Sample Questions Dent 2011 Biol

The Cell cycle consists of:

A M phase

B G1 phase

C S phase

D Interphase

E All of the above

E

Cyclin D plus cdk4 are necessary for what cell cycle phase transition?

A G1 to S

B S to G2

C G2 to M

D M to G1

E S to G1

A

(Cyclin D is always present however cdk4 is produced in G1 phase to allow transition)

A cell in G2 has exactly two:

A nuclei

B identical copies of its DNA

C S -phases

D mitotic spindles

E nucleosomes

B

(Mitosis -> G1 -> DNA synthesis -> mitosis)

Chromatin consists of

A DNA

B Histones

C cyclin D

D b and c

E a and b

E

(when chromatin bundles up = chromosomes)

What happens to the cell cycle when you treat cells with just caffeine

A nothing

B S-phase is longer

C the DNA failed to replicate

D cells arrest at S-phase

E cells arrest at mitosis

A

(need hydroxy urea to block DNA replication)

A stable kinetochore microtubule is attached to:

A centrosome

B kinetochore

C histones

D a and b

E b and c

D

Crossing over occurs during:

A S phase 1

B Prophase 1

C Metaphase 1

D a and b

E b and c

B

Synapsis occurs between:

A identical chromatids

B centrosomes

C homologs

D b and c

E a and b

C

Anaphase 1 of Meiosis 1 separates:

A identical chromatids

B homologs

C centrosomes

D a and b

E b and c

B

A triploid cannot complete:

A M-phase

B Meiosis 1

C Meiosis 2

D DNA synthesis

E all of the above

B

A true breeding 5' tall bezenia plant is crossed to a true breeding 3' tall bezenia. The F1 progeny are 3/4 5' tall and 1/4 3' tall. In bezenias, tallness is:

a) a discretely varying trait

b) a continuously varying trait

c) incompletely dominant

d) a and c

e) b and c

A

(Discrete -> only 5' or 3' is the option, incompletely dominant = half trait e.g. 4')

In geranias, the purple flower trait is dominant and the white flower trait is recessive. You want to determine the genotype of a purple flower. You will do a:

a) karyotype

b) a test cross

c) a dihybrid cross

d) a southern cross

e) none of the above

B

(Test cross = when you cross an unknown with a known homozygous recessive)

In gryphons, coat color is determined by a single gene with a completely dominant black (B) allele and a recessive (b) white allele. The cross-eyed character is determined by a single gene with a completely dominant normal-eyed allele (C) and a recessive cross-eyed allele (c). You cross two gryphons of unknown parentage: a black normal-eyed with a white normal-eyed. The F1 progeny are 3/4 black normal-eyed and 1/4 black cross-eyed. The genotypes of the parents are:

a) BBCc x bbCc

b) BbCc x bbCc

c) BBCC x bbCC

d) a or b

e) b or c

A

In gryphons, coat color is determined by a single gene with a completely dominant black (B) allele and a recessive (b) white allele. A feral male gryphon of unknown phenotype sneaks into your gryphon pen and mates with your prized black female gryphon. She produces 4 black F1 progeny and 2 white F1 progeny. You can confidently conclude that the genotype of the unknown male is:

a) BB

b) Bb

c) bb

d) a or b

e) b or c

E

You cross a true-breeding snap-poppy with blue flowers with a true-breeding snap-poppy with white flowers. The F1 progeny all have purple flowers. When you self the F1 progeny, to you get 1/4 white progeny, 3/16 blue progeny 3/16 red progeny and 3/8 purple progeny. You can conclude that in snap-poppies:

a) white is epistatic to pigmented

b) pigmented is epistatic to white

c) color is multigenic

d) a and c

e) b and c

D

Height in merigonias is determined by 3 genes that act additively. Each has two alleles, one completely dominant allele (A B or C) that makes plants taller and one recessive allele that makes plants shorter. You do the cross Aa BbCc x AaBbCc. The F1 progeny will be:

a) all as tall as the parents

b) all shorter than the parents

c) 1/64 will be as tall as the parents; 63/64 will be shorter than the parents

d) 9/64 will be as tall as the parents; 55/64 will be shorter than the parents

e) 27/64 will be as tall as the parents; 37/64 will be shorter than the parents

E

Petal number is controlled by a single gene in merigonias. The gene has a completely dominant wild type allele (F) that makes a plant have five petals and a mutant recessive six petal allele (f). However the six petal trait is only 50% penetrant. You do the cross Ff x Ff. What fraction of the progeny do you expect to have 6 petals?

a) 1/2

b) 3/8

c) 1/4

d) 1/8

e) 1/16

D

Gargoyles are tetraploid. A single gene determines whether they have wings or not. This gene has a completely dominant winged allele (W) and a recessive wingless allele (w). You cross two winged gargoyles with genotypes WWWw x WWWw. You expect the progeny to be:

a) all winged

b) 3/4 winged; 1/4 wingless

c) 5/8 winged; 3/8 wingless

d) 7/8 winged; 1/8 wingless

e) 1/16 winged; 15/16 wingless

A

In Gryphons, coat color is determined by a single gene with a dominant black (B) allele and a recessive (b) white allele. The cross-eyed character is determined by a single gene with a dominant normal-eyed allele (C) and a recessive cross-eyed allele (c). The two genes are linked with a map distance of 25 cM. You have developed a test to determine the genotype of individual sperm. You look at the sperm of a dihybrid produced by mating a true-breeding black normal-eyed with a true-breeding white cross-eyed. What fraction of sperm genotypes do you expect to find?

a) 1/2 BC; 1/2 bc

b) 1/2 Bc; 1/2 bC

c) 1/4 BC; 1/4 Bc; 1/4 bC; 1/4 bc

d) 3/4 BC; 1/4 bc

e) 3/8 BC; 3/8 bc; 1/8 Bc; 1/8 bC

E

Height in gryphons is determined by a single locus. You cross a two 9' tall gryphons and get 1/4 10'; 1/2 9'; 1/4 8' tall gryphons. If you cross two 7' tall gryphons you get 1/4 8'; 1/2 7'; 1/4 6' tall gryphons. In gryphons, alleles of the height locus:

a) are semidominant

b) form an allelic series

c) are codominant

d) a and b

e) b and c

D

1) The transforming principle is:

a) DNA

b) RNA

c) The substance from the S strain that makes the R strain virulent

d) a and b

e) a and c

E

If Hershey and Chase labeled T2 phage with heavy nitrogen (15 N) and radioactive sulfur (35 S) and infected bacteria, the new phage that emerge from the infected bacteria would have:

a) heavy DNA

b) heavy protein

c) radioactive DNA

d) a and b

e) b and c

A

If one strand of DNA is 5' CATTAGTAC 3', the complementary strand will be:

a) 5' CATGATTAC 3'

b) 5' CATTAGTAC 3'

f) 5' GTAATCATG 3'

g) 5' GTACTAATG 3'

h) none of the above

A

If I have the following piece of partially double stranded DNA:

5' ATCG 3'

3' TAGCGGCATCCG 5'

And I add DNA polymerase, dTTP, dGTP and dCTP, what will be the sequence of the nucleotides that will be added?

a) 5 'ATCGCCGTAGGC 3'

b) 5' GGCATCCG 3'

c) 5' CCGT 3'

d) 5' CCGTAGGC 3'

e) 5' GGC 3'

C

If you find the following stretch of nucleic acid in a cell in S-phase, at what step in DNA synthesis is it?

5' ...TTTT TTTTTTTT... 3'

3' ...AAAAAAAAAAAAAAAA... 5'

a) Before the primase acts

b) Before polymerase III acts

c) Before polymerase I acts

d) Before ligase acts

e) Synthesis is completed

D

In Neurospora, enzyme B converts ornithine to citrulline and enzyme C converts citrulline to arginine. I cross a mutant that is homozygous for the recessive alleles b and c (which are nonfunctional versions of the genes encoding enzymes B and C respectively) with a homozygous recessive b mutant. The F1 progeny will be able to grow on:

a) arginine

b) citrulline

c) ornithine

d) a and b

e) a and c

D

Reverse transcriptase synthesizes a:

a) DNA strand from a DNA template

b) RNA strand from a DNA template

c) RNA strand from a RNA template

d) DNA strand from a RNA template

e) None of the above

D

A transcript from this strand of DNA:

5' AGCTGGATACG 3'

might be:

a) 5' ACCUAUGC 3'

b) 5' UAUCCAGCU 3'

c) 5' CCAGCU 3'

d) a and b

e) b and c

E

If I take the tRNA that recognizes the codon 5' GCU 3' and mutate the 3' C of the anticodon of to a G, the tRNA will recognize the codon:

a) CCC

b) UGC

c) GCG

d) GCU

e) CGU

A

An aminoacyl tRNA synthase will add what amino acid to the 3' hydroxyl of the wild type tRNA with the anticodon 5' GUC 3':

a) leucine

b) aspartic acid

c) glutamine

d) valine

e) none of the above

B

A ribosome would translate the following mRNA:

5' CGAUGGAUACCUAAA 3'

a) arg-trp-ile-pro-lys

b) asp-gly-tyr-leu

c) met-val-thr

d) met-val-thr-tyr

e) none of the above

C

If a chromosome with the sequence of genes A B C D E F G, underwent an inversion, the sequence of genes might become:

a) ABCDEFEFG

b) ABCDEFGFG

c) ABCDFG

d) ABCCCFG

e) ABCFEDG

E

If a mutation in a gene changed its corresponding transcript from :

5' AUGCCCAAGUAA 3' to 5' AUGCCCAACUAA 3'

what kind of mutation would this be?

a) silent

b) missense

c) nonsense

d) inversion

e) frame shift

B

A silent mutation would result from changing what nucleotide in the mRNA:

5' AUGAACUGG 3'

a) A

b) T

c) G

d) C

e) Any of the above

D

Which of the following is a tRNA anticodon:

a) 5' UUA 3'

b) 5' CUA 3'

c) 5' UCA 3'

d) 5' CCA 3'

e) all of the above

D

The lysogenic cycle involves

a) Injection of viral DNA into the bacterium

b) Replication of viral DNA

c) Integration of prophage into bacterial chromosome

d) Lysis of the infected bacterial cell

e) All of the above

E

All animal viruses have:

a) DNA

b) Protein coat or capsid

c) Membrane envelope

d) a and b

e) a, b and c

B

A wild type Hfr strain conjugates with a auxotroph that requires arginine, tryptophan and lysine to grow. After twenty minutes the mating is interrupted by a Waring blender. Of the recombinant progeny, 50% do not require tryptophan, 30% do not require lysine and 20 % do not require arginine. The order of the genes encoding proteins necessary for the production of lysine (lys), tryptophan (trp) and arginine (arg) on the chromosome of the Hfr strain is:

a) F episome, lys, arg, trp

b) F episome, trp, lys, arg

c) trp, lys, arg, F episome

d) a and b

e) b and c

E

Ways that bacteria can exchange DNA include:

a) specialized transduction

b) integration

c) conjugation

d) a and b

e) a and c

E

A nonsense mutation in the third codon of the lac repressor will:

a) prevent transcription of the lac operon in the presence of lactose

b) prevent binding of lac repressor to lac operator

c) prevent translation of b-galactosidase

d) prevent transcription of the lac operon in the absence of lactose

e) none of the above

B

Examples of transposable elements include:

a) Retrotransposons

b) Alu elements

c) Telomeres

d) a and b

e) b and c

D

To move around the genome, retrotransposons require:

a) Transposase

b) Reverse transcriptase

c) Telomerase

d) Primase

e) None of the above

B

If telomerase is inactivated:

a) Chromosomes become shorter with every cell division

b) Telomeres become longer with every cell division

c) Introns will not be spliced

d) Transposons become longer with every cells division

e) None of the above

A

mRNA processing includes:

a) Transcription

b) Translation

c) Addition of a poly A tail

d) a and b

e) a and c

C

If a pre-mRNA has the exons 5' A^B^C^D^E 3' (where ^ represents and intron), alternative splicing might produce an mRNA with the exons:

a) 5' ABC 3'

b) 5' AABCDE 3'

c) 5' ABDCE 3'

d) 5' ABCBA 3'

e) none of the above

A

If I put a eukaryotic gene in a prokaryotic cell, the protein it encodes will not be expressed if:

a) The eukaryotic gene has a termination codon.

b) The eukaryotic gene has introns.

c) The eukaryotic gene has a wobble base.

d) The eukaryotic gene has a start codon.

e) All of the above

B

TFIID binds:

a) Enhancers

b) Promoters

c) TATA box

d) a and b

e) b and c

E

A triplo-X female (XXX) will have:

a) 1 Barr body

b) 2 Barr bodies

c) 3 Barr bodies

d) a or b

e) none of the above

B

Degradation of Cyclin D at the end of G1 is an example of:

a) Transcriptional regulation

b) Alternative splicing

c) Translational control

d) Posttranslational control

e) All of the above

D

The heat shock factor (HSF) is a transcription factor that binds to an enhancer element that is present in front of each of three genes, hspA, hspB and hspC that are transcriptionally activated by heat shock. If I delete the enhancer element in front of hspA and heat shock the cell:

a) HSF will not be transcribed

b) hspA will not be transcribed

c) hspB will not be transcribed

d) hspC will not be transcribed

e) All of the above

B

The restriction enzyme Hha I cuts the restriction site 5' GCGC 3'. If I cut the double stranded DNA fragment with the sequence:

5' ATGCGCGTTAAGCGAGCGCTTA 3'

with Hha I, how many DNA fragments will result?

a) 1

b) 2

c) 3

d) 4

e) 5

C

The restriction enzyme Spe I cuts the restriction site 5' ACTAGT 3' at the phosphodiester bond between the A and the C. Xba I cuts the restriction site

5' TCTAGA 3' at the TC phosphodiester bond. If I cut my gene with Spe I and my plasmid with Xba I and ligate the fragments together, the ligated junction between the gene and the plasmid will have the sequence:

a) 5' ACTAGT 3'

b) 5' ACTAGA 3'

c) 5' TCTAGA 3'

d) a or b

e) b or c

B

To make a cDNA requires:

a) A primer

b) Nucleotide triphosphates

c) Reverse transcriptase

d) RNA

e) All of the above

E

I have three pieces of double stranded DNA X, Y and Z and each has Bam HI sticky ends at both ends of the strand. If I ligate them together, which of the following ligation products is possible:

a) X-Y- Z

b) Y-X- Z

c) X-Z-Y

d) Z-Y-X

e) All of the above

E

If I have 1 microgram each of double stranded DNA X (3 kb), Y (2kb) and Z (7kb) and I mix them and run them on an agarose gel. The order of the fragments in the gel will be:

a) Anode(-) X Y Z cathode(+)

b) anode Z X Y cathode

c) anode X Z Y cathode

d) anode Y Z X cathode

e) all of the above

B

A human RFLP results in either a 5 kilobase (kb) band + a 4 kb band on a gel (restriction site present) or a 9 kb band (restriction site absent). If a man whose DNA yields 5 kb + 4 kb bands has children with a woman whose DNA yields a 9 kb band, the children's DNA will yield:

a. a 5 kb band + a 4 kb band

b. a 9 kb band

c. an 18 kb band

d. a and b

e. b and c

D

What is the evidence that the Cystic Fibrosis Transmembrane conductance Regulator (CFTR) gene is defective in individuals with Cystic Fibrosis (apart from the name):

a) The CFTR gene has a phenylalanine codon deletion in CF patients

b) The CFTR gene has a BamHI RFLP in its coding region in CF patients

c) The CFTR gene is tightly linked to the CF phenotype

d) a and b

e) a and c

E

If have the following DNA strand:

5' ATAGTCGAATTACAGC 3'

and I add the primers 5' GTCG 3' and 5' GTAA 3' as well as T. aquaticus DNA polymerase and dNTPs and do a PCR reaction, the resulting amplified DNA will have the sequence:

a) 5' ATAGTCGAATTACAGC 3'

b) 5' ATAGTCGAATTAC 3'

c) 5' GTCGAATTACAGC 3'

d) 5' GTCGAATTAC 3'

e) 5' GTCGTTAC 3'

D

If I have a plasmid with an ampicillin resistance gene as well as a tetracycline resistance gene that has a Bam HI site in it and I clone my favorite gene into that Bam HI site, then to find plasmids that contain my gene I will look for transformed bacteria that are:

a) ampicillin resistant

b) tetracycline resistant

c) ampicillin sensitive

d) a and b

e) b and c

A

In order to make large quantities of the protein encoded by a eukaryotic gene in bacteria, the bacterial plasmid you clone the gene into must have:

a) a bacterial promoter

b) the betagalactosidase gene

c) an origin of replication

d) a and b

e) a and c

E