exam 4 questions Flashcards
(95 cards)
1
Q
Binary fission in prokaryotes does not require the
A
assembly of the nuclear envelope
2
Q
Chromatin is composed of
A
DNA and protein
3
Q
What is a nucleosome?
A
a region of DNA wound around histone proteins
4
Q
what is the role of cohesin proteins in cell division?
A
they hold the DNA of the sister chromatids together
5
Q
the kinetochore is a structure that functions to
A
connect the centromere to microtubules
6
Q
separation of the sister chromatids occurs during
A
anaphase
7
Q
why is cytokinesis an important part of cell division?
A
it is responsible for the proper separation of the cytoplasmic contents
8
Q
what steps in the cell cycle represent irreversible commitments?
A
both b and c
b. The G1/S checkpoint
c. Anaphase
9
Q
Cyclin-dependent kinases(Cdks) are regulated by
A
the periodic destruction of cyclins
10
Q
The bacterial SMC proteins, eukaryotic cohesin proteins, and condensin proteins share a similar structure. Functionally they all
A
interact with DNA to compact or hold strands together
11
Q
Genetically, proto-oncogenes act in a dominant fashion. this is because
A
they act in a gain-of-function fashion to turn on the cell cycle
12
Q
The metaphase to anaphase transition involves
A
loss of cohesion between sister chromatids
13
Q
the main difference between bacterial cell division and eukaryotic cell division is that
A
none of the above is correct
14
Q
in animal cells, cytokinesis is accomplished by a contractile ring containing actin. The related process in bacteria is
A
septation via a ring of FtsZ protein, which is a tubulin-like protein
15
Q
in comparing somatic cells and gametes, somatic cells are
A
diploid with twice the number of chromosomes
16
Q
what are homologous chromosomes?
A
two genetically similar chromosomes, one from each parent
17
Q
chiasmata form
A
between homologous chromosomes
18
Q
crossing over involves each of the following with the exception of
A
the transfer of DNA between two sister chromatids
19
Q
during anaphase I
A
homologous chromosomes move to opposite poles
20
Q
at metaphase I the kinetochores of sister chromatids are
A
attached to microtubules from the same pole
21
Q
what occurs during anaphase of meiosis II?
A
sister chromatids are pulled to opposite poles
22
Q
which of the following does NOT contribute to genetic diversity?
A
metaphase of meiosis II
23
Q
how does DNA replication differ between mitosis and meiosis?
A
during meiosis, there is only one round of replication for two divisions
24
Q
which of the following is NOT a distinct feature of meiosis?
A
attachment of sister kinetochores to spindle microtubules
25
which phase of meiosis I is most similar to the comparable phase in mitosis?
telophase I
26
structurally, meiotic cohesins have different components than mitotic cohesins. This leads to the following functional difference:
centromeres remain attached during anaphase I of meiosis
27
mutations that affect DNA repair often also affect the accuracy of meiosis. This is because
the proteins involved in the repair of double-stranded breaks are also involved in crossing over
28
what property distinguished Mendel's investigation from previous studies?
Mendel quantified his results
29
the F1 generation of the monohybrid cross purple (PP) x white (pp) flower pea plants should
all have purple flowers
30
the F1 plants from the previous question are allowed to self-fertilize. The phenotypic ratio for the F2 should be
3 purple: 1 white
31
which of the following is NOT a part of Mendel's five-element model?
if an allele is present it will be expressed
32
an organism's __ is/are determined by its __
phenotype; genotype
33
phenotypes like height in humans, which show a continuous distribution, are usually the result of
the action of multiple genes on a single phenotype
34
Japanese four o'clocks that are red and tall are crossed to white short ones, producing an F1 that is pink and tall. If these genes assort independently, and the F1 is self-crossed, what would you predict for the ratio of F2 phenotypes?
3 red tall: 6 pink tall: 3 white tall: 1 red short: 2 pink short: 1 white short
35
If the two genes in the previous question showed complete linkage, what would you predict for an F2 phenotypic ratio?
1 red tall: 2 pink tall: 1 white short
36
what is the probability of obtaining an individual with the genotype bb from a cross between two individuals with the genotype Bb?
1/4
37
In a cross of AaBbcc x AaBbCc, what is the probability of obtaining an individual with the genotype AABbCc?
1/16
38
when you cross true-breeding tall and short tobacco plants you get an F1 that is intermediate in height. When this F1 is self-crossed, it yields an F2 with a continuous distribution of heights. What is the best explanation for these data?
Height is determined by the additive effects of many genes
39
Mendel's model assumes that each trait is determined by a single factor with alternative forms. We now know that this is too simplistic and that
a single gene can affect more than one trait, and traits may be affected by more than one gene
40
Why is the white-eye phenotype always observed in males carrying the white-eye allele?
Because the allele is located on the X chromosome and males only have one X
41
In an organism’s genome, autosomes are
all of the chromosomes other than sex chromosomes.
42
What cellular process is responsible for genetic recombination?
Crossing over between homologues
43
The map distance between two genes is determined by the
recombination frequency.
44
How many map units separate two alleles if the recombination frequency is 0.07?
7 cM
45
How does maternal inheritance of mitochondrial genes differ from sex linkage?
Since mitochondria are inherited from the mother, females and males are equally affected.
46
Which of the following genotypes due to nondisjunction of sex chromosomes is lethal?
OY
47
A recessive sex-linked gene in humans leads to a loss of sweat glands. A woman heterozygous for this will
have patches of skin with and without sweat glands.
48
As real genetic distance increases, the distance calculated by recombination frequency becomes an
underestimate due to multiple crossovers that cannot be scored.
49
Down syndrome is the result of trisomy for chromosome 21. Why is this trisomy viable and trisomy for most other chromosomes is not?
Chromosome 21 is a small chromosome with few genes so this does less to disrupt the genome.
50
Genes that are on the same chromosome can show independent assortment
when they are far enough apart that odd numbers of crossovers is about equal to even.
51
The A and B genes are 10 cM apart on a chromosome. If an A B/a b heterozygote is testcrossed to a b/a b, how many of each progeny class would you expect out of 100 total progeny?
45AB, 45ab, 5Ab, 5aB
52
During the process of spermatogenesis, a nondisjunction event that occurs during the second division would be
better than the first division because only two of the four meiotic products would be aneuploid.
53
What was the key finding from Griffith’s experiments using live and heat-killed pathogenic bacteria?
Genetic material can be transferred from dead to live bacteria.
54
Which of the following is NOT a component of DNA?
The pyrimidine uracil
55
Chargaff studied the composition of DNA from different sources and found that
the proportions of A equal that of T and G equals C.
56
The bonds that hold two complementary strands of DNA together are
hydrogen bonds.
57
The basic mechanism of DNA replication is semiconservative with two new molecules,
each with one new and one old strand.
58
One common feature of all DNA polymerases is that they
synthesize DNA in the 5′-to-3′ direction.
59
Which of the following is NOT part of the Watson–Crick model of the structure of DNA?
The two DNA strands are oriented in parallel (5′-to-3′).
60
If one strand of a DNA is 5′ATCGTTAAGCGAGTCA3′, then the complementary strand would be:
5′ TGACTCGCTTAACGAT 3′.
61
Hershey and Chase used radioactive phosphorus and sulfur to
differentially label DNA and protein.
62
The Meselson and Stahl experiment used a density label to be able to
distinguish between newly replicated and old strands.
63
The difference in leading versus lagging strand synthesis is a consequence of
both the physical structure of DNA and the action of
| polymerase enzyme.
64
if the activity of DNA ligase was removed from replication, this would have a greater effect on
synthesis on the lagging strand versus the leading strand.
65
Successful DNA synthesis requires all of the following except
endonuclease.
66
the synthesis of telomeres
requires telomerase, which uses an internal RNA as a
| template.
67
When mutations that affected DNA replication were isolated, two kinds were found. In cultures that were not synchronized (that is, not all dividing at the same time), one class put an immediate halt to replication, whereas the other put a much slower stop to the process. The first class affects functions at the replication fork like polymerase and primase. The second class affects functions necessary for
initiation: cells complete replication but cannot start a new round.
68
The experiments with nutritional mutants in Neurospora by Beadle and Tatum provided evidence that
genes specify enzymes.
69
What is the central dogma of molecular biology?
information passes from DNA to RNA to protein
70
In the genetic code, one codon
both a and b
a. consists of 3 bases
b. specifies a single amino acid
71
Eukaryotic transcription differs from prokaryotic in that
eukaryotes have three RNA polymerases
72
An anticodon would be found on which of the following types of RNA?
tRNA (transfer RNA)
73
RNA polymerase binds to a__ to initiate ___ .
promoter; transcription
74
During translation, the codon in mRNA is actually“read”by
the anticodon in a tRNA
75
You have mutants that all affect the same biochemical pathway. If feeding an intermediate in the pathway supports growth, this tells you that the enzyme encoded by the affected gene
acts before the intermediate used
76
the splicing process
can produce multiple mRNAs from the same transcript
77
The enzyme that forms peptide bonds is called peptidyl transferase because it transfers
the growing peptide from a tRNA to the next amino acid
78
In comparing gene expression in prokaryotes and eukaryotes
Both a and c are correct.
a. eukaryotic genes can produce more than one protein.
c. both produce mRNAs that are colinear with the protein.
79
The codon CCA could be mutated to produce
a silent mutation
80
An inversion will
only cause a mutant phenotype if the inversion breakpoints fall within a gene
81
What is the relationship between mutations and evolution?
mutations can create new alleles
82
In prokaryotes, control of gene expression usually occurs at the
initiation of transcription.
83
Regulatory proteins interact with DNA by
binding to the major groove of the double helix and interacting with base-pairs.
84
In E.coli, induction in the lac operon and repression in the trp operon are both examples of
negative control by a repressor.
85
The lac operon is controlled by two main proteins. These proteins
act in the opposite fashion, one negative and one positive.
86
In eukaryotes, binding of RNA polymerase to a promoter requires the action of
general transcription factors.
87
In eukaryotes, the regulation of gene expression occurs
at the level of transcription initiation, or post-transcriptionally.
88
In the trp operon, the repressor binds to DNA
in the presence of trp
89
The lac repressor, the trp repressor and CAP are all
allosteric proteins that bind to DNA and an effector.
90
Specific transcription factors in eukaryotes interact with enhancers, which may be a long distance from the promoter. These transcription factors then
can interact with the transcription apparatus via DNA looping.
91
Repression in the trp operon and induction in the lac operon are both mechanisms that
allow the cell to control the level of enzymes to fit
| environmental conditions.
92
Regulation by small RNAs and alternative splicing are similar in that both
Both a and b are correct.
a. act after transcription.
b. act via RNA/protein complexes.
93
Eukaryotic mRNAs differ from prokaryotic mRNAs in that they
are not colinear with the genes that encode them.
94
In the cell cycle, cyclin proteins are produced in concert with the cycle. This likely involves
control of initiation of transcription of cyclin genes, and ubiquitination of cyclin proteins.
95
A mechanism of control in E.coli not discussed in this chapter involves pausing of ribosomes allowing a transcription terminator to form in the mRNA. In eukaryotic fission yeast, this mechanism should
not occur as transcription occurs in the nucleus and translation in the cytoplasm.
