Multiple Choice: DARKEN the letter that best completes the statement
or answers the question.
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The pedigree below shows a family’s pedigree for Huntington’s
disease.
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1.
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Which family members are afflicted with Huntington’s disease?
A | I-1, II-2, II-3, II-7, III-3 | B | I-2, II-1, II-4, II-6, II-8 | C | II-2, II-3,
III-1 | D | II-3, II-7, II-8, III-4 |
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2.
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There are no carriers for Huntington’s disease. You either have it or you
don’t. With this in mind, what type of inheritance does Huntington’s disease
follow?
A | autosomal recessive | C | sex-linked | B | autosomal dominant | D | codominant |
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3.
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How many children did individuals I-1 and I-2 have?
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4.
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How many girls did II-1 and II-2 have? How many are affected?
A | 2 girls, none with Huntington’s disease | B | 2 girls, one with
Huntington’s disease | C | 3 girls, two with Huntington’s
disease | D | 3 girls, one with Huntington’s disease | E | 3 girls, all with
Huntington’s disease |
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5.
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How are individuals III-2 and II-4 related?
A | daughter and mother | B | nephew and aunt | C | grandson and
grandfather | D | niece and uncle | E | granddaughter and
grandfather |
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6.
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How are individuals I-2 and III-5 related?
A | grandma and granddaughter | D | father and
daughter | B | grandma and grandson | E | mother and daughter | C | mother and son |
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The pedigree below shows a family’s pedigree for colorblindness:
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7.
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Which sex can be carriers of colorblindness and not have it?
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8.
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What kind of trait is colorblindness?
A | autosomal dominant | D | codominant | B | autosomal recessive | E | sex-linked recessive | C | sex-linked
dominant |
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9.
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What is the genotype of individual IV-7?
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10.
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All of the following statements about colorblindness are true, EXCEPT:
A | Males are usually more severely affected than females. The trait may be lethal in
males. | B | Parents of affected children may be related. The rarer the trait in the general
population, the more likely a consanguineous mating is involved. | C | All sons of an
affected male and a normal female are normal. All daughters of an affected male and a normal female
are affected. | D | The trait is never passed from father to son. | E | Matings of affected
females and normal males produce 1/2 the sons affected and 1/2 the daughters
affected. |
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11.
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Why do all the daughters in generation II carry the colorblind gene?
A | Their mother is a carrier with the Xb gene | B | Their dad is affected
with only Xb gene to give to them | C | Their dad is homozygous dominant for the
disorder | D | Their mom is homozygous recessive for the disorder | E | Their mom is
heterozygous for the disorder |
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12.
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Name two fourth generation colorblind males.
A | IV-1 and IV-7 | D | IV-3 and Iv-6 | B | IV-5 and IV-7 | E | IV-1 and IV-5 | C | IV-1 and
IV-3 |
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In the pedigree below, different blood types are identified by the letters A,
B, AB, and O.
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13.
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Individuals II-4 and II-5 have just had identical twin girls. List the possible
blood types these infants may have based on the information provided in the pedigree
A | Types A, B, AB and O | D | Types A and AB | B | Types A and B | E | Type O | C | Types B and
AB |
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14.
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Individuals II-6 and II-7 have a second child with blood type O. What
does this tell you about II-6’s genotype?
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15.
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Could I-1 and I-2 have a child with the AB blood type?
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16.
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Which member of the family can only get blood from a type O donor?
A | I-2 | D | I-4, II-3, II-4, II-6 | B | I-3, II-1, II-2, II-5 | E | the twins | C | I-1 and
II-7 |
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17.
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Which blood type can receive from type A, B, AB and O?
A | type A | C | type AB | B | type B | D | type O |
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18.
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The Punnett square in Figure 7.1 shows a cross between two parents who have the
genotype Ss for a genetic disorder caused by a recessive allele. Which of the following will
have the genetic disorder?
A | Ss parent | B | Ss offspring | C | SS
offspring | D | ss offspring |
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19.
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Figure 8.1 shows a single strand of DNA. Choose the first three nucleotides of
the other DNA strand.
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20.
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Figure 8.2 shows a single strand of DNA. Choose the first three nucleotides of
the complementary RNA strand.
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21.
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Normal human males develop from fertilized eggs containing which of the
following sex chromosome combinations?
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22.
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Homologous chromosomes are pairs of chromosomes containing genes that code
for
A | different traits. | C | DNA. | B | the same traits. | D | sex. |
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23.
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In humans, gametes contain
A | 22 autosomes and 1 sex chromosome. | B | 1 autosome and 22 sex
chromosomes. | C | 45 autosomes and 1 sex chromosome. | D | 1 autosome and 45 sex
chromosomes. |
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24.
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In humans, the male determines the sex of the child because males have
A | two X chromosomes. | C | two Y chromosomes. | B | one X and one Y chromosome. | D | 46 chromosomes. |
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25.
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female : XX ::
A | female : gametes | C | male : YY | B | female : eggs | D | male : XY |
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In humans, having freckles ( F) is dominant to not having freckles
( f). The inheritance of these traits can be studied using a Punnett square similar to the one
shown below.
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26.
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Refer to the illustration above. The child represented in box 1 in the Punnett
square would
A | be homozygous for freckles. | B | have an extra freckles
chromosome. | C | be heterozygous for freckles. | D | not have
freckles. |
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27.
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Refer to the illustration above. The parents shown in the Punnett square could
have children with a phenotype ratio of
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28.
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Refer to the illustration above. Which box in the Punnett square represents a
child who does not have freckles?
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29.
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Refer to the illustration above. The child in box 3 of the Punnett square has
the genotype
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In rabbits, black fur ( B) is dominant to brown fur ( b). Consider
the following cross between two rabbits.
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30.
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Refer to the illustration above. The device shown, which is used to determine
the probable outcome of genetic crosses, is called a
A | Mendelian box. | C | genetic graph. | B | Punnett square. | D | phenotypic
paradox. |
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31.
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Refer to the illustration above. Both of the parents in the cross are
A | black. | C | homozygous dominant. | B | brown. | D | homozygous
recessive. |
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32.
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Refer to the illustration above. The phenotype of the offspring indicated by box
3 would be
A | brown. | C | a mixture of brown and black. | B | black. | D | white. |
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33.
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Refer to the illustration above. The genotypic ratio of the F1 generation would
be
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Hemophilia is a rare hereditary human disease of the blood. The blood of
individuals with this condition does not clot properly. Without the capacity for blood clotting, even
a small cut can be lethal. In a marriage of two non-hemophiliac parents, a bleeder son is
born.
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34.
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What is the genotype of the father?
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35.
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What is the genotype of the hemophiliac son?
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36.
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What is the genotype of the mother?
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37.
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What is the genotype of the hemophiliac daughter?
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Hemophilia is a sex-linked trait found on the X chromosome. To get this
disorder, a person must have a recessive copy of the gene (h) on every X chromosome. A woman who is a
carrier for the disease had a baby with a man who had the disease.
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38.
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What is the genotype of the woman?
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39.
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What is the genotype of the man?
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40.
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Predict the genotypic probabilities of the offspring in this marriage?
A | 25% XHY, 50% XHXh, 25%
XhY | B | 25% XhXh, 25% XHXh, 25% XHY,
25% XhY | C | 25% XHXH, 25%
XhXh, 25% XHY, 25% XhY | D | 50% XHY,
50% XHXh |
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41.
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Predict the phenotypic probabilities of the offspring in this marriage?
A | 50% of the offspring are affected, 50% of them are normal | B | 100% of the offspring
are affected | C | 100% of the offspring are normal | D | 25% of the offspring are affected, 75% of them
are normal |
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42.
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How many phenotypes would result from the following Punnett square, assuming the
trait has a pattern of complete dominance?
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The pedigree below shows a family’s pedigree for Hitchhiker’s thumb
versus having a straight thumb.
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43.
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Based on the pedigree, what inheritance pattern characterizes Hitchhiker’s
thumb?
A | autosomal dominant | D | codominant traits | B | autosomal recessive | E | polygenic inheritance | C | sex-linked
trait |
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44.
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What is the basis for the inheritance exhibited in Hitchhiker’s
thumb?
A | The genes are never passed from father to son. | B | About 1/2 of the
offspring of an affected individual are affected. | C | Males are usually more severely affected than
females | D | Each affected individual has an affected parent. | E | There is skipping of
generations. |
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45.
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How are individuals III-4 and III-5 related?
A | grandparents-grandchild | D | cousins | B | uncle - niece | E | siblings | C | father -
daughter |
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46.
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How would you name the two individuals that have hitchhiker’s
thumb?
A | IV-2 and IV-4 | D | III-4 and III-5 | B | I-1 and I-2 | E | III-1 and III-2 | C | II-3 and
II-4 |
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47.
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Is it possible for individual IV-3 to be a carrier?
A | yes | D | it depends on her father | B | no | E | only if it skips generation | C | highly
unlikely |
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48.
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Identify the two individuals that were carriers of Hitchhiker
thumb?
A | IV-2 and IV-4 | D | III-4 and III-5 | B | II-3 and II-4 | E | I-1 and I-2 | C | II-5 and
II-6 |
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49.
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How are individuals III-1 and III-2 related?
A | They are sisters | C | They are cousins | B | They are brothers | D | They are married |
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Figure A:
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50.
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What is the genotype of the father in Figure A?
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51.
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What is the genotype of the mother in Figure A?
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The energy pyramid below shows the flow of energy through the organisms in a
kelp forest ecosystem in the Pacific Ocean. Use the energy pyramid to answer the following
questions.
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52.
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What is the lowest level of the energy pyramid that contains
carnivores?
A | level 1 | B | level 2 | C | level
3 | D | level 4 |
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53.
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In humans, the allele for having a hairline with a widow’s peak (H) is
dominant to having a straight hairline (h). A woman that is
homozygous dominant for having a widow’s peak and a man that is heterozygous for having a
widow’s peak have a child. What is the probability that their child will have a widow’s
peak?
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Use the information and the figure below to answer the following
question.
The figure below shows the skeletal structure of a seal’s flipper and a
monkey’s arm. The skeletal structures of the flipper and
the arm are similar, even though they have different functions. Seals use their flippers for
swimming, while monkeys use their arms primarily for grasping and lifting.
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54.
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The seal’s flipper and the monkey’s arm differ in appearance. This
difference is the result of
A | migration | B | genetic engineering | C | succession | D | natural
selection |
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55.
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Which of these explains why the skeletal structures of the seal’s flipper
and the monkey’s arm are similar?
A | Seals and monkeys have a common ancestor. | B | Seals and monkeys
have identical DNA sequences. | C | All of the same genetic mutations occurred in
seals and monkeys. | D | All of the same vitamins are used for bone
formation in seals and monkeys |
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56.
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If a characteristic is sex-linked, the gene for it is found on
A | a sex chromosome. | C | a linked chromosome. | B | an autosome. | D | an allele. |
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57.
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Since the allele for colorblindness is located on the X chromosome,
colorblindness
A | cannot be inherited. | C | is sex-linked. | B | occurs only in adults. | D | occurs only in
females. |
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58.
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What would be the blood type of a person who inherited an A allele from
one parent and an O allele from the other?
A | type A | C | type AB | B | type B | D | type O |
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59.
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The ears of foxes help to regulate body heat. The fennec fox lives in the
North African desert and has large ears that release body heat. The Arctic fox lives in cold climates
and has small ears that conserve body heat. Which of these processes
led to the development of different ear sizes in foxes?
A | selective breeding | B | succession | C | natural
selection | D | mutualism |
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60.
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A student designed the chart below to classify different organisms into four
groups.
According to the student’s
classification chart, an organism with no wings and four legs would belong to which
group?
A | Group A | B | Group B | C | Group
C | D | Group D |
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61.
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In humans, hemophilia is a sex-linked trait. Females can be normal, carrier or
have the disease. Males will either have the disease or not. Which of the following shows the cross
between a female carrier and a hemophilic male?
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Use the the information and the pedigree below to answer the next three (3)
questions.
In humans, the allele for having feet with normal arches is dominant (A). The
allele for flat feet is recessive (a). The pedigree below shows the occurrence of normal arches
and flat feet in four generations of a family. In the pedigree, individuals are identified by the
generation and individual numbers. For example, Individual 2 in Generation I is identified as
I-2.
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62.
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Which of these individuals in the pedigree is a male with the genotype
aa?
A | Individual I-1 | C | Individual III-2 | B | Individual II-2 | D | Individual III-5 |
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63.
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Individuals III-6 and III-7 have two children and are expecting a third child.
Their two children have flat feet. What is the chance that the third child will have normal
arches?
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64.
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Which of these Punnett squares shows the cross between Individual II-4 and
Individual II-5?
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Use the information below to answer the following
item.
Scientists determined that excess fertilizer from farms entered a shallow lake.
The fertilizer caused an increase in aquatic plants in the lake and then a decrease in oxygen
in the water. Next, organic debris collected on the bottom of the lake. Over several years, the
lake gradually filled in with organic sediment.
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65.
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One species of aquatic plant found in the lake has 84 chromosomes in each cell.
As nutrient levels increased, the population of this species increased through vegetative
reproduction. How many chromosomes were in the cells of the offspring?
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66.
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A diagram in which several generations of a family and the occurrence of certain
genetic characteristics are shown is called a
A | Punnett square. | C | pedigree. | B | monohybrid cross. | D | family
karyotype. |
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Mrs. Clink is type A and Mr. Clink is type O. They have three children named
Matthew, Mark, and Luke. Mark is type O, Matthew is type A, and Luke is type AB.
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67.
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Based on the information above, Mr. Clink must have the genotype:
A | IAIA | D | IAi | B | IBIB | E | IBi | C | IAIB | F | ii |
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68.
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Based on the information above, Mrs. Clink must have the genotype:
A | IAIA | D | IAi | B | IBIB | E | IBi | C | IAIB | F | ii |
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69.
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Based on the information above, Luke must have the genotype:
A | IAIA | D | IAi | B | IBIB | E | IBi | C | IAIB | F | ii |
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70.
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_________ cannot be the child of these parents because neither parent has the
allele.
A | Luke | D | Luke and Mathew | B | Mathew | E | Mathew and Mark | C | Mark | F | Mark
and Luke |
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