BIOL133 A005 Sum 19 – APU Exam 3- Chapters 12-15
BIOL133 A005 Sum 19 – APU Exam 3- Chapters 12-15
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Part 1 of 1 – 100.0 Points
Question 1 of 40
2.5 Points
If genes are completely linked, all the offspring will be _________.
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A. hybrids
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B. dominant
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C. recombinants
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D. parental
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E. recessive
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Question 2 of 40
2.5 Points
The codon AUG specifies the amino acid methionine. What would the tRNA anitcodon be that recognizes this codon?
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A. UAC
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B. GUA
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C. AUG
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D. GCA
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Question 3 of 40
2.5 Points
Due to a random error by the aminoacyl tRNA synthetase, a tRNAVal was misloaded with Ile. What do you predict will result?
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A. One protein will get Val inserted where Ile was supposed to go, leading to minor structural change in the protein.
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B. One protein will get Ile inserted where Val was supposed to go, leading to major structural changes in the protein.
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C. One protein will get Ile inserted where Val was supposed to go, leading to minor structural change in the protein.
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D. The cell’s proteins will get Ile and Val inserted in the wrong places, leading to major structural changes in the proteins.
Feedback: Learning Objective: Analyze how genes determine the proteins found in all cells
Question 4 of 40
2.5 Points
A mutation causes a codon to change from UAC to UAU, both of which specify tyrosine. This is an example of
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A. a frameshift mutation
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B. missense mutation
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C. silent mutation
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D. nonsense mutation
Feedback: Learning Objective: Discuss the different types of mutations in DNA
Question 5 of 40
2.5 Points
Nondisjunction during meiosis II will result in ________.
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A. n
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B. n-1
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C. n+1
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D. n+1 and n-1
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E. n,n-1 and n+1
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Question 6 of 40
2.5 Points
There are a number of unique features to eukaryotic pre-mRNA splicing. Select all that apply.
A. Introns are thought to encode specific subunits of proteins.
B. Introns likely slow down gene expression.
C. Introns need to be removed precisely to prevent mutations.
D. There are often multiple introns in a gene.
Feedback: Learning Objective: Understand the significance of exons, introns, and splicing
Question 7 of 40
2.5 Points
The codon AUG specifies the amino acid methionine. What would the tRNA anitcodon be that recognizes this codon?
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A. UAC
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B. GUA
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C. AUG
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D. GCA
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Question 8 of 40
2.5 Points
Frameshift mutations occur when
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A. bases are added or deleted
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B. there is an amino acid substitution
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C. there is a transition substitution
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D. transversion substitution
Feedback: Incorrect-
Learning Objective Text: Discuss the different types of mutations in DNA
Question 9 of 40
2.5 Points
An experiment started with 15N15N DNA. After two generations in 14N medium, E. coli will have
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A.
25% 14N14N and 75% 15N15N
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B.
50% 14N14N and 50% 15N15N
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C.
75% 15N14N and 25% 14N14
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D.
50% 15N14N and 50% 14N14N
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E.
50% 15N14N and 50% 15N15N
Feedback:
Question 10 of 40
2.5 Points
Which amino acid can sometimes be present at the P site without first entering the A site of the ribosome?
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A. alanine
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B. methionine
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C. tryptophan
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D. leucine
Feedback: Learning Objective: Discuss the role of ribosomes in protein synthesis
Question 11 of 40
2.5 Points
A mutation changing an alanine to a glycine would likely have little effect on the protein function since they are both small, ________, amino acids.
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A. positive
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B. nonpolar
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C. negative
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D. charged
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E. polar
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Question 12 of 40
2.5 Points
What best describes the human karyotype?
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A. 42 pairs of autosomes
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B. one pair of sex chromosomes and 23 pairs of autosomes
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C. X chromosomes and 22 pairs of autosomes
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D. One pair of sex chromosomes and 22 pairs of autosomes
Feedback:
Question 13 of 40
2.5 Points
A karyogram that contains 23 pairs of chromosomes with a different pair for the 23rd pair would describe a(n)
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A. autosomal pair
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B. male
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C. female
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D. disorder
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Question 14 of 40
2.5 Points
Why is working with linked genes not simple?
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A. potential of crossing over in meiosis I
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B. potential of independent assortment in meiosis I
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C. they reside on different chromosomes
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D. one gene often masks the other gene
Feedback: Learning Objective: Explain the effect of linkage and recombination on gamete genotypes
Question 15 of 40
2.5 Points
For her science fair project, your little sister replicated one of Mendel’s experiments. From a P0 pea plant with purple flowers and a pea plant with white flowers, she got 1 purple F1 hybrid (she had forgotten to water, so just one survived). From this self-fertilized F1 she got 8 F2 offspring. 7 were purple and 1 was white, for a 7:1 ratio.
Did your little sister just disprove Mendel’s laws?
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A. Yes. This, and other examples of non-Mendelian genetics demonstrate that Mendel’s Laws are actually incorrect and misleading.
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B. No. When another pea pod forms on the plant, that will bring the numbers up to the 9:3, or 3:1 ratio
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C. No. Dehydrating the parent caused Lamarckian selection on the offspring.
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D. No. Random chance can cause deviations in the ratio of small numbers of progeny.
Feedback: Learning Objective: Analyze Mendelian and Non-Mendelian inheritance patterns
BIOL133 A005 Sum 19 – APU Exam 3- Chapters 12-15
Question 16 of 40
2.5 Points
A karyogram that contains 23 pairs of chromosomes with a different pair for the 23rd pair would describe a(n)
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A. autosomal pair
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B. male
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C. female
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D. disorder
Feedback:
Question 17 of 40
2.5 Points
The most common time individual assortment occurs is when genes are ________?
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A. close together
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B. unlinked
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C. linked
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D. on non-sister chromatids of homologues
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Question 18 of 40
2.5 Points
Low pitched male voices (RR) and a high pitched male voice (rr). Heterozygotes have a baritone voice. This is an example of _________.
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A. incomplete dominance
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B. codominance
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C. Mendelian genetics (dominance and recessive
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D. multiple alleles
Feedback: Learning Objective: Identify non-Mendelian inheritance patterns such as incomplete dominance, codominance, recessive lethals, multiple alleles, and sex linkage
Question 19 of 40
2.5 Points
Nondisjunction during meiosis I will result in ________.
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A. n
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B. n-1
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C. n+1
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D. n+1 and n-1
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E. n, n-1, and n+1
Feedback: Learning Objective: Explain how nondisjunction leads to disorders in chromosome number
Question 20 of 40
2.5 Points
What is the order of DNA compaction in eukaryotes?
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A. Helix, chromatin fiber, nucleosomes, duplicated chromosome, condensation of chromatin
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B. Condensation of chromatin, nucleosomes, chromatin fiber, helix, duplicated chromosome
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C. Chromatin fiber, condensation of chromatin, duplicated chromosome, nucleosome, helix
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D. Helix, nucleosome, chromatin fiber, condensation of chromatin, duplicated chromosome
Feedback:
Question 21 of 40
2.5 Points
What best describes the difference between paracentric and pericentric inversion?
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A. the number of genes inverted
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B. the number of centromeres inverted
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C. the placement of centromeres in the inversions
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D. if the inversion is on autosomal of sex chromosomes
Feedback: Learning Objective: Describe how errors in chromosome structure occur through inversions and translocations
Question 22 of 40
2.5 Points
What does a spliceosome remove from pre-mRNA during the process of splicing?
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A. introns
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B. exons
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C. spliceons
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D. 7-methlyguanosine
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E. poly-A tail
Feedback: Learning Objective: Describe the different steps in RNA processing
Question 23 of 40
2.5 Points
In co-dominant traits, what is the phenotypic ratio when two heterozygotes cross?
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A. 1:2:1
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B. 3:1
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C. 2:2 (1:1)
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D. 1:3
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E. 4:0
Feedback: Learning Objective: Identify non-Mendelian inheritance patterns such as incomplete dominance, codominance, recessive lethals, multiple alleles, and sex linkage
Question 24 of 40
2.5 Points
Combination of alleles that independently assort is usually higher than the number of chromosomes because
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A. crossing over
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B. translocation
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C. segregation
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D. jumping genes
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E. gene linkage
Feedback:
Question 25 of 40
2.5 Points
Which is likely to disrupt the function of the most genes: a 500kb deletion, a 500kb inversion, or a 500kb translocation?
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A. Deletion
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B. Inversion
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C. Translocation
Feedback: The deletion will disrupt all the genes contained within it, but the inversion and translocation will likely only affect the genes on the ends.
Question 26 of 40
2.5 Points
How can an UV induced mutation affect DNA?
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A. Creating guanine dimers between guanines next to each other
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B. Creating thymine dimers between thymines next to each other
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C. Creating adenine dimers between adenines next to each other
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D. Creating cytosine dimers between cytosines next to each other
Feedback:!
Question 27 of 40
2.5 Points
If a human has straight hair or curly hair they are homozygous. An individual with curly hair is heterozygte. This is an example of what type of genetic expression?
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A. Mendelian genetics (dominance and recessive)
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B. incomplete dominance
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C. multiple alleles
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D. codominance
Feedback: Incorrect-
Learning Objective Text: Identify non-Mendelian inheritance patterns such as incomplete dominance, codominance, recessive lethals, multiple alleles, and sex linkage
Question 28 of 40
2.5 Points
What is the process of ribosomes making proteins?
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A. translation
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B. replication
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C. transcription
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D. mutation
Feedback: Incorrect-Learning Objective Text: Explain the “central dogma” of protein synthesis
Question 29 of 40
2.5 Points
The fourth codon has a mutation changing a UAC to a UAG. What effect will this have on the protein?
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A. None due to the degeneracy of the code
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B. It wouldn’t be made due to coding for a nonsense codon
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C. None due to the universality of the code
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D. It would be made since there would no longer be a start codon.
Feedback:
Question 30 of 40
2.5 Points
A child contains what proportion of their parent’s genetic makeup?
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A. 50:50
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B. 40:60
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C. 33:67
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D. 25:75
Feedback:
BIOL133 A005 Sum 19 – APU Exam 3- Chapters 12-15
Question 31 of 40
2.5 Points
________ found in the -10 sequence makes it easier for the DNA to separate.
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A. G-C base pairs
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B. Template strands
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C. Non-template strands
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D. A-T base pairs
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E. Start codons
Feedback: Incorrect-Learning Objective Text: Discuss the role of promoters in prokaryotic transcription
Question 32 of 40
2.5 Points
How can an UV induced mutation affect DNA?
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A. Creating guanine dimers between guanines next to each other
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B. Creating thymine dimers between thymines next to each other
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C. Creating adenine dimers between adenines next to each other
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D. Creating cytosine dimers between cytosines next to each other
Feedback: Incorrect-
Learning Objective Text: Discuss the different types of mutations in DNA
Question 33 of 40
2.5 Points
What type of enzyme cuts the pre-mRNA between the AAUAAA consensus sequence and a GU-rich sequence?
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A. exonuclease
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B. cleavase
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C. endonuclease
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D. sequenase
Feedback:
Question 34 of 40
2.5 Points
What is a major difference between DNA polymerase I and DNA polymerase III?
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A. DNA polymerase I synthesizes DNA on leading strands and DNA polymerase III synthesizes DNA on lagging strands
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B. DNA polymerase I synthesizes DNA on lagging strands and DNA polymerase III synthesizes DNA on leading strands
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C. DNA polymerase I repairs DNA and DNA polymerase III synthesizes DNA in the 3? to 5? direction
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D. DNA polymerase I synthesizes DNA in the 5? to 3? direction and DNA polymerase III synthesizes on lagging strands
Feedback:
Question 35 of 40
2.5 Points
If genes are completely linked, all the offspring will be _________.
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A. dominant
•
B. parental
•
C. recombinants
•
D. recessive
•
E. hybrids
Feedback:
Question 36 of 40
2.5 Points
To determine the order of genes on a chromosome, one can perform a
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A. three point cross
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B. punnet square
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C. ideogram
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D. two point cross
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E. test cross
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F. karyotype
Feedback:
Question 37 of 40
2.5 Points
Where is the promoter sequence TATAAT located?
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A. +35
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B. +10
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C. +1
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D. -1
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E. -10
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F. -35
Feedback:
Question 38 of 40
2.5 Points
What is a major difference between DNA polymerase I and DNA polymerase III?
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A. DNA polymerase I synthesizes DNA on leading strands and DNA polymerase III synthesizes DNA on lagging strands
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B. DNA polymerase I synthesizes DNA on lagging strands and DNA polymerase III synthesizes DNA on leading strands
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C. DNA polymerase I repairs DNA and DNA polymerase III synthesizes DNA in the 3? to 5? direction
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D. DNA polymerase I synthesizes DNA in the 5? to 3? direction and DNA polymerase III synthesizes on lagging strands
Question 39 of 40
2.5 Points
A child contains what proportion of their parent’s genetic makeup?
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A. 50:50
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B. 33:67
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C. 40:60
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D. 25:75
Feedback:
BIOL133 A005 Sum 19 – APU Exam 3- Chapters 12-15
Question 40 of 40
2.5 Points
The allele for widows peak (H) is dominant for the allele for no widows peak (h). At a different gene locus, the allele for hitchhikers thumb (D) is dominant to the allele for non-hitchhikers thumb (d). A man is heterozygote for the traits and marries a woman who has no widows peak and is heterozygote for hitchhikers thumb. What is the man’s genotype?
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A. HHDD
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B. HhDd
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C. HHdd
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D. hhDD
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E. hhdd
