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Multiple Choice
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1
Haploinsufficiency describes a situation where one wild-type copy of a gene is not enough for normal development to occur.
A)True.
B)False.
2
RNAi can be used to create a phenocopy that mimics a loss-of-function mutation.
A)True.
B)False.
3
A gain-of-function mutation is always dominant to the wild-type allele.
A)True.
B)False.
4
The ectopic phenotype is the one that is expressed in the double mutant.
A)True.
B)False.
5
In situ RNA hybridization can be used to determine where in a developing organism a gene is expressed.
A)True.
B)False.
6
A good model organism exhibits all the characteristics below except:
A)ease of cultivation
B)long gestation period
C)small size
D)a sequenced genome
E)a stock center that maintains and distributes mutants
7
If one of the two cells of a two-cell ________ embryo is removed or destroyed, a complete organism cannot develop. If, however, the two cells of a two-cell ________ embryo are separated from each other, viable twins will develop.
A)C. elegans; human
B)human; C. elegans
C)C. elegans; C. elegans
D)human; human
E)none of the above
8
Mutations that give rise to protein with diminished or no biochemical activity are referred to as:
A)hypomorphic
B)dominant-negative
C)loss-of-function
D)null
E)conditional
9
Mutations that remove all function of a protein are referred to as:
A)hypomorphic
B)dominant-negative
C)loss-of-function
D)null
E)conditional
10
Mutations that retain partial activity are referred to as:
A)hypomorphic
B)dominant-negative
C)loss-of-function
D)null
E)conditional
11
Mutations that express a phenotype only under specific circumstances are referred to as:
A)hypomorphic
B)dominant-negative
C)loss-of-function
D)null
E)conditional
12
Mutations that result in an inactive product that "poisons" or otherwise counteracts the wild-type protein are referred to as:
A)hypomorphic
B)dominant-negative
C)loss-of-function
D)null
E)conditional
13
You are trying to determine when during development a particular C. elegans protein is required. This can be accomplished by determining the temperature sensitive period (tsp) of a protein. First, you isolate a temperature-sensitive (ts) allele of the gene that encodes this protein. In this particular case, the ts allele is sensitive to higher temperatures. Thus, the permissive temperature is 20°C and the non-permissive temperature is 25°C. To determine the tsp, you need to perform two temperature-shift experiments, a downshift and an upshift experiment. A downshift experiment involves fertilizing eggs at the non-permissive temperature, shifting them in batches at various times to the permissive temperature, and scoring the resulting progeny for the mutant phenotype. If the embryos were downshifted before the tsp began, then they will not exhibit the mutant phenotype. A downshift experiment defines the beginning of the tsp window. An upshift experiment involves fertilizing eggs at the permissive temperature, shifting them in batches at various times to the non-permissive temperature, and then scoring the resulting progeny for the mutant phenotype. If the embryos were upshifted after the end of the tsp, then they will not exhibit the mutant phenotype. An upshift experiment defines the end of the tsp window. When you perform your downshift and upshift experiments, you obtain the following results:

 % Mutant Animals
Time of Shift (hrs after fert.)Downshift ExperimentUpshift Experiment
0 0 100
1 0 100
2 0 100
3 0 100
4 5 100
5 35 100
6 90 94
7 92 90
8 94 85
9 96 80
10 100 40
11 100 5
12 100 0
13 100 0
14 100 0
15 100 0

What is the tsp for this protein?
A)5-15 hrs post-fertilization
B)0-10 hrs post-fertilization
C)5-10 hrs post-fertilization
D)0-4 hrs post-fertilization
E)11-15 hrs post-fertilization
14
Your favorite gene encodes a subunit of a multimeric transcription factor. This transcription factor binds as a tetramer and all four subunits must be functional to retain binding activity. A particular dominant-negative loss-of-function mutation of this gene prevents binding to DNA but not to other subunits. In an animal that is heterozygous for the wild-type and the dominant-negative alleles, what fraction of tetramers will be functional?
A)1/2
B)1/4
C)1/8
D)1/16
E)1/64
15
You are trying to identify the genes required for proper movement in the nematode. your first step is to expose wild-type animals to mutagen, then screen for animals that no longer move properly (uncoordinated or Unc). After many weeks of hard work, you isolate a total of 20 independent Unc mutants. You then do a complementation analysis of your mutant strains (1-20) and get the following results ("-" indicates failure to complement):

  1,3,6,17,20    
1,3,6,17,20 - 2,8,12,13,18   
2,8,12,13,18 + - 4  
4 + + - 5,9,10,14,19 
5,9,10,14,19 + + + - 7,11,15,16
7,11,15,16 + + + + -

How many different genes did you identify in your screen and is the screen saturated?
A)5; yes
B)5; no
C)20; yes
D)20; no
E)20; can't tell
16
You are studying a protein in yeast that you suspect interacts directly with another one to form a complex, but you don't know the identity of the other protein. To try to find it, you carry out a suppressor screen of a mutation in your known gene to find a rare compensating mutation in an unknown gene. Which type of mutant allele of your gene would be most suitable for such a suppressor screen?
A)null
B)conditional
C)promoter mutation that reduces the level of transcription
D)missense mutation that changes the shape of your protein
E)frameshift
17
You have isolated four developmental mutant strains that each arrest at a different stage of development (shown below). Give the likely order in which the genes function.

MutantArrests at
A four-cell stage
B one-cell stage
C blastula stage
D two-cell stage

A)CADB
B)BDAC
C)BDCA
D)ACDB
E)ABCD
18
Double mutant (or epistasis) analysis can often help you determine the order of function of genes in a developmental pathway. To correctly interpret the results, however, it is important to understand the type of pathway you are studying. In the yeast secretory pathway, for example, many genes act sequentially in order to secrete molecules from the cell. Mutations in these genes will cause molecules that are supposed to be secreted to accumulate at particular stages of secretion. In a strain that is mutant for gene A, secretory vesicles do not get loaded with molecules to be secreted. You can see molecules accumulating in the endoplasmic reticulum. In a strain that is mutant for gene B, vesicles are loaded, but cannot fuse to the cell membrane and release their contents. You will see secretory molecules accumulating within vesicles. In the doubly mutant strain, the accumulation of molecules that are supposed to be secreted in the endoplasmic reticulum tells you that the genes act in the order:
A)A, then B
B)B, then A
C)A and B at the same time
D)can't tell
E)none of the above
19
In C. elegans, the signaling pathways that leads to the development of the hermaphrodite vulva includes three genes, let-60, let-45, and mek-2. In wild-type animals, a signal from a nearby cell activates the LET-60 protein in particular cells destined to produce the vulva. The activation of LET-60 results in the activation of LET-45, which results in the activation of MEK-2, which eventually results in the formation of a vulva. In a mek-2 loss-of-function (lf) mutant, the signal is blocked at the mek-2 step and no vulva is formed (vulvaless). In a let-60 gain-of-function (gf) mutant, the LET-60 protein is always active, even in cells that have not received signal to "make a vulva". Those cells then try to make vulvas, resulting in an animal with too many vulvas (multivulva). What is the phenotype of the let-60(gf); mek-2(lf) double mutant?
A)wild-type
B)vulvaless
C)multivulva
D)dead
E)can't tell
20
Proteins isolated from jellyfish that allow other proteins to be located within a living organism.
A)homeodomains
B)GFP proteins
C)endocrine factors
D)juxtacrine factors
E)paracrine factors
21
Motifs that specify a DNA-binding domain.
A)homeodomains
B)GFP proteins
C)endocrine factors
D)juxtacrine factors
E)paracrine factors
22
Cell surface molecules that require direct contact to initiate cell signaling.
A)homeodomains
B)GFP proteins
C)endocrine factors
D)juxtacrine factors
E)paracrine factors
23
Ligands that are secreted by the signaling cell and typically received by a nearby cell.
A)homeodomains
B)GFP proteins
C)endocrine factors
D)juxtacrine factors
E)paracrine factors
24
Ligands that circulate throughout the body in the blood and can affect tissues far removed from the gland that produces them.
A)homeodomains
B)GFP proteins
C)endocrine factors
D)juxtacrine factors
E)paracrine factors
25
The level of the C. elegans LIN-14 protein is known to be regulated by the expression of the lin-4 gene, but this gene does not encode a protein. Animals carrying a loss-of-function mutation in the lin-4 gene fail to downregulate the translation of the lin-14 mRNA at the proper time during development. A suppressor screen was carried out in these animals and a suppressor mutation was identified within the lin-14 gene. When this lin-14 mutation was separated from the lin-4 mutation and crossed into a strain with a wild-type lin-4 gene, it was discovered that wild-type lin-4 was unable to downregulate the mutant lin-14 mRNA. Where is the most likely location of the lin-14 mutation obtained in the suppressor screen?
A)promoter
B)coding region
C)intron
D)splice site
E)3' UTR
26
Mutations affecting the direction of shell coiling in the snail Limnaea peregra are known to exhibit maternal effects. That is, the phenotypes of the progeny are determined by the genotype of the mother. An F1 snail produced by a cross between two individuals has a shell with a right-hand twist (dextral-coiling). This snail produces only left-hand (sinistral) progeny on selfing. If "D" represents the dominant dextral allele and "d" represents the recessive sinistral allele, what is the genotype of the F1 snail?
A)DD
B)Dd
C)dd
D)Dd or dd
E)DD or Dd
27
You are a new graduate student in a Drosophila lab and your thesis advisor has given you a new mutation to characterize. Your mutation causes fly larvae to have missing head structures (headless). Because this is a lethal mutation, you obviously cannot create a true-breeding strain. Instead, you cross animals that you know are heterozygous for the same mutation (don't worry about how you know they are heterozygous). Surprisingly, all the progeny from the headless X headless cross are fine (wild type). Thinking you must have messed the cross up, you set the vial containing the cross aside and work on other experiments for awhile. About 14 days (one generation) later, you notice some dead larvae in your headless X headless cross vial. After looking at the dead larvae in the microscope, you discover that they are headless! You also notice that there are some wild-type larvae in the vial also. Scratching your head, you talk to an older (and wiser) graduate student and explain your findings. He tells you to go read your genetics textbook. After pouring over your textbook for a few hours, you come upon a possible explanation. What do you think might be going on?
A)X-inactivation
B)cytoplasmic inheritance
C)incomplete penetrance
D)maternal effect lethality
E)epistasis
28
The mRNA of the bicoid (bcd) gene is transcribed in the nuclei of Drosophila nurse cells and then transported into the egg from its anterior end. As the bcd mRNA enters the developing oocyte, it becomes attached to the cortex at the anterior end of the egg cell. Then, as the mRNA is translated, a gradient of Bicoid protein is formed, thus establishing the anterior/posterior axis of the Drosophila embryo. Cells in regions containing high levels of Bicoid protein develop head structures, whereas cells in regions of low Bicoid protein develop abdominal structures. Since this event occurs within the mother in developing oocytes prior to fertilization, bcd mutations affecting this process are maternal-effect mutations. Thus, embryos produced by bcd/bcd mothers are headless. If bcd/+ flies are crossed together and individual F1 females from that cross are crossed to wild-type males, what proportion of the crosses will yield all headless larvae?
A)1/2
B)1/4
C)1/8
D)1/16
E)0
29
Place the following cell signaling events in order:
  1. Notch intracellular domain enters nucleus
  2. Notch protein changes shape
  3. Delta ligand on one cell binds Notch receptor on adjacent cell
  4. Notch intracellular domain activates transcription factor
  5. Notch protein is cleaved

A)CBEAD
B)CEBAD
C)CEDBA
D)ADBCE
E)ADBEC





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