Download Genetics Study Guide Answers

Genetics Study Guide
Things you need to know/ be able to answer:
Cell Cycle & Cell Division
1. Haploid vs. Diploid cells
2. Cell Cycle (Parts of interphase)
3. Cell Cycle Control (Cancer & examples
of internal and external control)
4. Mitosis vs. Meiosis
5. What processes in meiosis cause
variation?
6. Chromosomal abnormalities & Disorders
DNA to Protein
1. History of DNA
a. Scientists & their experiments
2. Structure of DNA
a. Monomers
b. Purines vs. pyrimidines
c. DNA in pro vs. euk
3. DNA replication
a. Why & when it happens
b. How it happens
4. Transcription (pro and euk)
5. Translation (pro and euk)
6. Mutations
7. Regulation of Gene expression (pro and
euk) and it’s relation to cell communication
a. Lac & tryp operon
Biotechnology
1. Tools
a. Restriction Enzymes
b. Vectors (plasmids)
c. cDNA
d. PCR
e. Gel electrophoresis
2. Applications
a. Cloning
b. Gene therapy
c. Pharmaceutical products
d. GMOs
Mendelian Genetics
1. Mendel’s Laws
2. Basic Rules
a. Monohybrids & dihybrids
b. Test cross
3. Complex Rules
a. Incomplete vs. codominance
b. Multiple alleles vs. polygenic
c. Blood types & Sex-linked traits
d. Pleiotropy vs. epistasis
4. Pedigrees
a. Predicting genotypes/phenotypes
5. Human Genetics & Disorders
a. Cause of linked genes
Example Questions:
Which of the following is true of a species that
has a chromosome number of 2n = 16?
A) The species is diploid with 32 chromosomes
per cell.
B) The species has 16 sets of chromosomes
per cell.
C) Each cell has eight homologous pairs.
D) During the S phase of the cell cycle there
will be 32 separate chromosomes.
E) A gamete from this species has four
chromosomes.
Chromatids are separated from each other.
A) The statement is true for mitosis only.
B) The statement is true for meiosis I only.
C) The statement is true for meiosis II only.
D) The statement is true for mitosis and
meiosis I.
E) The statement is true for mitosis and
meiosis II.
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%
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.
A particular triplet of bases in the template
strand of DNA is 5' AGT 3'. The corresponding
codon for the mRNA transcribed is
A) 3' UCA 5'.
B) 3' UGA 5'.
C) 5' TCA 3'.
D) 3' ACU 5'.
E) either UCA or TCA, depending on wobble in
the first base.
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.
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.
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.
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
Cinnabar eyes is a sex-linked recessive
characteristic in fruit flies. If a female having
cinnabar eyes is crossed with a wild-type male,
what percentage of the F1 males will have
cinnabar eyes?
A) 0%
B) 25%
C) 50%
D) 75%
E) 100%
Recombination between linked genes comes about for what reason?
A) Mutation on one homolog is different from that on the other homolog.
B) Independent assortment sometimes fails because Mendel had not calculated appropriately.
C) When genes are linked they always "travel" together at anaphase.
D) Crossovers between these genes result in chromosomal exchange.
E) Nonrecombinant chromosomes break and then re-join with one another.
What is the source of the extra chromosome 21 in an individual with Down syndrome? Nondisjunction
in the gametes of one of the parents
Each group of questions below consists of five lettered headings followed by a list of numbered phrases or
sentences. For each numbered phrase or sentence, select the one heading to which it is most closely related
and fill in the corresponding circle on the answer sheet. Each heading may be used once, more than once, or
not at all in each group.
Questions 53-56 refer to the following genetic crosses, in which R and T represent dominant alleles
and r and t represent recessive alleles for two genes that are located on different chromosomes.
(A) ttrr x ttrr
(B) TtRrxTtRr
(C) TTRrxTtRR
(D) TtRr x ttrr
(E) TtRrx ttRr
53. Offspring from this cross are expected to exhibit a 1 : 1 : 1 : 1 phenotypic ratio. D
54. Offspring from this cross are expected to exhibit a 3 : 3 : 1 : 1 phenotypic ratio. E
55. All offspring from this cross are expected to exhibit both dominant traits. C
56. This cross is an example of a dihybrid cross in which both parents are heterozygous. B
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The figure above represents a generalized hormone-signaling pathway. Briefly explain the role of each
numbered step in regulating target gene expression.
The following questions refer to this table of codons.
A possible sequence of nucleotides in
the template strand of DNA that would
code for the polypeptide sequence
phe-leu-ile-val would be
A) 5' TTG-CTA-CAG-TAG 3'.
B) 3' AAC-GAC-GUC-AUA 5'.
C) 5' AUG-CTG-CAG-TAT 3'.
D) 3' AAA-AAT-ATA-ACA 5'.
E) 3' AAA-GAA-TAA-CAA 5'.
UUU-CUU-AUU-GUU