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Exam 3 Biology

front 1

Character

back 1

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

front 2

Traits

back 2

Character variants (such as purple or white flowers)

front 3

True breeding

back 3

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

front 4

P generation

back 4

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

front 5

F1

back 5

Offspring. Hybrid offspring

front 6

F2 generation

back 6

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

F1*F1=F2 generation

front 7

Hybridization

back 7

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

front 8

Alleles are always found where?

back 8

On the homologous chromosome located on the locus

front 9

Allele

back 9

Alternative form of gene

front 10

Dominant allele

back 10

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

front 11

Recessive allele

back 11

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

front 12

Law of Segregation

back 12

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

front 13

Homozygous

back 13

An organism with two identical alleles for a character

front 14

Heterozygous

back 14

An organism that has two different alleles for a gene

front 15

Phenotype

back 15

physical appearance

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

front 16

Genotypes

back 16

Genetic makeup

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

front 17

Monohybrids

back 17

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

front 18

Monohybrid cross

back 18

A cross between monohybrids

front 19

Dihybrids

back 19

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

front 20

Law of independent assortment

back 20

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

front 21

Complete dominance

back 21

occurs when phenotypes of the heterozygote and dominant homozygote are identical

front 22

Multiplication rule

back 22

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

front 23

Incomplete Dominance

back 23

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.

front 24

CoDominant

back 24

two dominant alleles affect the phenotype in separate, distinguishable ways

front 25

Pleitropy

back 25

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

front 26

Epistasis

back 26

***Several alleles affect one phenotype.

front 27

Dihybrid crosses always gives you a phenotypic ratio of what?

back 27

9:3:3:1

front 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.

back 28

Answer: C

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

front 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.

back 29

Answer: D

One phenotype was completely dominant over another

front 30

Monohybrid always gives you a phenotypical ratio of what?

back 30

3:1

front 31

Norm of Reaction

back 31

Influenced by the enviornment

front 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%

back 32

Answer: C

50%

front 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

back 33

Answer: B

8

front 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.

back 34

Answer: C

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

front 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

back 35

Answer: C

amniocentesis

front 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.

back 36

Answer: C

New allele combinations are acted upon by natural selection

front 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.

back 37

Answer: B

roan × roan

front 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.

back 38

Answer D

Crossovers between these genes result in chromosomal exchange.

front 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

back 39

Answer B

expression of inappropriate gene products

front 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.

back 40

Answer: C

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

front 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

back 41

Answer: D

translocation

front 42

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%

back 42

Answer: C

50%

front 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.

back 43

Answer: A

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

front 44

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

back 44

Answer: E

A and G

front 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

back 45

Answer: D

1/2

front 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

back 46

Answer: C

ww

front 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.

back 47

Answer: B

45 X is the only known human live-born monosomy

front 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.

back 48

Answer: D

One member of the couple underwent nondisjunction in gamete production

front 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.

back 49

Answer: A

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

front 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

back 50

Answer: C

pleiotropy

front 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

back 51

Answer: D

activation of SRY in male embryos and masculinization of the gonads

front 52

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.

back 52

Answer: A

mitochondrial

front 53

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

back 53

Answer: E

1

front 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

back 54

Answer: A

47, +21

front 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%

back 55

Answer: C

50%

front 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.

back 56

Answer: D

males are hemizygous for the X chromosome

front 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

back 57

Answer: C

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

front 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

back 58

Answer: C

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

front 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

back 59

Answer: B

8

front 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.

back 60

Answer B

The gene involved is on the X chromosome

front 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

back 61

Answer: C

the probability that either one of two independent events will occur

front 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.

back 62

Answer: D

One phenotype was completely dominant over another.

front 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

back 63

Answer: D

deletion and duplication

front 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

back 64

Answer: E

0

front 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.

back 65

Answer: B

front 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.

back 66

Answer: D

It is pleiotropic

front 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.

back 67

Answer: A

They are located close together on the same chromosome

front 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

back 68

Answer: B

1% frequency of recombination between two genes

front 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

back 69

Answer: A

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

front 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.

back 70

Answer: D

nondisjunction or translocation in either parent

front 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.

back 71

Answer: C

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

front 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

back 72

Answer: E

skin pigmentation in humans

front 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

back 73

Answer: E

late prophase or metaphase

front 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.

back 74

Answer: B

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

front 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.

back 75

Answer: A

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

front 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

back 76

Answer: D

that the parents were both heterozygous for a single trait

front 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

back 77

Answer: E

XCXc and XCY

front 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

back 78

Answer: C

alignment of tetrads at the equator

front 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

back 79

Answer: A

a human chromosome 22 that has had a specific translocation

front 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.

back 80

Answer: B

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

front 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.

back 81

Answer: E

either an insertion or a deletion of a base.

front 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.

back 82

Answer: C

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

front 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.

back 83

Answer: B

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

front 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

back 84

Answer: D

single-strand binding proteins

front 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.

back 85

Answer: E

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

front 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.

back 86

Answer: A

8%

front 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.

back 87

Answer: C

It joins Okazaki fragments together

front 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

back 88

Answer: A

translating polypeptides directly from DNA

front 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.

back 89

Answer: B

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

front 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

back 90

Answer: D

a reduction in chromosome length in gametes

front 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

back 91

Answer: D

UUC

front 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

back 92

Answer: A

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

front 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

back 93

Answer: E

endonuclease, DNA polymerase I, DNA ligase

front 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

back 94

Answer: D

Erwin Chargaff

front 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

back 95

Answer: D

several transcription factors (TFs)

front 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.

back 96

Answer: B

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

front 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

back 97

Answer: B

an RNA with enzymatic activity

front 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.

back 98

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.

front 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.

back 99

Answer: C

It introduces a premature stop codon into the mRNA

front 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.

back 100

Answer: D

They cannot repair thymine dimers.

front 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.

back 101

Answer: E

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

front 102

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

back 102

Answer: D

met-ser-ser-leu-ser-leu

front 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.

back 103

Answer: B

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

front 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.

back 104

Answer: A

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

front 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'

back 105

Answer: D

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

front 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

back 106

Answer: A

sequence of bases

front 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.

back 107

Answer: A

Histones are positively charged, and DNA is negatively charged.

front 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.

back 108

Answer: C

UUU.

front 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

back 109

Answer: D

single-strand binding proteins

front 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.

back 110

Answer: A

many noncoding stretches of nucleotides are present in mRNA

front 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

back 111

Answer: A

the diameter of the helix

front 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

back 112

Answer: C

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

front 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.

back 113

Answer: D

It might substitute an amino acid in the active site

front 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.

back 114

Answer: E

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

front 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.

back 115

Answer: B

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

front 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.

back 116

Answer D

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

front 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.

back 117

Answer: E

either an insertion or a deletion of a base.

front 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.

back 118

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.

front 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

back 119

Answer: A

an accumulation of A and no production of B and C

front 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.

back 120

Answer: E

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

front 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

back 121

Answer: D

to translocate polypeptides across the ER membrane

front 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.

back 122

Answer: C

Heterochromatin is highly condensed, whereas euchromatin is less compact

front 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

back 123

Answer: B

replication without separation

front 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

back 124

Answer: C

ATP, RNA, and DNA

front 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

back 125

Answer: E

assimilation of external DNA into a cell

front 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.

back 126

Answer: B picture

The amino acid binds covalently

front 127

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

back 127

Answer: C

RNA

front 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

back 128

Answer: A

relieving strain in the DNA ahead of the replication fork

front 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

back 129

Answer: C

DNA polymerase III

front 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

back 130

Answer: D

a triplet in the same reading frame as an upstream AUG

front 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'

back 131

Answer: C

DNA polymerase I, DNA polymerase III

front 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.

back 132

Answer: C

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

front 133

Wild type

back 133

found normally in nature

front 134

Where is SRY gene found?

back 134

Y chromosome only

front 135

Recombination frequency equation

back 135

Add Recombiant offspring(last 2) x 100

Total offspring

17% crossover will occur ***Not linked

16% under ***Linked

Cant go over 50

front 136

Topoisomerase

back 136

Untwists the DNA

front 137

Leading Strand

back 137

Continuous

front 138

Helicase

back 138

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

front 139

Single Strand Binding Protein

back 139

Protector

front 140

Primase

back 140

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

front 141

Dna Pol III

back 141

5

front 142

DNA ligase

back 142

its going to come in and seal in the gaps.