Title | NYU Classes Principles of Biology Lab, Section 001 Tests & Quizzes 2 |
---|---|
Author | Anonymous User |
Course | Principles of Biology I, II |
Institution | New York University |
Pages | 16 |
File Size | 644.9 KB |
File Type | |
Total Downloads | 75 |
Total Views | 157 |
Download NYU Classes Principles of Biology Lab, Section 001 Tests & Quizzes 2 PDF
8/13/2020
NYU Classes : Principles of Biology Lab, Section 001 : Tests & Quizzes
Students: you may be asked to join remote classes via NYU Zoom (available in the left tool menu of your course site). View this Student Zoom Guide.
Your computer's time zone America/Chicago (GMT -5:00) does not match your account's time zone of America/New_York (GMT -4:00). Click here to update your time zone preferences.
Principles of Biology Lab, Section 001
Tests & Quizzes
Tests & Quizzes
Quiz 2 Return to Assessment List
Part 1 of 1 -
40.0 Points
Question 1 of 40 Which is NOT an advantage of using C. elegans in genetic screens, compared to the
1.0 Points
other model organisms (compared to Horse, Pig, Mouse)?
A. Short generation time
B. Large number of offspring
C. Easy to breed and maintain in a laboratory setup
D. They are multicellular organisms
Answer Key: D Question 2 of 40 Which is NOT a method of introducing RNAi to an organism?
A. Soak worms in dsRNA
B. dsRNA are injected directly into the body
C. Microinject RNAi into the organs
D. Feed worms with bacteria carrying gene for dsRNA
1.0 Points
https://newclasses.nyu.edu/portal/site/cd378193-70f6-4c7f-909e-157524012eef/tool/8c67dc67-77ff-496c-9e23-8f2d28ca5f99/jsf/select/selectIndex
1/16
8/13/2020
NYU Classes : Principles of Biology Lab, Section 001 : Tests & Quizzes
Answer Key: C Question 3 of 40 During the genetic screen in c. elegans, if EMS creates a recessive mutation that is
1.0 Points
viable and homozygous. What would you expect to see if we let a mutant F2 hermaphrodite self-fertilize?
A. 50% of F3 worms carry the mutation
B. 25% of F3 worms carry the mutation
C. None of the F3 worms carry the mutation
D. All F3 worms carry the mutation
Answer Key: D Question 4 of 40 How do germ-line mutations differ from somatic mutations?
1.0 Points
A. Germ-line mutations result in mutant gametes, while somatic mutations
do not.
B. Germ-line mutations are reversible, while somatic mutations are not.
C. Germ-line mutations involve small changes to DNA such as base-pair
substitutions, while somatic mutations usually involve large deletions.
D. Germ-line mutations occur during DNA replication, while somatic
mutations do not.
Answer Key: A Question 5 of 40 Why do you need to know the sequence of a gene before you can target it with RNAi?
A. The sequence of the gene will determine the size of the siRNA
B. This sequence of the gene will allow you to mutate it within the
1.0 Points
chromosome
https://newclasses.nyu.edu/portal/site/cd378193-70f6-4c7f-909e-157524012eef/tool/8c67dc67-77ff-496c-9e23-8f2d28ca5f99/jsf/select/selectIndex
2/16
8/13/2020
NYU Classes : Principles of Biology Lab, Section 001 : Tests & Quizzes
C. This will allow us to design a dsRNA that is complementary to the target mRNA
D. The sequence of the gene will provide information on the Dicer protein
Answer Key: C Question 6 of 40 Scientists wanted to determine the levels of pollution that a fish could withstand. They placed fish in tanks with differing pollution levels and timed how long they could live
1.0 Points
under those conditions. What is a controlled variable?
A. Amount of time survived
B. Levels of pollution
C. Type of pollution
D. Number of fish dead
Answer Key: C Question 7 of 40 You randomly created mutations in the genome of drosophila flies to identify genes involved in
1.0 Points
the development of a variation of wing shape. This is an example of forward genetics A. True
B. False
Answer Key: True Question 8 of 40 A scientist testing the effects of a chemical on apple yield sprays an orchard with the
1.0 Points
chemical. A second orchard does not receive the chemical. In the fall, the number of apples harvested from each forest is counted. Which of the following is the independent variable in the experiment?
A. the number of apples
B. the chemical
C. the first orchard
https://newclasses.nyu.edu/portal/site/cd378193-70f6-4c7f-909e-157524012eef/tool/8c67dc67-77ff-496c-9e23-8f2d28ca5f99/jsf/select/selectIndex
3/16
8/13/2020
NYU Classes : Principles of Biology Lab, Section 001 : Tests & Quizzes
D. the second orchard
Answer Key: B Question 9 of 40 RNA interference is a mechanism for silencing gene expression at the
A. level of replication.
B. Post- transcriptional level
C. level of translation.
D. post-translational level.
1.0 Points
Answer Key: B Question 10 of 40 How does the RNA induced silencing complex (RISC) bind to mRNA?
A. Based on base pairing of target mRNA and siRNA
B. Translation inhibited and mRNA degraded.
C. One strand remains bound to the protein.
D. By using the enzymatic function of Dicer to degrade siRNA
1.0 Points
Answer Key: A Question 11 of 40 1.0 Points The image below represents a cell death assay done on a human cell line. The fluorescent red dye stains
https://newclasses.nyu.edu/portal/site/cd378193-70f6-4c7f-909e-157524012eef/tool/8c67dc67-77ff-496c-9e23-8f2d28ca5f99/jsf/select/selectIndex
4/16
8/13/2020
NYU Classes : Principles of Biology Lab, Section 001 : Tests & Quizzes
A. Cells that have receptors for the dye molecule
B. Cells that are dead because their cell membrane is degraded and lets the dye in
C. Cells not fixed by paraformaldehyde
D. All cells that are alive
Answer Key: B Question 12 of 40 In C. elegans RNAi , L4 hermaphrodites are usually required for effective gene silencing in the
1.0 Points
progeny
A. True B. False
Answer Key: True Question 13 of 40 https://newclasses.nyu.edu/portal/site/cd378193-70f6-4c7f-909e-157524012eef/tool/8c67dc67-77ff-496c-9e23-8f2d28ca5f99/jsf/select/selectIndex
5/16
8/13/2020
NYU Classes : Principles of Biology Lab, Section 001 : Tests & Quizzes
For the RNAi treatment in C.elegans, the long dsRNA molecules are produced by the worm cell 1.0 Points True
False
Answer Key: False Question 14 of 40 Male c. elegans have
A. Two X chromosomes (XX)
B. one X chromosome (XO)
C. one X and one Y chromosome (XY)
D. no sex chromosome
1.0 Points
Answer Key: B Question 15 of 40 Why are two T7 promoters needed in the plasmid to create working RNAi?
A. One promoter is needed for the transcription of T7 polymerase
B. This allows transcription of the two complementary strands of the
1.0 Points
dsRNA
C. Because the promoters are not very effective
D. Because at least two molecules of dsRNA are needed to produce RNAi
Answer Key: B Question 16 of 40 You transferred an L4 male worm onto a plate with Roller RNAi. What do you expect to
1.0 Points
see on the plate after one week at 12C?
A. The plate should contain some worms that have the mutation as a
result of incubation with the gene
B. After one week, the offspring of the L4 will express the roller phenotype
https://newclasses.nyu.edu/portal/site/cd378193-70f6-4c7f-909e-157524012eef/tool/8c67dc67-77ff-496c-9e23-8f2d28ca5f99/jsf/select/selectIndex
6/16
8/13/2020
NYU Classes : Principles of Biology Lab, Section 001 : Tests & Quizzes
C. I would expect to see the L4 male worm growing into an adult male
worm
D. The progeny should move in corkscrew fashion staying in the same regions.
Answer Key: C Question 17 of 40 In a controlled cell death assay, what would be your conclusion if the negative control
1.0 Points
turned out positive?
A. All cells survived as expected
B. Our experimental setup is flawed
C. The negative control confirms the validity of the experiment
D. No conclusion can be made
Answer Key: B Question 18 of 40 In an RNAi experiment in c.elegans, someone accidentally forgot to add IPTG to a plate
1.0 Points
with dumpy RNAi plasmids. What phenotype should we expect after one week of treatment?
A. The worms will not feed on the bacteria as there is no IPTG, so we will
have normal worms
B. The IPTG does not influence the expression of the dumpy dsRNA, so
we will expect short and fat worms
C. The lacZ promoter will be inactive and T7 polymerase will not express, so we expect normal worms
D. The dumpy dsRNA gene will not be expressed, only normal worms will be on the plate
Answer Key: D Question 19 of 40 https://newclasses.nyu.edu/portal/site/cd378193-70f6-4c7f-909e-157524012eef/tool/8c67dc67-77ff-496c-9e23-8f2d28ca5f99/jsf/select/selectIndex
7/16
8/13/2020
NYU Classes : Principles of Biology Lab, Section 001 : Tests & Quizzes
How could you increase the ratio of males to hermaphrodites in a wild type stock of C.
1.0 Points
elegans?
A. By decreasing the loss of O chromosomes during reproduction
B. By decreasing the loss of X chromosomes during reproduction
C. By increasing nondisjunction of X chromosomes during reproduction
D. By decreasing cross fertilization in the culture
Answer Key: C Question 20 of 40 EMS is a mutagenic compound, that introduces point mutations into the DNA
1.0 Points
A. True
B. False
Answer Key: True Question 21 of 40 TWIST is a transcription factor that is found to be mutated in some SCS (Saethre-
1.0 Points
Chotzen syndrome) patients. Mutations in which other genes could potentially be a cause for SCS?
A. Genes that are expressed in adults
B. Genes that are not expressed in the same cells as TWIST
C. TWIST target genes upstream expressed in other cells
D. Genes that are controlled by the TWIST gene
Answer Key: D Question 22 of 40 A student set up an experiment to test if plants give off water vapor. Fifty pea plants, growing in pots, were covered with individual glass containers and left overnight. The
1.0 Points
next morning, the inside of each lid was covered in droplets of water. The lab student concluded that plants generally give off water vapor. What critique would you make of the experimental design? https://newclasses.nyu.edu/portal/site/cd378193-70f6-4c7f-909e-157524012eef/tool/8c67dc67-77ff-496c-9e23-8f2d28ca5f99/jsf/select/selectIndex
8/16
8/13/2020
NYU Classes : Principles of Biology Lab, Section 001 : Tests & Quizzes
A. The experiment was not precise, meaning it was not reproducible.
B. The student did not have a clearly stated hypothesis before beginning
the experiment.
C. There was not a large enough sample of pea plants used to get
adequate data. D. There was no control so the water could have come from other sources such as air in the jar or the soil.
Answer Key: D Question 23 of 40 During the process of RNA -interference, DICER is _______?
A. An enzyme that forms a complex with the mRNA
B. An enzyme that breaks long strands of single-stranded DNA into
1.0 Points
siRNA.
C. An enzyme that breaks long strands of double-stranded RNA into
siRNA.
D. An enzyme that inactivates mRNA after siRNA binds to it.
E. An enzyme that takes siRNA and makes it into mRNA.
Answer Key: C Question 24 of 40 How many independent variables can you have in one experiment?
A. 2
B. 1
C. 3
D. As many as you want
1.0 Points
Answer Key: B
https://newclasses.nyu.edu/portal/site/cd378193-70f6-4c7f-909e-157524012eef/tool/8c67dc67-77ff-496c-9e23-8f2d28ca5f99/jsf/select/selectIndex
9/16
8/13/2020
NYU Classes : Principles of Biology Lab, Section 001 : Tests & Quizzes
Question 25 of 40 You mutate a male worm with EMS and cross it to a hermaphrodite expressing GFP in
1.0 Points
the M lineage and find that all the F1 worms are wild type. What do you conclude?
A. The mutation is recessive
B. There are no recessive mutations in the M lineage
C. The mutation is dominant
D. There are neither recessive nor dominant mutations in the M lineage
Answer Key: A Question 26 of 40 You want to find out if temperature has an effect on the growth of bread mold. You decide
1.0 Points
to grow the mold in nine Petri dishes containing the same amount and type of nutrients. Three were kept at 0C, three were kept at 90C, and three were kept at room temperature, 27C. The containers were examined, and the growth of the bread mold was recorded each Friday for five weeks. Which of the following is your hypothesis?
A. If the number of containers change, then the amount of bread mold will change.
B. If the amount of nutrients is changed, then the amount of bread mold will change.
C. If the temperature changes then the amount of bread mold will change.
D. If there are different types of nutrient used, then that will cause differences in the amount of bread mold.
Answer Key: C Question 27 of 40 You are doing a small pilot study to determine the phenotypic effects of knocking out a 1.0 Points regulatory gene expressed early in development by using dsRNA. This an example of forward genetics
A. True B. False
Answer Key: False https://newclasses.nyu.edu/portal/site/cd378193-70f6-4c7f-909e-157524012eef/tool/8c67dc67-77ff-496c-9e23-8f2d28ca5f99/jsf/select/selectIndex
10/16
8/13/2020
NYU Classes : Principles of Biology Lab, Section 001 : Tests & Quizzes
Question 28 of 40 During a cross between a male worm and a hermaphrodite worm, you screened a total of
1.0 Points
2050 F1 progeny under the microscope and found only two males. What can you conclude?
A. Cross-fertilization occurred because we have ~0.1% males in F1
B. Self-fertilization occurred because we have ~0.1% males in F1
C. Self-fertilization occurred because we have ~50% males in F1
D. Cross-fertilization occurred because we have ~50% males in F1
Answer Key: B Question 29 of 40 With regards to siRNA, the most efficient siRNA duplexes will contain which of the
1.0 Points
following features?
A. Will be 21-23 nucleotides in length
B. Comprise a sense strand and an antisense strand of RNA
C. The antisense strand harbors 100% complementarity to the mRNA
target sequence
D. a and b only
E. a, b, and c
Answer Key: E Question 30 of 40 Why do you think wild type hermaphrodites do NOT produce genetically identical
1.0 Points
progeny when crossed to males?
A. Because hermaphrodite sperm outcompetes the sperm from males
B. Because male sperm does not reach the hermaphrodites' eggs
C. Because male sperm outcompetes the sperm from hermaphrodites
D. Because male sperm is not functional
https://newclasses.nyu.edu/portal/site/cd378193-70f6-4c7f-909e-157524012eef/tool/8c67dc67-77ff-496c-9e23-8f2d28ca5f99/jsf/select/selectIndex
11/16
8/13/2020
NYU Classes : Principles of Biology Lab, Section 001 : Tests & Quizzes
Answer Key: C Question 31 of 40 You created a gene knockout of a fibroblast protein in the genome of c elegans to identify genes 1.0 Points involved in the development of cuticle shape. This is an example of forward genetics A. True
B. False
Answer Key: False Question 32 of 40 Which protein, associated with the RNA-inducing silencing complex (RISC), activates
1.0 Points
and cleaves mRNA in RNAi?
A. None of these
B. RNase III Dicer
C. Endonuclease
D. Argonaute
E. RNA helicase
Answer Key: D Question 33 of 40 1.0 Points The following c. elegans is a
A. Cannot tell
B. Hermaphrodite
C. Female
https://newclasses.nyu.edu/portal/site/cd378193-70f6-4c7f-909e-157524012eef/tool/8c67dc67-77ff-496c-9e23-8f2d28ca5f99/jsf/select/selectIndex
12/16
8/13/2020
NYU Classes : Principles of Biology Lab, Section 001 : Tests & Quizzes
D. Male
Answer Key: B Question 34 of 40 What is the structure of the RNAi moiety that targets and binds to specific mRNAs?
A. DNA-protein complex
B. Single stranded RNA
C. Single sheet polypeptide
D. Double stranded RNA
E. Double stranded DNA
1.0...