Multiple Choice
Identify the choice that best completes the statement or
answers the question.
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1.
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Mendel verified true-breeding pea plants for certain traits
before undertaking his experiments. The term "true-breeding" refers to:
a. | genetically pure lines. | b. | organisms that have a high rate of reproduction. | c. | organisms that will produce identical copies of themselves upon
reproduction. | d. | organisms that are
heterozygous for a given trait. | e. | organisms that are
homozygous for all possible traits. |
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2.
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Mating a true-breeding pink rose plant with a true-breeding
pink rose plant will produce:
a. | plants with pink, red, and white
roses. | b. | only plants with pink roses. | c. | plants with red or white roses in a 3:1 ratio. | d. | plants with white or red roses in a 3:1 ratio. | e. | None of the above. |
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3.
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The term "dominant" means that:
a. | both alleles can be expressed in a
hybrid. | b. | all members of the F2 generation of a hybrid cross
exhibit the dominant phenotype. | c. | one allele can mask the
expression of another in a hybrid. | d. | the dominant phenotype
shows up in 100% of the offspring in all generations. | e. | the dominant phenotype is more beneficial than the recessive
phenotype. |
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4.
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Mendel's principle of segregation states
that:
a. | alleles from one parent mask the expression of alleles from the
other parent. | b. | alleles separate from each
other before forming gametes. | c. | hybrids will express a
phenotype intermediate between the two parental phenotypes. | d. | true-breeding parents produce offspring of the same
phenotype. | e. | different loci separate
from each other. |
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5.
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A pear plant with the genotype Aa can produce gametes
containing:
a. | either A or Aa. | b. | only the dominant A. | c. | only the recessive
a. | d. | either A or a. | e. | either AA, Aa or aa. |
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6.
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The physical appearance of an organism for a given trait is
termed:
a. | genetics. | b. | dominance. | c. | synapsis. | d. | genotype. | e. | phenotype. |
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7.
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The physical location of a particular gene on a chromosome
is called:
a. | an allele. | b. | a
locus. | c. | a trait. | d. | a
chromatid. | e. | None of the
above. |
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8.
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____________ are alternative forms of a gene that govern the
same feature, such as eye color, and occupy corresponding positions on homologous
chromosomes.
a. | Alleles | b. | Loci | c. | Homozygotes | d. | Coupled
traits | e. | None of the above. |
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9.
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Using standard conventions for naming alleles, which of the
following pairs is correct?
a. | Tt—recessive phenotype | b. | TT—heterozygous | c. | tt—homozygous | d. | tt—dominant
phenotype | e. | All of the above. |
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10.
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The separation of alleles of a gene takes place
during:
a. | anaphase of mitosis. | b. | cytokinesis of mitosis. | c. | anaphase I of
meiosis. | d. | telophase II of meiosis. | e. | cytokinesis of meiosis. |
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11.
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Which of the following represents the possible genotype(s)
resulting from a cross between an individual homozygous (BB) and one heterozygous (Bb)
individual?
a. | BB and Bb | b. | BB, Bb, and
bb | c. | BB only | d. | Bb
only | e. | bb only |
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12.
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Which of the following represents the possible genotype(s)
resulting from a cross between two individuals that are heterozygous (Bb)?
a. | BB and Bb | b. | BB, Bb, and
bb | c. | BB only | d. | Bb
only | e. | bb only |
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13.
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Which of the following represents the possible genotype(s)
resulting from a cross between an individual heterozygous (Bb) and one that is homozygous
(bb)?
a. | BB and Bb | b. | Bb, and
bb | c. | BB only | d. | Bb
only | e. | bb only |
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14.
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Mating an individual expressing a dominant phenotype, but
whose genotype is unknown, with an individual expressing the corresponding recessive phenotype is an
example of:
a. | a heterozygous cross. | b. | an F1 cross. | c. | an F2
cross. | d. | a parental cross. | e. | a test cross. |
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15.
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The genotype for a pea plant that is homozygous recessive
for both height and pea color would be:
a. | tt. | b. | YY. | c. | TtYy. | d. | ttyy. | e. | TTYY. |
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16.
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In peas, Mendel found that tall plants and yellow peas are
dominant. The phenotype for a pea plant with the genotype TTyy would be:
a. | heterozygous. | b. | Ty. | c. | short with yellow
peas. | d. | tall with green peas. | e. | tall with yellow peas. |
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17.
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In peas, Mendel found that tall plants and yellow peas are
dominant. The phenotype for a pea plant with the genotype TtYy would be:
a. | heterozygous. | b. | Ty. | c. | short with yellow
peas. | d. | tall with green peas. | e. | tall with yellow peas. |
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18.
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The height of pea plants from a cross between parent plants
heterozygous for height, in which tall is dominant, would be:
a. | all short. | b. | all
tall. | c. | 1 tall : 3 short. | d. | 2 short : 2 tall. | e. | 3 tall : 1
short. |
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19.
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Which of the following represents the possible genotype(s)
resulting from a cross between an individual homozygous for black hair (BB) and an individual
homozygous for blonde hair (bb)?
a. | BB and Bb | b. | BB, Bb, and
bb | c. | BB only | d. | Bb
only | e. | bb only |
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20.
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What is the probability that two lizards that are
heterozygous for stripes on their tails (Ss) will produce an offspring that is homozygous for no
stripes (ss)?
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21.
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In humans, assume that brown eyes is dominant and blue eyes
is recessive. If two brown-eyed individuals have a child with blue eyes, that means:
a. | both parents are homozygous for brown
eyes. | b. | both parents are heterozygous for eye
color. | c. | there is a 1/4 chance that their second child will have brown
eyes. | d. | there is a 50/50 chance that their second child will have blue
eyes. | e. | None of the above. |
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22.
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If a couple is planning on having two children, what is the
probability that both will be male?
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23.
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If a couple is planning on having three children, what is
the probability that only one will be male?
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24.
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A brown-eyed couple already has a child with blue eyes. What
is the probability that their next child will have blue eyes, assuming that brown eyes is dominant
and blue eyes is recessive?
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25.
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A couple has already had 3 girls with cystic fibrosis, and
were hoping to have a normal child for their fourth. What are the chances that the fourth child will
be a normal male?
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26.
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A brown-eyed couple heterozygous for eye color are planning
on having two children. What is the probability that both children will have blue eyes, assuming
brown eyes is dominant and blue eyes is recessive?
a. | 0 | b. | 1/32 | c. | 1/16 | d. | 1/4 | e. | 1/2 |
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27.
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Two Martians fall in love and marry. One Martian is
homozygous for red eyes and the other is heterozygous. The recessive eye color is purple. What is the
probability that they will have a child with purple eyes?
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28.
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Two Martians fall in love and marry. One Martian is
homozygous for red eyes and the other is heterozygous. The recessive eye color is purple. What are
the chances that the alien couple will have a child with red eyes?
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29.
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Two Martians fall in love and marry. One Martian is
homozygous for red eyes and the other is heterozygous. The recessive eye color is purple. What is the
probability that the alien couple will have a child that is heterozygous for eye color?
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30.
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The principle of independent assortment is not true
for:
a. | incomplete dominance. | b. | mutations. | c. | heterozygotes. | d. | homozygotes. | e. | X-linked
genes. |
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31.
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The probability that two genes will be separated by
crossing-over is related to:
a. | the phenotype that they control. | b. | how far the two genes are from the centromere. | c. | the distance between the two genes on the chromosome. | d. | whether the two genes are located on a sex chromosome. | e. | how far the genes are from the kinetochore. |
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32.
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Genes that tend to be inherited together are said to
be:
a. | associated. | b. | related. | c. | similar. | d. | linked. | e. | alleles. |
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33.
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A ______________ is best used to demonstrate the linkage of
two genes.
a. | monohybrid cross | b. | dihybrid cross | c. | monohybrid test
cross | d. | two-allele test cross | e. | two-point test cross |
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34.
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In genetics, map units express the distance
between:
a. | chromosomes during metaphase. | b. | two loci on a chromosome. | c. | alleles. | d. | polar
bodies. | e. | homologous chromosomes. |
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35.
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In a two-point test cross, 36 of the offspring were
recombinant types. The remaining 64 offspring were parental types. How many map units separate the
two loci?
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36.
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The sex of most mammals, birds, and insects is determined
by:
a. | the temperature. | b. | the environment. | c. | sex
chromosomes. | d. | chance. | e. | None of the
above. |
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37.
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The sex of a human is determined by:
a. | the number of chromosomes. | b. | the number of autosomes. | c. | the presence of only one X
chromosome. | d. | the number of sex
chromosomes. | e. | the presence of Y
chromosome. |
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38.
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The offspring of two heterozygous gray-bodied, normal-winged
flies should be 50% gray-bodied/normal wings (BbRr) and 50% black-bodied/vestigial wings (bbrr)
because these alleles are linked. If a small number, say 15%, of the offspring are instead
black-bodied with normal wings, this is most likely the result of:
a. | crossing-over. | b. | incomplete dominance. | c. | codominance. | d. | an error in
meiosis. | e. | mutation. |
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39.
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An organism with the genotype of AaXx can produce gametes
containing _________ if the two genes are unlinked.
a. | either Aa or Xx | b. | either AX, Ax, aX, ax | c. | AaXx | d. | AX or
ax | e. | None of the above. |
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40.
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A lizard with a striped tail is crossed with one having a
spotted head, producing normal looking (no stripes or spots) lizard progeny. What progeny would be
expected to be produced by mating these progeny with each other, if the genes conferring stripes and
spots were on different chromosomes?
a. | equal numbers of normal, striped, spotted and striped and
spotted | b. | 3 striped : 1 spotted | c. | 9 striped and spotted : 3 spotted : 3 striped : 1 normal | d. | 9 striped : 3 spotted : 1 striped or spotted | e. | 9 normal : 3 striped : 3 spotted : 1 striped and
spotted |
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41.
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A lizard with striped tails is crossed with one having a
spotted head, producing normal looking (no stripes or spots) progeny. What progeny would be expected
to be produced by mating one of these lizards with another that had a striped tail and spotted head,
if the genes conferring stripes and spots were close together on the same chromosome?
a. | equal numbers of normal, striped, spotted and striped and
spotted | b. | 3 striped : 1 spotted | c. | mostly progeny that are striped or spotted | d. | mostly progeny that are normal or striped and spotted | e. | 9 normal : 3 striped : 3 spotted : 1 striped and
spotted |
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42.
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Why is color-blindness more common in males than in
females?
a. | Because females would have to receive two copies of the recessive
color blindness gene to actually express the trait. | b. | Because a male only needs to receive the recessive gene from his mother to be
color-blind. | c. | Because color-blindness is
an X-linked trait. | d. | All of the
above. | e. | None of the above. |
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43.
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What are the possible phenotypes of the children if the
mother has Type A blood and the father has type AB blood? (Use the Punnett square to verify your
answer.)
a. | all AB | b. | A,
B | c. | A, AB | d. | A, B,
AB | e. | A, B, O |
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44.
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What are the possible phenotypes of the children if the
mother's genotype is IAi for blood type and the father is
IBi? (Use the Punnett square to verify your answer.)
a. | all AB | b. | A,
B | c. | A, AB | d. | A, B,
O | e. | A, B, AB, O |
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45.
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What are the predicted phenotypes of the male children from
the union of a woman who is heterozygous for hemophilia and a man who has normal blood clotting
characteristics? (Use the Punnett square to verify your answer.)
a. | all normal | b. | 3 normal : 1
hemophilia | c. | 1 hemophilia : 1
normal | d. | 1 hemophilia : 3 normal | e. | all hemophiliacs |
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46.
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What are the predicted phenotypes of the female children
from the union of a woman who is heterozygous for hemophilia and a man who has normal blood clotting
characteristics? (Use the Punnett square to verify your answer.)
a. | all carriers | b. | 3 homozygous normal : 1 carrier | c. | 1 homozygous normal : 1
carrier | d. | 1 hemophilia: 2 homozygous normal : 1
carrier | e. | all hemophiliacs |
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47.
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What are the possible genotypes of a female child from the
union of a woman who is heterozygous for hemophilia and a man who has normal blood clotting
characteristics?
a. | XHXH or
XHXh | b. | XHXh | c. | XHYH | d. | Hh | e. | HH |
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48.
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A Barr body in a mammalian female cell
represents:
a. | an inactivated oocyte. | b. | a polar body. | c. | a degenerate
nucleus. | d. | an inactivated X chromosome. | e. | an inactivated Y chromosome. |
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49.
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Calico cats are never male because:
a. | recessive genes are not usually expressed on their X
chromosome. | b. | male hormones prevent
expression of the calico phenotype. | c. | two different X chromosomes
are needed for the expression of the calico phenotype. | d. | two different Y chromosomes are needed for the expression of the calico
phenotype. | e. | one X chromosome and one Y
chromosome are needed for the expression of the calico
phenotype. |
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50.
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A diploid individual has a maximum of ____________ different
alleles for a particular locus.
a. | one | b. | two | c. | three | d. | four | e. | more than
four |
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51.
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A particular gene that controls seed coat color in peas also
determines the susceptibility of these peas to a particular disease. This situation is referred to
as:
a. | variegation. | b. | additive dominance. | c. | codominance. | d. | pleiotropy. | e. | incomplete
dominance. |
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52.
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When certain medium height hybrid plants were crossed, they
produced offspring that were dwarf, medium, and tall in a ratio of 1 : 2 : 1. This is an example
of:
a. | variegation. | b. | hybrid vigor. | c. | incomplete
dominance. | d. | epistasis. | e. | a polygenic
trait. |
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53.
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Breeding a yellow dog with a brown dog produced puppies with
both yellow and brown hairs intermixed. This is an example of:
a. | variegation. | b. | codominance. | c. | incomplete
dominance. | d. | epistasis. | e. | a polygenic
trait. |
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54.
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A gene that affects, prevents, or masks the expression of a
gene at another locus is a(n) _________ gene.
a. | recessive | b. | dominant | c. | epistatic | d. | codominant | e. | plieotropic |
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55.
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The range of phenotypic possibilities that can develop from
a certain dog genotype under different environmental conditions is called the:
a. | epistatic interaction. | b. | norm of reaction. | c. | nurture
limit. | d. | genotype range. | e. | maximum phenotype. |
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56.
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________________________ refers to multiple independent
pairs of genes having similar and additive effects on the same characteristic.
a. | Codominance | b. | Epistasis | c. | Polygenic
inheritance | d. | Complete
dominance | e. | Additive dominance |
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57.
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In the experiments of Griffith, the conversion of non-lethal
R-strain bacteria to lethal S-strain bacteria:
a. | was the result of genetic mutation. | b. | was an example of the genetic exchange known as
transformation. | c. | supported the case for
proteins as the genetic material. | d. | could not be reproduced by
other researchers. | e. | was an example of
conjugation. |
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58.
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The first experimenters to use Griffith's
transformation assay to identify the genetic material were:
a. | Meselson and Stahl. | b. | Watson and Crick. | c. | Franklin and
Wilkins. | d. | Avery, MacLeod, and McCarty. | e. | Hershey and Chase. |
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59.
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The bacteriophages used in Alfred Hershey's and Martha
Chase's experiments showed that:
a. | DNA was injected into bacteria. | b. | DNA and protein were injected into bacteria. | c. | DNA remained on the outer coat of bacteria. | d. | proteins were injected into bacteria. | e. | proteins were responsible for the production of new viruses within the
bacteria. |
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60.
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The main reason scientists thought that proteins, rather
than DNA, were the carriers of genetic material in the cell was:
a. | their presence within the nucleus. | b. | their abundance within the cell. | c. | the large number of
possible amino acid combinations. | d. | their ability to self
replicate within the cytoplasm. | e. | their ability to be
exported from the cell. |
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61.
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Which of the following statements about DNA is
false?
a. | DNA is capable of forming many different
sequences. | b. | DNA contains thymine
instead of uracil. | c. | DNA is double-stranded
rather than single-stranded. | d. | DNA is only found in
eukaryotic cells. | e. | DNA contains the sugar
deoxyribose. |
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62.
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The information carried by DNA is incorporated in a code
specified by the:
a. | phosphodiester bonds of the DNA
strand. | b. | number of separate strands of DNA. | c. | size of a particular chromosome. | d. | specific nucleotide
sequence of the DNA molecule. | e. | number of bases in a DNA
strand. |
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63.
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X-ray diffraction studies are used to
determine:
a. | the sequence of amino acids in protein
molecules. | b. | the sequence of nucleic
acids in nucleic acid molecules. | c. | the distances between atoms
of molecules. | d. | the type of chemical under
investigation. | e. | the wavelength of light
emitted by chemicals. |
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64.
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The two molecules that alternate to form the backbone of a
polynucleotide chain are:
a. | adenine and thymine. | b. | cytosine and guanine. | c. | sugar and
phosphate. | d. | base and
sugar. | e. | base and phosphate. |
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65.
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______________________ used x-ray diffraction to provide
images of DNA.
a. | Watson and Crick | b. | Crick and Wilkins | c. | Franklin | d. | Franklin and
Crick | e. | Watson and Wilkins |
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66.
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______________________ determined the structure of the
molecule DNA.
a. | Crick and Wilkins | b. | Watson and Crick | c. | Franklin and
Crick | d. | Franklin | e. | Watson, Crick, and
Wilkins |
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67.
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Chargaff determined that DNA from any source contains about
the same amount of guanine as __________.
a. | uracil | b. | thymine | c. | adenine | d. | cytosine | e. | guanine |
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68.
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X-ray crystallography showed that DNA:
a. | had the bases in the center of the
molecule. | b. | had the sugars and phosphates on the outside of the
molecule. | c. | was a very long molecule. | d. | was made of 2 strands. | e. | was a
helix. |
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69.
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Why is DNA able to store large amounts of
information?
a. | It contains a large number of different
nucleotides. | b. | Its nucleotides can be
arranged in a large number of possible sequences. | c. | It is capable of assuming a wide variety of shapes. | d. | The sugar and phosphates can be arranged in many different
sequences. | e. | The nucleotides can be
altered to form many different letters in the sequence. |
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70.
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In DNA, the genetic information:
a. | is contained in each strand. | b. | is contained in only one of the strands. | c. | is contained in both strands together. | d. | is contained in the differences between the 2 strands. | e. | None of the above. |
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71.
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Two chains of DNA must run in ____________ direction(s) and
must be ____________ if they are to bond with each other.
a. | the same; uncomplementary | b. | opposite; uncomplementary | c. | parallel;
uncomplementary | d. | parallel;
complementary | e. | antiparallel;
complementary |
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72.
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Hydrogen bonds can form between guanine and ____________,
and between adenine and ____________.
a. | phosphate; sugar | b. | thymine; cytosine | c. | cytosine;
thymine | d. | sugar; phosphate | e. | adenine; guanine |
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Figure 11-01
Use the figure below to answer the
corresponding questions.
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73.
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The portion of the molecule in box 5 of Figure 11-01
is:
a. | a hydrogen bond. | b. | a phosphate. | c. | a
nucleotide. | d. | a
pyrimidine. | e. | a
protein. |
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74.
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In Figure 11-01, the portion of the molecule in box
_________ is a pyrimidine.
a. | 1 | b. | 3 | c. | 4 | d. | 1 and
3 | e. | 3 and 4 |
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75.
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The portion of the molecule in box 3 of Figure 11-01
is:
a. | a sugar. | b. | a
protein. | c. | a pyrimidine. | d. | a purine. | e. | a
nucleotide. |
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76.
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Which of the following nucleotide sequences represents the
complement to the DNA strand 5´ - AGATCCG- 3´?
a. | 5´ – AGATCCG- 3´ | b. | 3´ – AGATCCG- 5´ | c. | 5´ – CTCGAAT-
3´ | d. | 3´ – CTCGAAT- 5´ | e. | 3´ – TCTAGGC- 5´ |
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77.
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How is a single strand of DNA able to serve as a template
for the synthesis of another strand?
a. | Nucleotides pair with those of the original strand to form a new
strand. | b. | Hydrogen bonds holding the two strands together are easy to
break, allowing one strand to be a template. | c. | A single strand of DNA is
not able to serve as a template. | d. | One strand of DNA directs
the synthesis of a new strand on its partner. | e. | Both A and
B. |
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78.
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Which of the following best describes semiconservative
replication?
a. | The translation of a DNA molecule into a complementary strand of
RNA. | b. | A DNA molecule consists of one parental strand and one new
strand. | c. | The number of DNA molecules is doubled with every other
replication. | d. | The replication of DNA
never takes place with 100% accuracy. | e. | The replication of DNA
takes place at a defined period in the cell cycle. |
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79.
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Who first confirmed that the replication of DNA was
semiconservative?
a. | Chargaff and Hershey | b. | Watson and Crick | c. | Avery and
Griffith | d. | Meselson and Stahl | e. | Watson, Crick, and Wilkins |
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80.
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If DNA replication rejoined the 2 parental strands, it would
be termed:
a. | dispersive. | b. | gradient. | c. | semiconservative. | d. | parental. | e. | conservative. |
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81.
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Meselson and Stahl separated DNA from different generations
using:
a. | density gradient centrifugation. | b. | gel electrophoresis. | c. | an electron
microscope. | d. | differential radioisotope
labeling. | e. | None of the above. |
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82.
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When a DNA molecule containing a wrong base at one location
in one strand is replicated:
a. | the mutation is corrected by the DNA polymerase
enzyme. | b. | the mutation is ignored by the DNA polymerase
enzyme. | c. | the mutation is copied into one of the two daughter
molecules. | d. | the mutation is copied into
both of the daughter molecules. | e. | the replication is
stopped. |
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83.
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What prevents knot formation in replicating
DNA?
a. | protosomes | b. | topoisomerases | c. | scaffolding
proteins | d. | chromatin | e. | histones |
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84.
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Why does DNA synthesis only proceed in the 5´to 3´
direction?
a. | Because DNA polymerases can only add nucleotides to the 3´
end of a polynucleotide strand. | b. | Because the 3´ end of
the polynucleotide molecule is more electronegative than the 5´ end. | c. | Because that is the direction in which the two strands of DNA
unzip. | d. | Because that is the only direction that the polymerase can be
oriented. | e. | Because the chromosomes are always aligned in the 5´ to
3´ direction in the nucleus. |
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85.
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How is the chromosome of a bacterial cell
replicated?
a. | The linear DNA molecule is replicated from multiple origins of
replication bidirectionally. | b. | The linear DNA molecule is
replicated from one origin of replication bidirectionally. | c. | The circular DNA molecule is replicated from multiple origins of replication
bidirectionally. | d. | The circular DNA molecule
is replicated from one origin of replication bidirectionally. | e. | The circular DNA molecule is replicated from one origin of replication
unidirectionally. |
|
|
|
86.
|
How are the chromosomes of a eukaryote cell
replicated?
a. | The linear DNA molecules are replicated from multiple origins of
replication bidirectionally. | b. | The linear DNA molecules
are replicated from one origin of replication bidirectionally. | c. | The circular DNA molecules are replicated from multiple origins of replication
bidirectionally. | d. | The circular DNA molecules
are replicated from one origin of replication bidirectionally. | e. | The linear DNA molecules are replicated from one origin of replication
unidirectionally. |
|
|
|
87.
|
Which of the following adds new nucleotides to a growing DNA
chain?
a. | DNA polymerase | b. | DNA helicase | c. | RNA
primer | d. | primase | e. | RNA
polymerase |
|
|
|
88.
|
Which of the following cause the unwinding of the DNA double
helix?
a. | DNA polymerase | b. | DNA helicase | c. | RNA
primer | d. | primosome | e. | RNA
polymerase |
|
|
|
89.
|
In DNA replication, the lagging strand:
a. | is synthesized as a series of Okazaki
fragments. | b. | is synthesized as a
complementary copy of the leading strand. | c. | pairs with the leading
strand by complementary base pairing. | d. | is made up entirely of RNA
primers. | e. | is not synthesized until the synthesis of the leading strand is
completed. |
|
|
|
Figure 11-02
Use the figure below to answer the
corresponding questions.
|
|
|
90.
|
The correct designation for the DNA strand labeled C in
Figure 11-02 is:
a. | the leading strand. | b. | 3´. | c. | Okazaki
fragments. | d. | polymerase. | e. | None of the
above. |
|
|
|
91.
|
The segments labeled F in Figure 11-02 are responsible
for:
a. | linking short DNA segments. | b. | synthesizing the leading strand. | c. | forming the replication
fork. | d. | initiating DNA synthesis. | e. | unwinding the DNA double helix. |
|
|
|
92.
|
The enzyme represented by the letter D in Figure 11-02 is
responsible for:
a. | linking short DNA segments. | b. | synthesizing the leading strand. | c. | forming the replication
fork. | d. | forming nucleosomes. | e. | unwinding the DNA double helix. |
|
|
|
93.
|
The structures represented by the letter E in Figure 11-02
are called:
a. | leading fragments | b. | Okazaki fragments. | c. | replication
forks. | d. | nucleosomes. | e. | DNA polymerases. |
|
|
|
94.
|
In replication, once the DNA strands have been separated,
reformation of the double helix is prevented by:
a. | DNA helicase enzyme. | b. | helix-destabilizing proteins. | c. | DNA
polymerases. | d. | ATP. | e. | GTP. |
|
|
|
95.
|
Enzymes called ____________ form nicks in the DNA molecules
to prevent the formation of knots in the DNA helix during replication.
a. | topoisomerases | b. | helix-destabilizing enzymes | c. | DNA
polymerases | d. | RNA
polymerases | e. | DNA
ligases |
|
|
|
96.
|
The DNA strand that is replicated smoothly and continuously
is called the:
a. | primary strand. | b. | first strand. | c. | leading
strand. | d. | alpha strand. | e. | lagging strand. |
|
|
|
97.
|
The final product of DNA replication is:
a. | mRNA, tRNA, and rRNA molecules. | b. | a wide variety of proteins. | c. | DNA
fragments. | d. | two DNA molecules, each of
which contains one new and one old DNA strand. | e. | the enzymes needed for
further processes, such as DNA polymerase. |
|
|
|
98.
|
A replication fork:
a. | is only seen in prokaryotic
chromosomes. | b. | is only seen in bacterial
cells. | c. | is a Y-shaped structure where both DNA strands are replicated
simultaneously. | d. | is a site where one DNA
strand serves as a template, but the other strand is not replicated. | e. | is created by the action of the enzyme RNA
polymerase. |
|
|
|
99.
|
The 5´ end of each Okazaki fragment begins
with:
a. | the same RNA primer that began synthesis on the leading
strand. | b. | a DNA primer binding to the template
DNA. | c. | DNA polymerase binding to the template
DNA. | d. | a separate RNA primer. | e. | a small DNA primer. |
|
|
|
100.
|
Okazaki fragments are joined together by:
a. | RNA polymerase. | b. | DNA ligase. | c. | DNA
polymerase. | d. | RNA
ligase. | e. | primase. |
|
|
|
101.
|
Primase is the enzyme responsible for:
a. | unwinding the DNA double strand to allow DNA polymerase access to
the template DNA. | b. | introducing nicks into the
DNA double strand in order to prevent the formation of knots. | c. | hydrolyzing ATP to facilitate DNA unwinding. | d. | making short strands of RNA at the site of replication
initiation. | e. | forming a replication fork
in the DNA double helix. |
|
|
|
102.
|
____________, the ends of eukaryotic chromosomes, shorten
with every cell replication event.
a. | Centromeres | b. | Telomeres | c. | Kinetochores | d. | Primosomes | e. | Nucleosomes |
|
|
|
103.
|
The ends of eukaryotic chromosomes can be lengthened
by:
a. | apoptosis. | b. | reverse
transcriptase. | c. | primase. | d. | telomerase. | e. | DNA
polymerase. |
|
|
|
104.
|
Cancer cells differ from noncancerous cells in
that:
a. | they have elevated levels of
telomerase. | b. | they are virtually
immortal. | c. | they have the ability to resist
apoptosis. | d. | they can maintain telomere
length as they divide. | e. | All of the
above. |
|
|
|
105.
|
In 1998, Bodnar and her colleagues found that by introducing
genes coding for telomeres into cultured human cells:
a. | the cells underwent more cell divisions than
normal. | b. | the cells underwent fewer cell divisions than
normal. | c. | the cells all lived indefinitely. | d. | the cells underwent gene expression more vigorously. | e. | the cell cycle shortened. |
|
|
|
106.
|
Which of the following is not a reason why
Neurospora is an ideal organism to study the effects of genetic mutations?
a. | Neurospora is easy to grow. | b. | Neurospora grows as a haploid organism. | c. | Neurospora is easy to genetically manipulate. | d. | Neurospora reproduces both sexually and asexually. | e. | Neurospora contains homologous chromosomes that are easily viewed with a light
microscope. |
|
|
|
107.
|
Garrod first proposed that:
a. | metabolic defects were due to a lack of an
enzyme. | b. | metabolic defects were due to excess
enzyme. | c. | metabolic defects were due to chromosomal
changes. | d. | mutations were inheritable. | e. | metabolic defects did not occur in humans. |
|
|
|
108.
|
Why was it important in the studies of Beadle and Tatum that
Neurospora is haploid?
a. | Because it is easier to grow haploid molds in the
laboratory. | b. | Because haploid molds have
simpler nutritional requirements than do diploid molds. | c. | Because a mutation that arises is not masked by a normal allele on a homologous
chromosome. | d. | Because haploid
Neurospora will always mutate. | e. | Because diploid
Neurospora will always mutate. |
|
|
|
109.
|
Beadle and Tatum began their studies with wild-type
Neurospora, which is:
a. | Neurospora that only grows in the
wild. | b. | a mutant strain that will only grow in the lab on complete
medium. | c. | a strain that will not grow in the
lab. | d. | a virulent strain of
Neurospora. | e. | a normal phenotype that
will grow on minimal medium. |
|
|
|
110.
|
Beadle and Tatum irradiated Neurospora and initially
grew the mutant strains on complete medium. How were they able to determine what type of mutation
each strain had?
a. | By growing the mold on a complete medium with extra vitamins and
nutrients. | b. | By growing the mold on
minimal media supplemented with different combinations of amino acids, vitamins,
etc. | c. | By growing the mold in its diploid form to see which traits were
masked. | d. | By comparing Neurospora to other species of
mold. | e. | By observing the marked differences in morphology between the
different strains. |
|
|
|
111.
|
What conclusions did Beadle and Tatum reach with their
studies of Neurospora?
a. | Each mutant gene affected several
enzymes. | b. | Each mutant gene affected a pair of
enzymes. | c. | Each mutant gene affected only one
enzyme. | d. | Mutant genes had no effect on the enzymes produced by the
cells. | e. | None of the above. |
|
|
|
112.
|
Linus Pauling demonstrated that:
a. | the structure of hemoglobin was altered by a mutation of a single
gene. | b. | mutations only caused defects in
enzymes. | c. | mutations alter the structure of RNA, but not
proteins. | d. | mutations were inherited. | e. | the structure of hemoglobin was altered by mutations in any of a dozen
genes. |
|
|
|
Figure 12-01
Use the figure below to answer the
corresponding questions.
|
|
|
113.
|
The experimental design in Figure 12-01 was used to
examine:
a. | the relationship between genetic changes and metabolic
enzymes. | b. | the mutation rate of
Neurospora. | c. | resistance of
Neurospora to genetic poisons. | d. | toxicity of arginine
metabolites. | e. | growth of Neurospora
in the presence of different antibiotics. |
|
|
|
114.
|
The conclusion associated with mutant strain III in Figure
12-01 was that:
a. | it contained all the enzymes needed for normal
metabolism. | b. | it was missing all the
enzymes for metabolism of amino acids. | c. | it was missing an enzyme
for metabolism and could not synthesize arginine. | d. | it was missing an enzyme for metabolism and could not synthesize
citrulline. | e. | it was missing an enzyme
for metabolism and could not synthesize ornithine. |
|
|
|
115.
|
One of the mRNA codons specifying the amino acid leucine is
5´-CUA-3´. Its corresponding anticodon is:
a. | 5´-GAT-3´. | b. | 3´-AUC-5´. | c. | 3´-GAU-5´. | d. | 3´-GAT-5´. | e. | 5´-GAU-3´. |
|
|
|
116.
|
Which of the following is a characteristic of
uracil?
a. | The ability to bond with adenine. | b. | The ability to bond with guanine. | c. | It is a
purine. | d. | The ability to bond with cytosine. | e. | It contains two nitrogenous rings. |
|
|
|
117.
|
RNA differs from DNA in all the following
except:
a. | RNA is single stranded and DNA is double
stranded. | b. | RNA is a larger molecule than DNA. | c. | RNA contains uracil and DNA contains thymine. | d. | RNA contains ribose and DNA contains deoxyribose. | e. | None of the above. |
|
|
|
118.
|
Ribose differs from deoxyribose by having:
a. | two attached bases. | b. | one less oxygen. | c. | an extra hydroxyl
group. | d. | an extra carbon in the ring. | e. | None of the above. |
|
|
|
119.
|
Uracil forms a complementary pair with ____________ in RNA
and _____________ in DNA.
a. | adenine; adenine | b. | adenine; thymine | c. | thymine;
thymine | d. | uracil; adenine | e. | adenine; uracil |
|
|
|
120.
|
RNA synthesis is also known as:
a. | elongation. | b. | reverse
transcription. | c. | termination. | d. | translation. | e. | transcription. |
|
|
|
121.
|
All RNA except for _________ is made from DNA.
a. | tRNA | b. | mRNA | c. | rRNA | d. | snRNA | e. | None of the
above. |
|
|
|
122.
|
The total number of different three-base combinations of the
four nucleic acid bases is:
a. | 12. | b. | 16. | c. | 20. | d. | 64. | e. | 256. |
|
|
|
123.
|
During protein synthesis, ribosomes:
a. | attach to the mRNA molecule and travel along its
length. | b. | attach to the DNA molecule and travel along its length to produce
an mRNA molecule. | c. | translate mRNA into
tRNA. | d. | transcribe mRNA to tRNA. | e. | translate mRNA into DNA. |
|
|
|
124.
|
How is the four-letter language of nucleic acids converted
into the 20-word language of amino acids?
a. | The 4 nucleic acid bases combine in 2-letter combinations that
define different amino acids. | b. | The 4 nucleic acid bases
combine in 3-letter sequences that define different amino acids. | c. | Triplets of the 2-letter nucleic acid bases are translated into the 20 different amino
acids. | d. | The 4 bases each specify 1 amino acid, which give rise to the
remaining 16 amino acids. | e. | The 4 bases are first
converted into tRNA molecules, which can each attach to 5 amino
acids. |
|
|
|
125.
|
Why is only one strand of DNA transcribed into
mRNA?
a. | Because mRNA is only required in small
quantities. | b. | Because transcribing both
DNA strands would produce different amino acid sequences. | c. | Because the other strand would produce the same amino acid sequence in reverse
order. | d. | Because all genes are located on the same DNA strand, while the
other strand acts as protection. | e. | Because the other strand is
transcribed directly into amino acids. |
|
|
|
126.
|
Initiation of transcription requires:
a. | a promoter sequence. | b. | DNA polymerase. | c. | an RNA
primer. | d. | a DNA primer. | e. | Okazaki fragments. |
|
|
|
127.
|
A sequence of bases located upstream from a reference
point occurs:
a. | towards the 3´ end of the amino acid
sequence. | b. | towards the 5´ end of the mRNA
sequence. | c. | towards the 3´ end of the mRNA
sequence. | d. | towards the 5´ end of the transcribed DNA
strand. | e. | towards the carboxyl end of the amino acid
sequence. |
|
|
|
128.
|
How does the first nucleotide at the 5´ end of a new
mRNA chain differ from the other nucleotides in the chain?
a. | The first nucleotide is always a
uracil. | b. | The first nucleotide is always a
cytosine. | c. | The first nucleotide retains its triphosphate group, while the
others do not. | d. | The first nucleotide does
not retain its triphosphate group, while the others in the chain do. | e. | The first nucleotide is always a modified
cytosine. |
|
|
|
Figure 12-02
Use the figure below to answer the
corresponding questions.
|
|
|
129.
|
In Figure 12-02, the transcription process begins at the
area labeled:
a. | A. | b. | D. | c. | E. | d. | G. | e. | None of the
above. |
|
|
|
130.
|
The component labeled B in Figure 12-02 is:
a. | DNase. | b. | DNA
polymerase. | c. | RNA
primase. | d. | RNA polymerase. | e. | reverse transcriptase. |
|
|
|
131.
|
The transcript in Figure 12-02 is labeled:
|
|
|
132.
|
The process illustrated in Figure 12-02 is:
a. | DNA synthesis. | b. | translation. | c. | transcription. | d. | a frame shift
mutation. | e. | protein synthesis. |
|
|
|
133.
|
Leader sequences contain signals that:
a. | prevent enzymes from degrading the newly synthesized
mRNA. | b. | inhibit ribosome binding until the appropriate
time. | c. | initiate chain termination. | d. | allow the ribosomes to be properly positioned to translate the
message. | e. | allow tRNA molecules to successfully bind to
mRNA. |
|
|
|
134.
|
Aminoacyl-tRNA synthetases ________ link ________ to their
respective tRNA molecules.
a. | ionically; mRNAs | b. | loosely; mRNAs | c. | terminally;
codons | d. | covalently; amino acids | e. | enzymatically; codons |
|
|
|
135.
|
Which of the following numbered terms represents the correct
order of sequences in a prokaryotic mRNA molecule as it was synthesized?
| 1. | 3´ trailing sequences | | 2. | coding sequences | | 3. | leader
sequences | | 4. | termination
signals | | |
a. | 1 ® 2
® 3 ® 4 | b. | 3 ® 2 ® 4 ® 1 | c. | 2 ® 1 ® 4 ®
3 | d. | 4 ® 2
® 1 ® 3 | e. | 3 ® 4 ® 2 ® 1 |
|
|
|
136.
|
Which of the following serves as an "adapter" in
protein synthesis and bridges the gap between mRNA and proteins?
a. | tRNA | b. | cDNA | c. | rRNA | d. | promoter
sequences | e. | DNA |
|
|
|
137.
|
The codon is found in the:
a. | template strand of DNA. | b. | non-template strand of DNA. | c. | mRNA. | d. | tRNA. | e. | rRNA. |
|
|
|
138.
|
The tRNA:
a. | must be recognized by ribosomes. | b. | must have an anticodon. | c. | must have an attachment
site for the amino acid. | d. | must be recognized by a
specific aminoacyl-tRNA synthetase that adds the correct amino acid. | e. | All of the above. |
|
|
|
139.
|
Which of the following numbered events represents the
correct sequence of events of prokaryotic translation initiation?
| 1. | large ribosomal subunit binds to initiation complex | | 2. | initiation tRNA binds small ribosomal subunit | | 3. | initiation complex binds to ribosome recognition sequence on
mRNA | | |
a. | 1 ® 2
® 3 | b. | 1 ® 3 ® 2 | c. | 2 ® 1 ® 3 | d. | 2 ® 3 ® 1 | e. | 3 ® 2 ® 1 |
|
|
|
140.
|
Where is the amino-acid binding site located on the tRNA
molecule?
a. | in the middle of the loop | b. | at the end of a "stem" that is the 3´ end of the
molecule | c. | in the first loop | d. | along the longest stretch of base pairing in the molecule | e. | on the 5´ end of the molecule |
|
|
|
Figure 12-03
Use the figure below to answer the
corresponding questions.
|
|
|
141.
|
In Figure 12-03, the portion of the molecule in the figure
that contains the anti-codon is:
|
|
|
142.
|
In Figure 12-03, the portion of the molecule labeled 5
is:
a. | the attached amino acid. | b. | a double-stranded region. | c. | a single-stranded
region. | d. | the anti-codon. | e. | the codon. |
|
|
|
143.
|
The enzyme peptidyl transferase, which catalyzes the
transfer of the polypeptide chain attached to the tRNA in the ____________ site to the aminoacyl-tRNA
in the ____________ site, is thought to be a(an) ____________ molecule and not a
protein.
a. | A; P; rDNA | b. | P; A;
tRNA | c. | A; P; mRNA | d. | P; A;
rRNA | e. | P; A; sugar |
|
|
|
144.
|
Translocation is the process whereby the __________ moves in
order to place the tRNA bound to the growing polypeptide chain in the __________ site, thereby
freeing the __________ site for a new aminoacyl-tRNA.
a. | mRNA; A; P | b. | ribosome; P;
A | c. | tRNA; P; A | d. | ribosome; A;
P | e. | tRNA; A; P |
|
|
|
145.
|
Following peptide bond formation between the amino acid in
the A site on the ribosome and the growing polypeptide chain, the tRNA in the A site:
a. | releases the growing polypeptide
chain. | b. | picks up another amino acid to add to the
chain. | c. | moves to the P site of the ribosome. | d. | forms a peptide bond with A site of the ribosome. | e. | forms a covalent bond with the P site of the
ribosome. |
|
|
|
146.
|
If a human gene mRNA were placed into a cell of yeast, it
would be:
a. | degraded immediately. | b. | translated into a repeating amino acid chain. | c. | translated into a chain of random amino acids not resembling the protein in
humans. | d. | translated into the protein that is found in
humans. | e. | integrated into the genome of the
yeast. |
|
|
|
147.
|
In all organisms, the AUG codon codes for:
a. | the initiation of translation. | b. | the termination of transcription. | c. | the termination of chain
elongation. | d. | the amino acid
valine. | e. | a termination tRNA
molecule. |
|
|
|
148.
|
A polyribosomes is:
a. | a complex of many ribosome and an
mRNA. | b. | a complex of many ribosomes in
eukaryotes. | c. | an initiation complex in
eukaryotes. | d. | an elongation complex in
eukaryotes. | e. | a complex of a ribosome
with its two subunits and several mRNAs. |
|
|
|
149.
|
Introns in pre mRNA are known to:
a. | code for specific protein domains. | b. | undergo excision, whereby they are spliced out of the
message. | c. | be able to move within the mRNA, thereby giving rise to new exon
combinations. | d. | protect pre mRNA from
enzyme degradation. | e. | code for important amino
acid sequences. |
|
|
|
150.
|
Proteins synthesized in E. coli have which of the
following at their amino terminal end?
a. | N-formyl-methionine | b. | N-acetyl-adenine | c. | adenine
triphosphate | d. | the AUG
codon | e. | the UUU codon |
|
|
|
151.
|
The wobble hypothesis states that:
a. | more than one ribosome can bind to an mRNA
molecule. | b. | some amino acids are coded for by more than one
codon. | c. | there is more than one stop codon in the genetic
code. | d. | a particular amino acid may be linked to more than one type of
tRNA molecule. | e. | certain tRNA anticodons can
pair with more than one codon sequence. |
|
|
|
152.
|
Binding of the appropriate aminoacyl-tRNA to the A site
requires:
a. | no additional energy. | b. | the input of two ATP molecules to supply the needed energy. | c. | energy supplied by GTP. | d. | activation of the A
site. | e. | phosphorylation of the tRNA
molecule. |
|
|
|
153.
|
Translocation in translation requires:
a. | no additional energy. | b. | activation of the P site. | c. | the input of two ATP
molecules to supply the needed energy. | d. | energy supplied by
GTP. | e. | phosphorylation of the mRNA
molecule. |
|
|
|
154.
|
Interrupted coding sequences include long sequences of bases
that do not code for amino acids. These noncoding sequences, called ____________, are found in
____________ cells.
a. | exons; prokaryotic | b. | introns; prokaryotic | c. | exons;
eukaryotic | d. | introns;
eukaryotic | e. | None of the
above. |
|
|
|
155.
|
An mRNA "5´ cap":
a. | prevents translation. | b. | facilitates binding of ribosomes. | c. | marks the mRNA for
degradation. | d. | decreases the half-life of
the mRNA. | e. | protects newly synthesized mRNA from
degradation. |
|
|
|
156.
|
The 3´ end of eukaryotic pre-mRNAs are changed
by:
a. | removing the last phosphate group. | b. | adding a "cap." | c. | copying the last few bases
so that it can form a duplex structure. | d. | cutting and adding 100-250
adenine nucleotides. | e. | phosphorylation of the mRNA
molecule. |
|
|
|
157.
|
Walter Gilbert proposed that exons are:
a. | remnants of older life forms. | b. | sequences that code for protein domains that are shuffled to form new
proteins. | c. | the result of mutation of introns. | d. | not present in prokaryotes. | e. | sequences that interrupt
the coding sequences of proteins. |
|
|
|
158.
|
Retroviruses or RNA tumor viruses use __________ to make
DNA:
a. | DNA polymerase | b. | DNA-dependent RNA polymerase | c. | RNA
polymerase | d. | primase | e. | reverse
transcriptase |
|
|
|
159.
|
Substitution of one base pair for another can result in a
____________ mutation that results in the conversion of an amino acid specifying codon to a
termination codon.
a. | nonsense | b. | frameshift | c. | chromosomal | d. | missense | e. | None of the
above. |
|
|
|
160.
|
A mutation that replaces one amino acid in a protein with
another is called a ____________ mutation.
a. | frameshift | b. | recombinant | c. | nonsense | d. | missense | e. | neutral |
|
|
|
161.
|
Frameshift mutations result from:
a. | the substitution of one base pair for
another. | b. | the substitution of more than one base
pair. | c. | the insertion or deletion of one or two base
pairs. | d. | the substitution of a stop codon for an amino acid-specifying
codon. | e. | the substitution of a start codon for an amino acid
codon. |
|
|
|
162.
|
A gene can now be defined as:
a. | a DNA sequence that carries information to produce a specific RNA
or protein product. | b. | a DNA nucleotide sequence
that carries information to produce a specific polypeptide. | c. | a DNA or RNA sequence that carries information to produce a single
polypeptide. | d. | a DNA nucleotide sequence
that carries information to produce an enzyme. | e. | a DNA or RNA sequence that
carries information to produce a specific polypeptide. |
|