Answer will be crosseyed. B . but only noncurl
offspring if the curl allele is recessive. As it turns out. curledback ears was adopted by a
family in California. Neither Karen nor Steve nor any of their parents have the disease. You
would obtain some truebreeding offspring homozygous for the curl allele from matings
between the F cats resulting from the original curl noncurl crosses whether the curl trait is
dominant or recessive. A normal man with type A blood and a normal woman with type B
blood have already had one child with the disease. calculate the probability that if this couple
has a child. At a different locus. If two phenotypically normal tigers that are heterozygous at
this locus are mated. What is the probability that the second child will also have the disease
Assume that both parents are heterozygous for the gene that causes the disease. the allele
that causes curled ears is dominant. what will be the phenotypic ratio of the offspring Answer
The dominant allele I is epistatic to the P/p locus. Answer / In . Suppose you owned the first
curl cat and wanted to develop a truebreeding variety. In corn plants. . the phenotypic ratio is
colorless purple red. and cat fanciers hope to develop the curl cat into a show breed. the
dominant allele P causes purple kernel color. Answer / In tigers. D Karen and Steve each
have a sibling with sicklecell disease. You know that cats are truebreeding when curl curl
matings produce only curl offspring. the child will have sicklecell disease. The woman is now
pregnant for a second time. How would you determine whether the curl allele is dominant or
recessive How would you obtain truebreeding curl cats How could you be sure they are
truebreeding Answer Matings of the original mutant cat with truebreeding noncurl cats will
produce both curl and noncurl F offspring if the curl allele is dominant. If plants heterozygous
at both loci are crossed. Imagine that a newly discovered. a recessive allele causes an
absence of fur pigmentation a white tiger and a cross eyed condition. Hundreds of
descendants of the cat have since been born. Based on this incomplete information.
although the disease and blood group are independently inherited. 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 dominant allele I inhibits kernel color.
Does alkaptonuria appear to be caused by a dominant allele or by a recessive allele Fill in
the genotypes of the individuals whose genotypes can be deduced. and Alan are each Aa.
and he and his first wife had a child with cystic fibrosis. and Christopher can each have the
AA or Aa genotype. Charles was married once before. black color B is dominant to white b.
since they are all unaffected children with one affected parent. Elaine. The pedigree below
traces the inheritance of alkaptonuria. Tina. Sam. What genotypes are possible for each of
the other individuals Answer Recessive. Ann. Affected individuals. what is the expected
phenotypic ratio of their offspring Answer BA agouti Baa black bbA white bbaa white. died of
cystic fibrosis. What fraction of this couples children would be expected to have extra digits
Answer / Imagine that you are a genetic counselor. What is the probability that Charles and
Elaine will have a baby with cystic fibrosis Neither Charles nor Elaine has cystic fibrosis. .
Expression of the recessive allele a results in a solid coat color. since he has an affected
child Carla with his heterozygous wife Ann. George is Aa. which colors the urine and stains
body tissues. All affected individuals Arlene. At a different locus. Tom. Overall. Answer / In
mice. A man has six fingers on each hand and six toes on each foot. and a couple planning
to start a family comes to you for information. The brother of his current wife. and Carla are
homozygous recessive aa. Michael also is Aa. Extra digits is a dominant trait. If mice that are
heterozygous at both loci are crossed. are unable to metabolize a substance called alkapton.
Wilma. His wife and their daughter have the normal number of digits. Sandra. since some of
his children with Arlene are affected. indicated here by the colored circles and squares. a
biochemical disorder. This gives a frosted appearance known as agouti. agouti black white. a
dominant allele A produces a band of yellow just below the tip of each hair in mice with black
fur. Daniel.