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Exam 3 Biology

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1

Character

Varieties with distinct heritable features such as the color of a flower. 2 alleles

2

Traits

Character variants (such as purple or white flowers)

3

True breeding

Producing offspring with the same trait. An organism must be homozygous. Referred to as P generation.

4

P generation

Parental (always F1=offspring true breeding). True breeding parents.

5

F1

Offspring. Hybrid offspring

6

F2 generation

When F1 individuals self-pollinate or cross- pollinate with other F1 hybrids.

F1*F1=F2 generation

7

Hybridization

The mating, or crossing, of two true breeding varieties.

8

Alleles are always found where?

On the homologous chromosome located on the locus

9

Allele

Alternative form of gene

10

Dominant allele

If two alleles at a locus differs then the dominant one determines the appearance of the organism. Example Dd, so the Bigger D will be the dominant

11

Recessive allele

Has no noticeable affect on the organisms appearance. Example Dd, so the smaller d will be recessive

12

Law of Segregation

The two alleles for a heritable character separate (segregate) during gamete formation and end up in different gametes.

13

Homozygous

An organism with two identical alleles for a character

14

Heterozygous

An organism that has two different alleles for a gene

15

Phenotype

physical appearance

  • PP and Pp plants have the same phenotype (purple) but different genotypes
16

Genotypes

Genetic makeup

  • PP and Pp plants have the same phenotype (purple) but different genotypes
17

Monohybrids

The F1 offspring produced in this cross were monohybrids, individuals that are heterozygous for one character

18

Monohybrid cross

A cross between monohybrids

19

Dihybrids

Crossing two true-breeding parents differing in two characters. in the F1 generation, heterozygous for both characters

20

Law of independent assortment

Two or more enes sort independent of one another. States that each pair of alleles segregates independently of each other pair of alleles during gamete formation

21

Complete dominance

occurs when phenotypes of the heterozygote and dominant homozygote are identical

22

Multiplication rule

the probability that two or more independent events will occur together is the product of their individual probabilities

23

Incomplete Dominance

the phenotype of F1 hybrids is somewhere between the phenotypes of the two parental varieties. This is basically a mixture. Genotype takes on a mixture of both the dominant allele and recessive allele one allele red and the other White...both colors mix to make pink.

24

CoDominant

two dominant alleles affect the phenotype in separate, distinguishable ways

25

Pleitropy

Most genes have a phenotypic effects. ***Gives out several phenotypes

26

Epistasis

***Several alleles affect one phenotype.

27

Dihybrid crosses always gives you a phenotypic ratio of what?

9:3:3:1

28

1) What do we mean when we use the terms monohybrid cross and 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 cross involves 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 9:3:3:1 ratio whereas a dihybrid cross gives a 3:1 ratio.

Answer: C

A dihybrid cross involves organisms that are heterozygous for two characters and a monohybrid cross involves only one.

29

Why did the F₁ offspring of Mendel's classic pea cross always look like one of the two parental varieties?
A) No genes interacted to produce the parental phenotype.
B) Each allele affected phenotypic expression.
C) The traits blended together during fertilization.
D) One phenotype was completely dominant over another.
E) Different genes interacted to produce the parental phenotype.

Answer: D

One phenotype was completely dominant over another

30

Monohybrid always gives you a phenotypical ratio of what?

3:1

31

Norm of Reaction

Influenced by the enviornment

32

When crossing an organism that is homozygous recessive for a single trait with a heterozygote, what is the chance of producing an offspring with the homozygous recessive phenotype?
A) 0%
B) 25%
C) 50%
D) 75%
E) 100%

Answer: C

50%

33

How many unique gametes could be produced through independent assortment by an individual with the genotype AaBbCCDdEE?
A) 4
B) 8
C) 16
D) 32
E) 64

Answer: B

8

34

Phenylketonuria (PKU) is a recessive human disorder in which an individual cannot appropriately metabolize a particular amino acid. The amino acid is not otherwise produced by humans. 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.
B) Transfuse the patients with blood from unaffected donors.
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, such as twice per week.
E) Feed the patients an excess of the missing product.

Answer: C

Regulate the diet of the affected persons to severely limit the uptake of the amino acid

35

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. The obstetrician would most reasonably offer which of the following procedures to her patient?
A) CVS
B) ultrasound imaging
C) amniocentesis
D) blood transfusion
E) X-ray

Answer: C

amniocentesis

36

Why does recombination between linked genes continue to occur?
A) Recombination is a requirement for independent assortment.
B) Recombination must occur or genes will not assort independently.
C) New allele combinations are acted upon by natural selection.
D) The forces on the cell during meiosis II always result in recombination.
E) Without recombination there would be an insufficient number of gametes.

Answer: C

New allele combinations are acted upon by natural selection

37

In cattle, roan coat color (mixed red and white hairs) occurs in the heterozygous (Rr) offspring of red (RR) and white (rr) homozygotes. Which of the following crosses would produce offspring in the ratio of 1 red:2 roan:1 white?

A) red × white
B) roan × roan
C) white × roan
D) red × roan
E) The answer cannot be determined from the information provided.

Answer: B

roan × roan

38

Recombination between linked genes comes about for what reason?

A) Nonrecombinant chromosomes break and then rejoin with one another.

B) Independent assortment sometimes fails.

C) Linked genes travel together at anaphase.

D) Crossovers between these genes result in chromosomal exchange.

Answer D

Crossovers between these genes result in chromosomal exchange.

39

Abnormal chromosomes are frequently found in malignant tumors. Errors such as translocations may place a gene in close proximity to different control regions. Which of the following might then occur to make the cancer worse?

A) an increase in nondisjunction

B) expression of inappropriate gene products

C) a decrease in mitotic frequency

D) failure of the cancer cells to multiply

Answer B

expression of inappropriate gene products

40

Which of the following provides an example of epistasis?

A) Recessive genotypes for each of two genes (aabb) results in an albino corn snake.
B) The allele b17 produces a dominant phenotype, although b1 through b16 do not.
C) In rabbits and many other mammals, one genotype (cc) prevents any fur color from developing.
D) In Drosophila (fruit flies), white eyes can be due to an X-linked gene or to a combination of other genes.
E) In cacti, there are several genes for the type of spines.

Answer: C

In rabbits and many other mammals, one genotype (cc) prevents any fur color from developing.

41

One possible result of chromosomal breakage is for a fragment to join a nonhomologous chromosome. What is this alteration called?
A) deletion
B) transversion
C) inversion
D) translocation
E) duplication

Answer: D

translocation

42
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The following question refer to the pedigree chart in Figure 14.2 for a family, some of whose members exhibit the dominant trait, W. Affected individuals are indicated by a dark square or circle.

41) What is the likelihood that the progeny of IV-3 and IV-4 will have the trait?
A) 0%
B) 25%
C) 50%
D) 75%
E) 100%

Answer: C

50%

43

Which of the following statements is true of linkage?
A) The closer two genes are on a chromosome, the lower the probability that a crossover will occur between them.
B) The observed frequency of recombination of two genes that are far apart from each other has a maximum value of 100%.
C) All of the traits that Mendel studied–seed color, pod shape, flower color, and others–are due to genes linked on the same chromosome.
D) Linked genes are found on different chromosomes.
E) Crossing over occurs during prophase II of meiosis.

Answer: A

The closer two genes are on a chromosome, the lower the probability that a crossover will occur between them.

44
card image

This a map of four genes on a chromosome (See Image)

51) Between which two genes would you expect the highest frequency of recombination?
A) A and W
B) W and E
C) E and G
D) A and E
E) A and G

Answer: E

A and G

45

Black fur in mice (B) is dominant to brown fur (b). Short tails (T) are dominant to long tails (t). What fraction of the progeny of crosses BbTt × BBtt will be expected to have black fur and long tails?

A) 1/16
B) 3/16
C) 3/8
D) 1/2
E) 9/16

Answer: D

1/2

46

The following question refer to the pedigree chart in Figure 14.2 for a family, some of whose members exhibit the dominant trait, W. Affected individuals are indicated by a dark square or circle.

40) What is the genotype of individual II-5?
A) WW
B) Ww
C) ww
D) WW or ww
E) ww or Ww

Answer: C

ww

47

Which of the following is true of aneuploidies in general?
A) A monosomy is more frequent than a trisomy.
B) 45 X is the only known human live-born monosomy.
C) Some human aneuploidies have selective advantage in some environments.
D) Of all human aneuploidies, only Down syndrome is associated with mental retardation.
E) An aneuploidy resulting in the deletion of a chromosome segment is less serious than a duplication.

Answer: B

45 X is the only known human live-born monosomy

48

A couple has a child with Down syndrome. The mother is 39 years old at the time of delivery. Which of the following is the most probable cause of the child's condition?
A) The woman inherited this tendency from her parents.
B) One member of the couple carried a translocation.
C) One member of the couple underwent nondisjunction in somatic cell production.
D) One member of the couple underwent nondisjunction in gamete production.
E) The mother had a chromosomal duplication.

Answer: D

One member of the couple underwent nondisjunction in gamete production

49

What does a frequency of recombination of 50% indicate?
A) The two genes are likely to be located on different chromosomes.
B) All of the offspring have combinations of traits that match one of the two parents.
C) The genes are located on sex chromosomes.
D) Abnormal meiosis has occurred.
E) Independent assortment is hindered.

Answer: A

The two genes are likely to be located on different chromosomes.

50

) Cystic fibrosis affects the lungs, the pancreas, the digestive system, and other organs, resulting in symptoms ranging from breathing difficulties to recurrent infections. Which of the following terms best describes this?
A) incomplete dominance
B) multiple alleles
C) pleiotropy
D) epistasis
E) codominance

Answer: C

pleiotropy

51

In humans, clear gender differentiation occurs, not at fertilization, but after the second month of gestation. What is the first event of this differentiation?
A) formation of testosterone in male embryos
B) formation of estrogens in female embryos
C) anatomical differentiation of a penis in male embryos
D) activation of SRY in male embryos and masculinization of the gonads
E) activation of SRY in females and feminization of the gonads

Answer: D

activation of SRY in male embryos and masculinization of the gonads

52
card image

The pedigree in Figure 15.3 shows the transmission of a trait in a particular family. Based on this pattern of transmission, the trait is most likely
A) mitochondrial.
B) autosomal recessive.
C) sex-linked dominant.
D) sex-linked recessive.
E) autosomal dominant.

Answer: A

mitochondrial

53
card image

The following question refer to the pedigree chart in Figure 14.2 for a family, some of whose members exhibit the dominant trait, W. Affected individuals are indicated by a dark square or circle.

42) What is the probability that individual III-1 is Ww?
A) 3/4
B) 1/4
C) 2/4
D) 2/3
E) 1

Answer: E

1

54

Of the following human aneuploidies, which is the one that generally has the most severe impact on the health of the individual?
A) 47, +21
B) 47, XXY
C) 47, XXX
D) 47, XYY
E) 45, X

Answer: A

47, +21

55

When crossing an organism that is homozygous recessive for a single trait with a heterozygote, what is the chance of producing an offspring with the homozygous recessive phenotype?
A) 0%
B) 25%
C) 50%
D) 75%
E) 100%

Answer: C

50%

56

Males are more often affected by sex-linked traits than females because

A) male hormones such as testosterone often alter the effects of mutations on the X chromosome.
B) female hormones such as estrogen often compensate for the effects of mutations on the X chromosome.
C) X chromosomes in males generally have more mutations than X chromosomes in females.
D) males are hemizygous for the X chromosome.
E) mutations on the Y chromosome often worsen the effects of X-linked mutations.

Answer: D

males are hemizygous for the X chromosome

57

SRY is best described in which of the following ways?

A) a gene present on the X chromosome that triggers female development
B) an autosomal gene that is required for the expression of genes on the Y chromosome
C) a gene region present on the Y chromosome that triggers male development
D) an autosomal gene that is required for the expression of genes on the X chromosome
E) a gene required for development, and males or females lacking the gene do not survive past early childhood

Answer: C

a gene region present on the Y chromosome that triggers male development

58

What is a syndrome?
A) a characteristic facial appearance
B) a group of traits, all of which must be present if an aneuploidy is to be diagnosed
C) a group of traits typically found in conjunction with a particular chromosomal aberration or gene mutation
D) a characteristic trait usually given the discoverer's name
E) a characteristic that only appears in conjunction with one specific aneuploidy

Answer: C

a group of traits typically found in conjunction with a particular chromosomal aberration or gene mutation

59

How many unique gametes could be produced through independent assortment by an individual with the genotype AaBbCCDdEE?
A) 4
B) 8
C) 16
D) 32
E) 64

Answer: B

8

60

When Thomas Hunt Morgan crossed his red-eyed F1 generation flies to each other, the F2 generation included both red- and white-eyed flies. Remarkably, all the white-eyed flies were male. What was the explanation for this result?

A) The gene involved is on the Y chromosome.

B) The gene involved is on the X chromosome.

C) The gene involved is on an autosome, but only in males.

D) Other male-specific factors influence eye color in flies.

Answer B

The gene involved is on the X chromosome

61

Which of the following is the best statement of the use of the addition rule of probability?
A) the probability that two or more independent events will both occur
B) the probability that two or more independent events will both occur in the offspring of one set of parents
C) the probability that either one of two independent events will occur
D) the probability of producing two or more heterozygous offspring
E) the likelihood that a trait is due to two or more meiotic events

Answer: C

the probability that either one of two independent events will occur

62

Why did the F₁ offspring of Mendel's classic pea cross always look like one of the two parental varieties?
A) No genes interacted to produce the parental phenotype.
B) Each allele affected phenotypic expression.
C) The traits blended together during fertilization.
D) One phenotype was completely dominant over another.
E) Different genes interacted to produce the parental phenotype.

Answer: D

One phenotype was completely dominant over another.

63

A nonreciprocal crossover causes which of the following products?
A) deletion only
B) duplication only
C) nondisjunction
D) deletion and duplication
E) duplication and nondisjunction

Answer: D

deletion and duplication

64

In certain plants, tall is dominant to short. If a heterozygous plant is crossed with a homozygous tall plant, what is the probability that the offspring will be short?
A) 1
B) 1/2
C) 1/4
D) 1/6
E) 0

Answer: E

0

65

Gene S controls the sharpness of spines in a type of cactus. Cactuses with the dominant allele, S, have sharp spines, whereas homozygous recessive ss cactuses have dull spines. At the same time, a second gene, N, determines whether or not cactuses have spines. Homozygous recessive nn cactuses have no spines at all.

61) The relationship between genes S and N is an example of

A) incomplete dominance.
B) epistasis.
C) complete dominance.
D) pleiotropy.
E) codominance.

Answer: B

66

Humanoids on the newly explored planet Brin (in a hypothetical galaxy in ~50 years from the present) have a gene structure similar to our own, but many very different plants and animals.

73) Marfan syndrome in humans is caused by an abnormality of the connective tissue protein fibrillin. Patients are usually very tall and thin, with long spindly fingers, curvature of the spine, sometimes weakened arterial walls, and sometimes ocular problems, such as lens dislocation. Which of the following would you conclude about Marfan syndrome from this information?
A) It is recessive.
B) It is dominant.
C) It has a late age of onset (> 60).
D) It is pleiotropic.
E) It is epistatic.

Answer: D

It is pleiotropic

67

What is the reason that linked genes are inherited together?
A) They are located close together on the same chromosome.
B) The number of genes in a cell is greater than the number of chromosomes.
C) Chromosomes are unbreakable.
D) Alleles are paired together during meiosis.
E) Genes align that way during metaphase I of meiosis.

Answer: A

They are located close together on the same chromosome

68

The centimorgan (cM) is a unit named in honor of Thomas Hunt Morgan. To what is it equal?
A) the physical distance between two linked genes
B) 1% frequency of recombination between two genes
C) 1 nanometer of distance between two genes
D) the distance between a pair of homologous chromosomes
E) the recombination frequency between two genes assorting independently

Answer: B

1% frequency of recombination between two genes

69

An ideal procedure for fetal testing in humans would have which of the following features?
A) the procedure that can be performed at the earliest time in the pregnancy
B) lowest risk procedure that would provide the most reliable information
C) the procedure that can test for the greatest number of traits at once
D) a procedure that provides a three-dimensional image of the fetus
E) a procedure that could test for the carrier status of the fetus

Answer: A

the procedure that can be performed at the earliest time in the pregnancy

70

What is the source of the extra chromosome 21 in an individual with Down syndrome?
A) nondisjunction in the mother only
B) nondisjunction in the father only
C) duplication of the chromosome
D) nondisjunction or translocation in either parent
E) It is impossible to detect with current technology.

Answer: D

nondisjunction or translocation in either parent

71

What do we mean when we use the terms monohybrid cross and 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 cross involves 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 9:3:3:1 ratio whereas a dihybrid cross gives a 3:1 ratio.

Answer: C

A dihybrid cross involves organisms that are heterozygous for two characters and a monohybrid cross involves only one

72

Which of the following is an example of polygenic inheritance?
A) pink flowers in snapdragons
B) the ABO blood group in humans
C) Huntington's disease in humans
D) white and purple flower color in peas
E) skin pigmentation in humans

Answer: E

skin pigmentation in humans

73

At which phase(s) is it preferable to obtain chromosomes to prepare a karyotype?
A) early prophase
B) late telophase
C) anaphase
D) late anaphase or early telophase
E) late prophase or metaphase

Answer: E

late prophase or metaphase

74

Which of the following is the meaning of the chromosome theory of inheritance as expressed in the early 20th century?

A) Individuals inherit particular chromosomes attached to genes.
B) Mendelian genes are at specific loci on the chromosome and in turn segregate during meiosis.
C) Homologous chromosomes give rise to some genes and crossover chromosomes to other genes.
D) No more than a single pair of chromosomes can be found in a healthy normal cell.
E) Natural selection acts on certain chromosome arrays rather than on genes.

Answer: B

Mendelian genes are at specific loci on the chromosome and in turn segregate during meiosis.

75

Gene S controls the sharpness of spines in a type of cactus. Cactuses with the dominant allele, S, have sharp spines, whereas homozygous recessive ss cactuses have dull spines. At the same time, a second gene, N, determines whether or not cactuses have spines. Homozygous recessive nn cactuses have no spines at all.

62) A cross between a true-breeding sharp-spined cactus and a spineless cactus would produce
A) all sharp-spined progeny.
B) 50% sharp-spined, 50% dull-spined progeny.
C) 25% sharp-spined, 50% dull-spined, 25% spineless progeny.
D) all spineless progeny.
E) It is impossible to determine the phenotypes of the progeny.

Answer: A

A cross between a true-breeding sharp-spined cactus and a spineless cactus would produce

76

Two plants are crossed, resulting in offspring with a 3:1 ratio for a particular trait. What does this suggest?
A) that the parents were true-breeding for contrasting traits
B) that the trait shows incomplete dominance
C) that a blending of traits has occurred
D) that the parents were both heterozygous for a single trait
E) that each offspring has the same alleles for each of two traits

Answer: D

that the parents were both heterozygous for a single trait

77

Red-green color blindness is a sex-linked recessive trait in humans. Two people with normal color vision have a color-blind son. What are the genotypes of the parents?
A) XcXc and XcY
B) XcXc and XCY
C) XCXC and XcY
D) XCXC and XCY
E) XCXc and XCY

Answer: E

XCXc and XCY

78

Mendel's 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

alignment of tetrads at the equator

79

Which of the following is known as a Philadelphia chromosome?

A) a human chromosome 22 that has had a specific translocation
B) a human chromosome 9 that is found only in one type of cancer
C) an animal chromosome found primarily in the mid-Atlantic area of the United States
D) an imprinted chromosome that always comes from the mother
E) a chromosome found not in the nucleus but in mitochondria

Answer: A

a human chromosome 22 that has had a specific translocation

80

Which of the following is a function of a poly-A signal sequence?
A) It adds the poly-A tail to the 3' end of the mRNA.
B) It codes for a sequence in eukaryotic transcripts that signals enzymatic cleavage ~1035 nucleotides away.
C) It allows the 3' end of the mRNA to attach to the ribosome.
D) It is a sequence that codes for the hydrolysis of the RNA polymerase.
E) It adds a 7-methylguanosine cap to the 3' end of the mRNA.

Answer: B

It codes for a sequence in eukaryotic transcripts that signals enzymatic cleavage ~1035 nucleotides away.

81

A frameshift mutation could result from
A) a base insertion only.
B) a base deletion only.
C) a base substitution only.
D) deletion of three consecutive bases.
E) either an insertion or a deletion of a base.

Answer: E

either an insertion or a deletion of a base.

82

Which of the following does not occur in prokaryotic eukaryotic gene expression, but does in eukaryotic gene expression?
A) mRNA, tRNA, and rRNA are transcribed.
B) RNA polymerase binds to the promoter.
C) A poly-A tail is added to the 3' end of an mRNA and a cap is added to the 5' end.
D) Transcription can begin as soon as translation has begun even a little.
E) RNA polymerase requires a primer to elongate the molecule.

Answer: C

A poly-A tail is added to the 3' end of an mRNA and a cap is added to the 5' end.

83

What is meant by the description "antiparallel" regarding the strands that make up DNA?
A) The twisting nature of DNA creates nonparallel strands.
B) The 5' to 3' direction of one strand runs counter to the 5' to 3' direction of the other strand.
C) Base pairings create unequal spacing between the two DNA strands.
D) One strand is positively charged and the other is negatively charged.
E) One strand contains only purines and the other contains only pyrimidines.

Answer: B

The 5' to 3' direction of one strand runs counter to the 5' to 3' direction of the other strand.

84

Which of the following help(s) to hold the DNA strands apart while they are being replicated?
A) primase
B) ligase
C) DNA polymerase
D) single-strand binding proteins
E) exonuclease

Answer: D

single-strand binding proteins

85

Accuracy in the translation of mRNA into the primary structure of a polypeptide depends on specificity in the
A) binding of ribosomes to mRNA.
B) shape of the A and P sites of ribosomes.
C) bonding of the anticodon to the codon.
D) attachment of amino acids to tRNAs.
E) bonding of the anticodon to the codon and the attachment of amino acids to tRNAs.

Answer: E

bonding of the anticodon to the codon and the attachment of amino acids to tRNAs

86

Cytosine makes up 42% of the nucleotides in a sample of DNA from an organism. Approximately what percentage of the nucleotides in this sample will be thymine?
A) 8%
B) 16%
C) 31%
D) 42%
E) It cannot be determined from the information provided.

Answer: A

8%

87

What is the role of DNA ligase in the elongation of the lagging strand during DNA replication?
A) It synthesizes RNA nucleotides to make a primer.
B) It catalyzes the lengthening of telomeres.
C) It joins Okazaki fragments together.
D) It unwinds the parental double helix.
E) It stabilizes the unwound parental DNA.

Answer: C

It joins Okazaki fragments together

88

Which of the following variations on translation would be most disadvantageous for a cell?
A) translating polypeptides directly from DNA
B) using fewer kinds of tRNA
C) having only one stop codon
D) lengthening the half-life of mRNA
E) having a second codon (besides AUG) as a start codon

Answer: A

translating polypeptides directly from DNA

89

There are 61 mRNA codons that specify an amino acid, but only 45 tRNAs. This is best explained by the fact that
A) some tRNAs have anticodons that recognize four or more different codons.
B) the rules for base pairing between the third base of a codon and tRNA are flexible.
C) many codons are never used, so the tRNAs that recognize them are dispensable.
D) the DNA codes for all 61 tRNAs but some are then destroyed.
E) competitive exclusion forces some tRNAs to be destroyed by nucleases.

Answer: B

the rules for base pairing between the third base of a codon and tRNA are flexible

90

Which of the following would you expect of a eukaryote lacking telomerase?
A) a high probability of somatic cells becoming cancerous
B) production of Okazaki fragments
C) inability to repair thymine dimers
D) a reduction in chromosome length in gametes
E) high sensitivity to sunlight

Answer: D

a reduction in chromosome length in gametes

91

The figure represents tRNA that recognizes and binds a particular amino acid (in this instance, phenylalanine). Which codon on the mRNA strand codes for this amino acid?
A) UGG
B) GUG
C) GUA
D) UUC
E) CAU

Answer: D

UUC

92

Which of the following sets of materials are required by both eukaryotes and prokaryotes for replication?
A) double-stranded DNA, four kinds of dNTPs, primers, origins
B) topoisomerases, telomerases, polymerases
C) G-C rich regions, polymerases, chromosome nicks
D) nucleosome loosening, four dNTPs, four rNTPs
E) ligase, primers, nucleases

Answer: A

double-stranded DNA, four kinds of dNTPs, primers, origins

93

To repair a thymine dimer by nucleotide excision repair, in which order do the necessary enzymes act?
A) exonuclease, DNA polymerase III, RNA primase
B) helicase, DNA polymerase I, DNA ligase
C) DNA ligase, nuclease, helicase
D) DNA polymerase I, DNA polymerase III, DNA ligase
E) endonuclease, DNA polymerase I, DNA ligase

Answer: E

endonuclease, DNA polymerase I, DNA ligase

94

Which of the following investigators was/were responsible for the following discovery?
In DNA from any species, the amount of adenine equals the amount of thymine, and the amount of guanine equals the amount of cytosine.
A) Frederick Griffith
B) Alfred Hershey and Martha Chase
C) Oswald Avery, Maclyn McCarty, and Colin MacLeod
D) Erwin Chargaff
E) Matthew Meselson and Franklin Stahl

Answer: D

Erwin Chargaff

95

Transcription in eukaryotes requires which of the following in addition to RNA polymerase?
A) the protein product of the promoter
B) start and stop codons
C) ribosomes and tRNA
D) several transcription factors (TFs)
E) aminoacyl synthetase

Answer: D

several transcription factors (TFs)

96

When the function of the newly made polypeptide is to be secreted from the cell where it has been made, what must occur?
A) It must be translated by a ribosome that remains free of attachment to the ER.
B) Its signal sequence must target it to the ER, from which it goes to the Golgi.
C) It has a signal sequence that must be cleaved off before it can enter the ER.
D) It has a signal sequence that targets it to the cell's plasma membrane where it causes exocytosis.
E) Its signal sequence causes it to be encased in a vesicle as soon as it is translated.

Answer: B

Its signal sequence must target it to the ER, from which it goes to the Golgi

97

What is a ribozyme?
A) an enzyme that uses RNA as a substrate
B) an RNA with enzymatic activity
C) an enzyme that catalyzes the association between the large and small ribosomal subunits
D) an enzyme that synthesizes RNA as part of the transcription process
E) an enzyme that synthesizes RNA primers during DNA replication

Answer: B

an RNA with enzymatic activity

98

In his transformation experiments, what did Griffith observe?
A) Mutant mice were resistant to bacterial infections.
B) Mixing a heat-killed pathogenic strain of bacteria with a living nonpathogenic strain can convert some of the living cells into the pathogenic form.
C) Mixing a heat-killed nonpathogenic strain of bacteria with a living pathogenic strain makes the pathogenic strain nonpathogenic.
D) Infecting mice with nonpathogenic strains of bacteria makes them resistant to pathogenic strains.
E) Mice infected with a pathogenic strain of bacteria can spread the infection to other mice.

Answer: B

Mixing a heat-killed pathogenic strain of bacteria with a living nonpathogenic strain can convert some of the living cells into the pathogenic form.

99

What is the effect of a nonsense mutation in a gene?
A) It changes an amino acid in the encoded protein.
B) It has no effect on the amino acid sequence of the encoded protein.
C) It introduces a premature stop codon into the mRNA.
D) It alters the reading frame of the mRNA.
E) It prevents introns from being excised.

Answer: C

It introduces a premature stop codon into the mRNA

100

Individuals with the disorder xeroderma pigmentosum are hypersensitive to sunlight. This occurs because their cells are impaired in what way?
A) They cannot replicate DNA.
B) They cannot undergo mitosis.
C) They cannot exchange DNA with other cells.
D) They cannot repair thymine dimers.
E) They do not recombine homologous chromosomes during meiosis.

Answer: D

They cannot repair thymine dimers.

101

For a science fair project, two students decided to repeat the Hershey and Chase experiment, with modifications. They decided to label the nitrogen of the DNA, rather than the phosphate. They reasoned that each nucleotide has only one phosphate and two to five nitrogens. Thus, labeling the nitrogens would provide a stronger signal than labeling the phosphates. Why won't this experiment work?

A) There is no radioactive isotope of nitrogen.
B) Radioactive nitrogen has a half-life of 100,000 years, and the material would be too dangerous for too long.
C) Avery et al. have already concluded that this experiment showed inconclusive results.
D) Although there are more nitrogens in a nucleotide, labeled phosphates actually have 16 extra neutrons; therefore, they are more radioactive.
E) Amino acids (and thus proteins) also have nitrogen atoms; thus, the radioactivity would not distinguish between DNA and proteins.

Answer: E

Amino acids (and thus proteins) also have nitrogen atoms; thus, the radioactivity would not distinguish between DNA and proteins.

102
card image

65) What amino acid sequence will be generated, based on the following mRNA codon sequence?
5' AUG-UCU-UCG-UUA-UCC-UUG 3'
A) met-arg-glu-arg-glu-arg
B) met-glu-arg-arg-glu-leu
C) met-ser-leu-ser-leu-ser
D) met-ser-ser-leu-ser-leu
E) met-leu-phe-arg-glu-glu

Answer: D

met-ser-ser-leu-ser-leu

103

What is the function of the release factor (RF)?
A) It separates tRNA in the A site from the growing polypeptide.
B) It binds to the stop codon in the A site in place of a tRNA.
C) It releases the amino acid from its tRNA to allow the amino acid to form a peptide bond.
D) It supplies a source of energy for termination of translation.
E) It releases the ribosome from the ER to allow polypeptides into the cytosol.

Answer: B

It binds to the stop codon in the A site in place of a tRNA

104

What is the function of GTP in translation?
A) GTP energizes the formation of the initiation complex, using initiation factors.
B) GTP hydrolyzes to provide phosphate groups for tRNA binding.
C) GTP hydrolyzes to provide energy for making peptide bonds.
D) GTP supplies phosphates and energy to make ATP from ADP.
E) GTP separates the small and large subunits of the ribosome at the stop codon.

Answer: A

GTP energizes the formation of the initiation complex, using initiation factors

105

At a specific area of a chromosome, the sequence of nucleotides below is present where the chain opens to form a replication fork:
3' C C T A G G C T G C A A T C C 5'
An RNA primer is formed starting at the underlined T (T) of the template. Which of the following represents the primer sequence?
A) 5' G C C T A G G 3'
B) 3' G C C T A G G 5'
C) 5' A C G T T A G G 3'
D) 5' A C G U U A G G 3'
E) 5' G C C U A G G 3'

Answer: D

5' A C G U U A G G 3'

106

It became apparent to Watson and Crick after completion of their model that the DNA molecule could carry a vast amount of hereditary information in which of the following?
A) sequence of bases
B) phosphate-sugar backbones
C) complementary pairing of bases
D) side groups of nitrogenous bases
E) different five-carbon sugars

Answer: A

sequence of bases

107

Why do histones bind tightly to DNA?
A) Histones are positively charged, and DNA is negatively charged.
B) Histones are negatively charged, and DNA is positively charged.
C) Both histones and DNA are strongly hydrophobic.
D) Histones are covalently linked to the DNA.
E) Histones are highly hydrophobic, and DNA is hydrophilic.

Answer: A

Histones are positively charged, and DNA is negatively charged.

108

A particular triplet of bases in the coding sequence of DNA is AAA. The anticodon on the tRNA that binds the mRNA codon is
A) TTT.
B) UUA.
C) UUU.
D) AAA.
E) either UAA or TAA, depending on first base wobble.

Answer: C

UUU.

109

Which of the following help(s) to hold the DNA strands apart while they are being replicated?
A) primase
B) ligase
C) DNA polymerase
D) single-strand binding proteins
E) exonuclease

Answer: D

single-strand binding proteins

110

A transcription unit that is 8,000 nucleotides long may use 1,200 nucleotides to make a protein consisting of approximately 400 amino acids. This is best explained by the fact that
A) many noncoding stretches of nucleotides are present in mRNA.
B) there is redundancy and ambiguity in the genetic code.
C) many nucleotides are needed to code for each amino acid.
D) nucleotides break off and are lost during the transcription process.
E) there are termination exons near the beginning of mRNA.

Answer: A

many noncoding stretches of nucleotides are present in mRNA

111

Which of the following can be determined directly from X-ray diffraction photographs of crystallized DNA?
A) the diameter of the helix
B) the rate of replication
C) the sequence of nucleotides
D) the bond angles of the subunits
E) the frequency of A vs. T nucleotides

Answer: A

the diameter of the helix

112

What is the function of DNA polymerase III?
A) to unwind the DNA helix during replication
B) to seal together the broken ends of DNA strands
C) to add nucleotides to the 3' end of a growing DNA strand
D) to degrade damaged DNA molecules
E) to rejoin the two DNA strands (one new and one old) after replication

Answer: C

to add nucleotides to the 3' end of a growing DNA strand

113

Why might a point mutation in DNA make a difference in the level of protein's activity?
A) It might result in a chromosomal translocation.
B) It might exchange one stop codon for another stop codon.
C) It might exchange one serine codon for a different serine codon.
D) It might substitute an amino acid in the active site.
E) It might substitute the N-terminus of the polypeptide for the C-terminus.

Answer: D

It might substitute an amino acid in the active site

114

Accuracy in the translation of mRNA into the primary structure of a polypeptide depends on specificity in the
A) binding of ribosomes to mRNA.
B) shape of the A and P sites of ribosomes.
C) bonding of the anticodon to the codon.
D) attachment of amino acids to tRNAs.
E) bonding of the anticodon to the codon and the attachment of amino acids to tRNAs.

Answer: E

bonding of the anticodon to the codon and the attachment of amino acids to tRNAs.

115

If a cell were unable to produce histone proteins, which of the following would be a likely effect?

A) There would be an increase in the amount of "satellite" DNA produced during centrifugation.
B) The cell's DNA couldn't be packed into its nucleus.
C) Spindle fibers would not form during prophase.
D) Amplification of other genes would compensate for the lack of histones.
E) Pseudogenes would be transcribed to compensate for the decreased protein in the cell.

Answer: B

The cell's DNA couldn't be packed into its nucleus.

116

In an experimental situation, a student researcher inserts an mRNA molecule into a eukaryotic cell after he has removed its 5' cap and poly-A tail. Which of the following would you expect him to find?
A) The mRNA could not exit the nucleus to be translated.
B) The cell recognizes the absence of the tail and polyadenylates the mRNA.
C) The molecule is digested by restriction enzymes in the nucleus.
D) The molecule is digested by exonucleases since it is no longer protected at the 5' end.
E) The molecule attaches to a ribosome and is translated, but more slowly.

Answer D

The molecule is digested by exonucleases since it is no longer protected at the 5' end

117

A frameshift mutation could result from
A) a base insertion only.
B) a base deletion only.
C) a base substitution only.
D) deletion of three consecutive bases.
E) either an insertion or a deletion of a base.

Answer: E

either an insertion or a deletion of a base.

118

The leading and the lagging strands differ in that
A) the leading strand is synthesized in the same direction as the movement of the replication fork, and the lagging strand is synthesized in the opposite direction.
B) the leading strand is synthesized by adding nucleotides to the 3' end of the growing strand, and the lagging strand is synthesized by adding nucleotides to the 5' end.
C) the lagging strand is synthesized continuously, whereas the leading strand is synthesized in short fragments that are ultimately stitched together.
D) the leading strand is synthesized at twice the rate of the lagging strand.

Answer: A

the leading strand is synthesized in the same direction as the movement of the replication fork, and the lagging strand is synthesized in the opposite direction.

119

A mutation results in a defective enzyme A. Which of the following would be a consequence of that mutation?
A) an accumulation of A and no production of B and C
B) an accumulation of A and B and no production of C
C) an accumulation of B and no production of A and C
D) an accumulation of B and C and no production of A
E) an accumulation of C and no production of A and B

Answer: A

an accumulation of A and no production of B and C

120

A new DNA strand elongates only in the 5' to 3' direction because
A) DNA polymerase begins adding nucleotides at the 5' end of the template.
B) Okazaki fragments prevent elongation in the 3' to 5' direction.
C) the polarity of the DNA molecule prevents addition of nucleotides at the 3' end.
D) replication must progress toward the replication fork.
E) DNA polymerase can only add nucleotides to the free 3' end.

Answer: E

DNA polymerase can only add nucleotides to the free 3' end.

121

Which of the following is a function of a signal peptide?
A) to direct an mRNA molecule into the cisternal space of the ER
B) to bind RNA polymerase to DNA and initiate transcription
C) to terminate translation of the messenger RNA
D) to translocate polypeptides across the ER membrane
E) to signal the initiation of transcription

Answer: D

to translocate polypeptides across the ER membrane

122

Which of the following statements describes chromatin?
A) Heterochromatin is composed of DNA, whereas euchromatin is made of DNA and RNA.
B) Both heterochromatin and euchromatin are found in the cytoplasm.
C) Heterochromatin is highly condensed, whereas euchromatin is less compact.
D) Euchromatin is not transcribed, whereas heterochromatin is transcribed.
E) Only euchromatin is visible under the light microscope.

Answer: C

Heterochromatin is highly condensed, whereas euchromatin is less compact

123

Polytene chromosomes of Drosophila salivary glands each consist of multiple identical DNA strands that are aligned in parallel arrays. How could these arise?
A) replication followed by mitosis
B) replication without separation
C) meiosis followed by mitosis
D) fertilization by multiple sperm
E) special association with histone proteins

Answer: B

replication without separation

124

The nitrogenous base adenine is found in all members of which group?
A) proteins, triglycerides, and testosterone
B) proteins, ATP, and DNA
C) ATP, RNA, and DNA
D) α glucose, ATP, and DNA
E) proteins, carbohydrates, and ATP

Answer: C

ATP, RNA, and DNA

125

How do we describe transformation in bacteria?
A) the creation of a strand of DNA from an RNA molecule
B) the creation of a strand of RNA from a DNA molecule
C) the infection of cells by a phage DNA molecule
D) the type of semiconservative replication shown by DNA
E) assimilation of external DNA into a cell

Answer: E

assimilation of external DNA into a cell

126

The tRNA shown in the figure has its 3' end projecting beyond its 5' end. What will occur at this 3' end?
A) The codon and anticodon complement one another.
B) The amino acid binds covalently.
C) The excess nucleotides (ACCA) will be cleaved off at the ribosome.
D) The small and large subunits of the ribosome will attach to it.
E) The 5' cap of the mRNA will become covalently bound.

Answer: B picture

The amino acid binds covalently

127

During splicing, which molecular component of the spliceosome catalyzes the excision reaction?
A) protein
B) DNA
C) RNA
D) lipid
E) sugar

Answer: C

RNA

128

What is the function of topoisomerase?
A) relieving strain in the DNA ahead of the replication fork
B) elongating new DNA at a replication fork by adding nucleotides to the existing chain
C) adding methyl groups to bases of DNA
D) unwinding of the double helix
E) stabilizing single-stranded DNA at the replication fork

Answer: A

relieving strain in the DNA ahead of the replication fork

129

Which enzyme catalyzes the elongation of a DNA strand in the 5' → 3' direction?
A) primase
B) DNA ligase
C) DNA polymerase III
D) topoisomerase
E) helicase

Answer: C

DNA polymerase III

130

Which of the following nucleotide triplets best represents a codon?
A) a triplet separated spatially from other triplets
B) a triplet that has no corresponding amino acid
C) a triplet at the opposite end of tRNA from the attachment site of the amino acid
D) a triplet in the same reading frame as an upstream AUG
E) a sequence in tRNA at the 3' end

Answer: D

a triplet in the same reading frame as an upstream AUG

131

An Okazaki fragment has which of the following arrangements?
A) primase, polymerase, ligase
B) 3' RNA nucleotides, DNA nucleotides 5'
C) 5' RNA nucleotides, DNA nucleotides 3'
D) DNA polymerase I, DNA polymerase III
E) 5' DNA to 3'

Answer: C

DNA polymerase I, DNA polymerase III

132

Which of the following does not occur in prokaryotic eukaryotic gene expression, but does in eukaryotic gene expression?
A) mRNA, tRNA, and rRNA are transcribed.
B) RNA polymerase binds to the promoter.
C) A poly-A tail is added to the 3' end of an mRNA and a cap is added to the 5' end.
D) Transcription can begin as soon as translation has begun even a little.
E) RNA polymerase requires a primer to elongate the molecule.

Answer: C

A poly-A tail is added to the 3' end of an mRNA and a cap is added to the 5' end.

133

Wild type

found normally in nature

134

Where is SRY gene found?

Y chromosome only

135

Recombination frequency equation

Add Recombiant offspring(last 2) x 100

Total offspring

17% crossover will occur ***Not linked

16% under ***Linked

Cant go over 50

136

Topoisomerase

Untwists the DNA

137

Leading Strand

Continuous

138

Helicase

Unzips DNA-breaks hydrogen bond between the two strands of DNA

139

Single Strand Binding Protein

Protector

140

Primase

primer (RNA mucleotides) needs the primer to start DNA replication

141

Dna Pol III

5

142

DNA ligase

its going to come in and seal in the gaps.


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