Principles of Genetics: Final Exam

In mice, a, a recessive gene that determines albinism (A is normal), is linked to the beta globin gene which has two codominant alleles, r and m. If males who are homozygous for both a and m are to be used in mattings, what genotype would you suggest is most appropriate for the females in order to detect recombinants?
a. Aamr
b. aamr
c. Aarr
d. aamm
e. aarr

A

In a certain species, the A locus and the D locus are so tightly linked that no recombination is ever observed between them. Alleles A and D are dominant to a and d at these two loci, respectively. If a cross was made in coupling, what phenotypic ratio will be observed in the F2?
a. 9:3:3:1
b. 1:1
c. 1:2:1
d. 3:1
e. 1:0

D

In crosses involving two loci, why can't we detect linkage of more than 50 map units?
a. because there is too much crossing over when more than 50 map units separate genes.
b. because the math gets too difficult.
c. because chromosomes are only 50 map units long.
d. because linkage with more than 50 map units gives the same results as independent assortment.
e. because when genes are more than 50 map units apart, they cross over to non homologous chromosomes.

D

A cross is made between BBRR and bbrr, and the F1 is testcrossed with a completely recessive individual. If the loci are closely linked, which two genotypes will be in the minority in the testcross progeny?

a. bbRr and bbrr

b. BbRr and Bbrr

c. bbRr and BbRr

d. Bbrr and bbRr

e. BbRr and bbrr

D

A and a are alleles of the A gene, and B and b are alleles for the B gene. Loci for genes A and B are on the same chromosome. Two kinds of crosses are made, one in coupling phase and the other in repulsion phase. Linkage distance between the two loci:
a. is greater in repulsion.
b. is less in repulsion.
c. is the same in both crosses.
d. is greater in coupling.
e. does not exist.

C

You make the cross Ar/aR X ar/ar. Out of a large number of progeny, 4% are aarr. The distance between loci is:
a. 8 map units.
b. 4 map units.
c. 6 map units.
d. 10 map units.
e. 12 map units.

A

In snowbirds, a migratory bird found in South Florida, obnoxious is dominant to well‑mannered, and yellow‑bellied is dominant to red‑bellied. Since we all know the well‑mannered allele is now extinct, the following data are from an ancient manuscript. A testcross of a heterozygous snowbird gave the following:
-obnoxious, yellow‑bellied 183
-obnoxious, red‑bellied 13
-well‑mannered, yellow‑bellied 17
-well‑mannered, red‑bellied 187

The percent recombination was:
a. 12.5%
b. 1.5%
c. 0.75%
d. 7.5%
e. 15.0%

D

In a herd of 100 cattle, 25 have red coat (RR), 36 have white coats (rr) and 39 have roan coats (Rr). What is the frequency of the R allele?

a. 0.445

b. 0.5

c. 0.545

d. 0.4

e. 0.25

A

In a certain human population, the frequency of the recessive sex-linked allele for hemophilia is 0.01. What proportion of males in this population would be expected to have hemophilia? (assume genetic equilibrium in the population)
a. 0.0198
b. 0.99
c. 0.9999
d. 0.01
e. 0.0001

D

In a population in equilibrium, 36% of the population show the dominant trait. What is the frequency of the recessive allele?
a. 0.20
b. 0.60
c. 0.80
d. 0.64
e. 0.40

C

In a certain breed of dogs, DD makes dogs obedient to their master. Dd makes them reluctantly listen to their master, and dd makes them independent and downright obnoxious. If the allelic frequency of D is 0.4, what will be the frequency of reluctantly listening dogs in a population in equilibrium for this locus?
a. 0.48
b. 0.16
c. 0.6
d. 0.36
e. 0.4

A

Studies on liver color in bulldogs have shown that the lily color (L) is dominant to white (l). If the frequency of the L allele is 0.6, how many lily-livered bulldogs would be expected out of 100, assuming genetic equilibrium?
a. 36
b. 84
c. 100
d. 18
e. 60

B

Over generations, a population of wild rats obeying the Hardy-Weinberg equilibrium with respect to coat color would be expected to:
a. increase the proportion of dominant phenotypes.
b. eliminate heterozygosity.
c. increase the proportion of recessive phenotypes.
d. increase the proportion of heterozygotes.
e. maintain a constant proportion of dominant and recessive phenotypes.

E

All of the alleles of the members of a population are known as __________.
a. genetic drift
b. phylogeny
c. the gene pool
d. speciation

C

In a certain human population in genetic equilibrium for the typical X-linked red-green color blindness locus, 6 males in 100 are color blind. What is the percentage of color-blind females in the population?
a. 0.36%
b. 11.3%
c. 88.4%
d. 94%

A

If a recessive disease is found in 50 out of 100,000 individuals, what is the frequency of the heterozygote carriers for this disease?
a. 0.0005
b. 0.022
c. 0.043
d. 0.956

C

A wild population of a butterfly species in equilibrium consists of 99% orange and 1% yellow. Yellow is due to an autosomal recessive gene. The percent of homozygotes in this population is:
a. 9%
b. 18%
c. 1%
d. 82%
e. 31%

D

Five samples of nucleic acids were analyzed for base composition with the following results. Which one was most probably a single stranded DNA sample?
a. 13%=A, 21%=G, 36%=T, and 30%=C
b. 23%=A. 23%=T, 27%=C, and 27%=G
c. 15%=A, 15%=C, 34%=G, and 369%=U
d. 22%=A, 29%=U, 24%=C, and 25%=G
e. 19%=A, 31%=C, 31%=G, and 19%=U

A

Deoxyribonucleotides are added to each other forming a polynucleotide chain in a 5’ to 3’ direction. How are these nucleotides linked to each other?

a. covalent bonding between sugars.

b. hydrogen bonding between bases.

c. covalent bonding between phosphates.

d. covalent bonding between phosphate and sugar.

e. hydrogen bonding between sugars.

D

In class, we learned that the two strands of a double helix have a constant distance between them. What single feature of the structure of a DNA molecule is most important for ensuring that this is true?
a. DNA contains thymine rather than uracil.
b. the two sides are antiparallel.
c. only one phosphate unit is found in a nucleotide, rather than three.
d. a purine always pairs with a pyrimidine and vice-versa.
e. the sides or stands are wound in a double helix configuration.

D

Based on the Watson Crick model of DNA structure, which of the following statements is (are) true?
a. none of the above.
b. the double helix contains 40% purines and 60% pyrimidines.
c. the double helix contains two strands of DNA that exhibit a parallel nature.
d. if the percentage of one nitrogenous base is known, the percentage of the other nitrogenous bases are known.
e. for every adenine, there is a corresponding guanine.

D

DNA is extracted from the blood contained in a fossilized mosquito (uh oh!). The genetic material from the blood sample was found to be DNA and constructed according to the Watson‑Crick model. This sample was determined to contain 20% guanine. How much adenine should it contain?
a. 15%
b. 20%
c. 35%
d. 30%
e. I don't know, but it is right behind you!

D

There was reluctance in the scientific community to accept DNA as the genetic material because:
a. proteins had more different amino acids than DNA did.
b. DNA contained sugars which could not exist in hereditary material.
c. people were opposed to antiparallel molecules.
d. DNA had too many different combinations of nitrogen bases.
e. DNA was considered to be too simple a molecule.

E

The Central Dogma is best described by which of the following?
a. DNA is converted to RNA that is then converted to protein.
b. DNA information is passed to RNA which, in turn, passes the information to protein.
c. DNA is double stranded and RNA is single stranded.
d. RNA information is passed to DNA that is then passed to protein.

B

In the bacterial transformation experiments of Avery, MacLeod and McCarty, when the transforming principle was treated with DNAase, the result was:
a. no transformation occurred.
b. the accidental creation of the teenage mutant ninja turtles.
c. increased transformation compared to a sample treated with RNAase.
d. increased transformation compared to a sample treated with protease.
e. same transformation as the control or any other treatment.

A

An experiment by Hershey and Chase provided evidence that DNA is the genetic material. Which of the following facts allowed them to document the function of DNA?
a. only protein could be labeled.
b. DNA contains sulphur but no phosphorus, and protein has phosphorus but no sulphur.
c. DNA contains phosphorus but no sulphur, and protein contains sulphur but no phosphorous.
d. results obtained by Mendel.
e. only DNA could be labeled.

C

Which type of RNA is the least stable?
a. wRNA
b. uRNA
c. mRNA
d. tRNA
e. rRNA

C

What would have happened in Griffith's transformation experiment with virulent and avirulent strains of bacteria if he had mixed heat-killed virulent bacteria with living avirulent bacteria and injected this mixture into live mice?
a. the same thing as if he had just injected heat-killed virulent bacteria.
b. he would have produced a peculiar mouse the world would come to know as Mighty Mouse.
c. the same thing as if he had just injected living virulent bacteria.
d. the same thing as if he had just injected heat-killed avirulent bacteria.
e. the same thing as if he had just injected living avirulent bacteria.

C

In bacteria, DNA replication is __________ and ____________ , and the replicon constitutes _____________.

a. semi-conservative, unidirectional; the entire chromosome.

b. conservative, unidirectional; the entire chromosome.

c. semi-conservative, bidirectional; the entire chromosome.

d. conservative, bidirectional; the entire chromosome.

e. hard to see, takes days to accomplish; a gaggle of leprechauns.

C

Meselson and Stahl grew bacteria in medium containing heavy nitrogen (15N) and then switched to medium containing light nitrogen (14N). If DNA replicated by a conservative mechanism, what would they have observed?

a. after one replication event the DNA molecules are of two types, half are light and half are heavy.

b. one replication event produced all light DNA.

c. two replication events produced all heavy DNA molecules.

d. after two replication events, the DNA was of two types, half was light and half was intermediate.

e. after one replication event, all the DNA molecules were intermediate in density.

A

During DNA replication, nucleotides are:

a. added to the 5’ end of the growing DNA strands.

b. added to both the 3’ end and the 5’ end of growing strands.

c. added to the 3’ end of the growing DNA strands.

d. added to the 3’ end of one strand and the 5’ end of the other strand.

e. added at both ends of the strand as well as at interior sites.

C

DNA polymerase III has proofreading capability which increases the accuracy of replication. Which of the following is correct with respect to this function of the enzyme?

a. it reads each base pair twice to ensure that a mistake has not occurred.

b. identification of a mistake is based simply on whether hydrogen bonds are rapidly formed.

c. it randomly replaces a base every 100 base pairs since that is its normal error rate.

d. it reads the newly synthesized strand after replication is complete in order to identify mistakes.

e. because it only can cleave bases in a 5' 3' direction, it is very limited in its proofreading capabilities.

B

DNA ligase is required in DNA synthesis for:

a. the addition of each new nucleotide in a growing DNA strand.

b. the joining of Okazaki fragments.

c. the unwinding of the double helix for synthesis.

d. the stabilization of unwound DNA strands.

e. the settlement of litigation between contiguous genes for available nucleotides.

B

Life on the distant planet Omicron 6 Delta has DNA synthesis like that on earth, except for the enzyme DNA polymerase IV. This enzyme can synthesize new strands of DNA in both a 5' ⟶⟶ 3' direction and a 3' ⟶⟶ 5' direction. What other difference would you find in DNA replication on this planet as compared to earth?

a. there would be no Okazaki fragments.
b. there would be no RNA primer needed for replication.
c. all synthesis would be discontinuous.
d. replication would follow a dispersive mode.
e. synthesis would be much slower.

A

Differences in DNA replication exist between eukaryotic and prokaryotic organisms. Which of the following is not true with respect to these differences?

a. DNA replication is said to be more complex in eukaryotes than in prokaryotes.

b. DNA polymerase III is the primary enzyme in DNA replication for both prokaryotes and eukaryotes.

c. triphosphate nucleosides are required in DNA replication of both prokaryotic and eukaryotic organisms.

d. template DNA is required in both prokaryotes and eukaryotes.

e. metallic ions are required for enzyme activation in both prokaryotes and eukaryotes.

B

In E. coli, a secondary proofreading mechanism exists which is separate from the capabilities of DNA polymerase III. How does this system differentiate the template strand from the newly synthesized strand?

a. the enzymes recognize the template strand by the presence of methylated adenines in GATC sequences.

b. the enzymes receive a molecular signal from DNA poly III informing them which strand is the template strand.

c. the enzymes only correct those errors involving cytosine, regardless of the strand involved.

d. the enzymes identify the strand to which DNA poly III is attached, and correct mistakes in the other strand.

e. the enzymes randomly select a strand, so a new mutation is produced ½ the time.

A

In prokaryotes, why do DNA polymerase molecules require a RNA primer to synthesize a strand?

a. these enzymes require a free 3' -OH group to attach a nucleotide.

b. these enzymes require a free 3' -PO4 group to attach a nucleotide.

c. these enzymes require a free 5' -OH group to attach a nucleotide.

d. these enzymes require a free 5' -PO4 group to attach a nucleotide.

e. they don't require a primer.

A

Where has the telomerase enzyme been found to be in high abundance and with the highest level of activity?

a. in skin cells

b. where chromosomes are short

c. close to hairpin loops

d. in reproductive cells

e. in somatic tissues

D

If the normal DNA sequence is AAAAA and the mutant is AATAA, this mutation would be called a:

a. frameshift.

b. transversion.

c. inversion.

d. transition.

e. mistake (technically correct, but no credit for this answer)

B

Based on our "bunny trail" lecture, how would you categorize a mutation in a plant species that caused a reduction in height due to greatly reduced activity in the growth hormone?
a. behavioral mutation.
b. conditional mutation.
c. lethal mutation.
d. bad mutation.
e. morphological mutation.

E

Which of the following is correct with respect to tautomeric shifts?
a. a shift in the atoms (proton shift) arrangement in the bases, result in base pair changes during DNA replication.
b. these only occur when the bacteria is exposed to blue light.
c. these are the primary source of adaptive mutations.
d. these are caused in bacteria when neither DNA poly III or the secondary proofreading system corrected a base mismatch.
e. these occur only when a base analogue was substituted for a normal base during DNA replication.

A

Certain repair systems operate to correct problems in DNA. Which of the following statements is correct for the photoreactivation repair system?
a. it adds 2 bases to a single base addition mutation in order to correct the reading frame.
b. it breaks the bonds between adjacent thymine (T) bases to restore proper bonding between strands.
c. it is found only in prokaryotic organisms.
d. small segments of DNA are cleaved from the strand, then DNA poly I replaces the missing bases.
e. it corrects base substitutions by replacing the incorrect base with the correct base.

B

In lecture we discussed mutation frequencies both within and among species. Which of the following is correct with respect to that discussion?
a. mutation frequencies in humans are so low they can't be measured.
b. mutation frequencies are always higher in prokaryotes compared to eukaryotes.
c. different genes in the same species have different mutation frequencies.
d. mutation frequencies are the same for different genes within a species.
e. the mutation frequency of a specific gene is the same in all species that have that gene.

C

Although rare, mutations in DNA do occur. One such mutation affected a UGA termination codon and changed it to UGG. What effect would this mutation most likely cause?

a. the polypeptide produced will have extra amino acids.

b. no RNA will be transcribed.

c. the RNA transcribed from this gene will be much longer than normal.

d. no polypeptide will be produced.

e. DNA replication will not be turned off at the appropriate location

A

A strand of DNA with the sequence 5' ATGCCAGGAis transcribed. What will be the sequence of the transcription product?

a. 5' UCCUGGCAU

b. 5' TACGGTCCT

c. 5' AUGCCAGGA

d. 5' TCCTGGCAT

e. 5' UACGGUCCU

A

The wobble hypothesis says:
a. there is loose pairing between the 5' end of a single anticodon and the 3' end of more than one codon.
b. there is loose pairing between the 5' end of a single codon and the 3' end of more than one anticodon.
c. there is loose pairing between the 3' end of a single anticodon and the 5' end of more than one codon.
d. there is loose pairing between the 3' end of a single codon and the 5' end of more than one anticodon.
e. after several hours in a bar, one will wobble upon departure.

A

The wobble hypothesis is necessary to explain.
a. how the same tRNA can recognize several different codons.
b. why tryptophan has only one codon while leucine has six.
c. how the codon 5' AUG 3' can code for both N formyl methionine and for the regular methionine.
d. how several different codons can code for the same amino acid.
e. why students wobble upon leaving bars after taking a genetics exam.

A

The following relationships are known: CGG specifies arginine, GGC specifies glycine and GCG specifies alanine. Based on your knowledge about the genetic code, it would be most likely that CGAspecifies______________.

a. arginine

b. glycine

c. alanine

d. proline

e. none of the above; it is a stop codon.

A

What function does the sigma subunit play in RNA polymerase activity?
a. it recognizes and facilitates binding to the promoter.
b. it allows the enzyme to move from the 5' end to the 3' end of DNA.
c. it is necessary for the catalytic activity of the enzyme.
d. it recognizes the terminator sequence.
e. it is the subunit that distinguishes the RNA polymerase for rRNA from the one for tRNA.

A

Promoters in prokaryotes usually have a ‑35 box. What is the primary function of this sequence?
a. it is involved in promoter recognition by RNA polymerase.
b. we do not know the function of this segment.
c. it bonds with a rRNA strand of the ribosomal small subunit during the initiation of translation.
d. it is used to identify the chromosome from other genetic material in the cell.
e. it is involved in the binding of RNA polymerase to the DNA template.

A

The initial mRNA transcribed from a prokaryotic gene is approximately the same size as the final transcript, while the transcript from a eukaryotic gene is much smaller in its final version than the original transcript. Why?
a. eukaryotic genes contain sequences that are transcribed but not translated.
b. eukaryotic genes contain information that is not transcribed.
c. prokaryotic proteins are all smaller than eukaryotic proteins.
d. eukaryotic genes include information for tRNA and rRNA which are not specified by prokaryotic genes.
e. prokaryotic genes are nested, so sometimes a smaller gene is transcribed, rather than the larger one.

A

What is the function of the poly-A tail of a mRNA?
a. it greatly retards degradation of the strand.
b. it is involved in the termination of translation since AAA is a termination codon.
c. it is involved in the specific binding of the mRNA to the ribosome to initiate translation.
d. it is simply an archaic remnant of molecular processes that were needed prior to evolving into eukaryotes.
e. it really has no known function, but is simply there.

A

We discussed how transfer RNA's undergo substantial post‑transcriptional base modification. Which of the following is correct relative to this phenomenon?
a. without base modification a tRNA will not function.
b. most of the bases in the stems are changed such that they can no longer base pair with each other.
c. bases in the anticodon are always replaced with rare or unusual bases.
d. bases in the 3' tail are highly modified so they can be recognized by certain codons.
e. base modification prevents a tRNA from functioning.

A

Transfer RNAs that do not contain rare (unusual) bases:
a. do not function.
b. have an abnormal cloverleaf structure.
c. function normally.
d. can bind to amino acids, but not mRNA.
e. are very dull molecules.

A

A particular tRNA was known to be 90 nucleotides in length and a mRNA was 300 nucleotides in length. Assume that f‑methionine, the initiator, is hydrolyzed off and that only the last codon is a terminator. If these RNA's were involved in a system actively synthesizing protein, what would be the number of amino acids in the functional protein synthesized?
a. 98
b. 300
c. 90
d. 88
e. 30

A

In our discussion of the genetic code, we learned that three codons are known as termination or stop codons. How many tRNA's normally recognize these codons?
a. 0
b. 1
c. 2
d. 3
e. 4

A

The Shine‑Dalgarno sequence:
a. is complimentary to 3' end of the 16S rRNA strand.
b. is complimentary to promoter regions.
c. is responsible for termination of translation.
d. is responsible for termination of transcription.
e. is yet another feeble attempt for scientists to spread their fame by naming an esoteric entity after themselves.

A

_________catalyses the formation of the peptide bond linking two amino acids during translation.
a. 23S RNA.
b. elongation factor proteins.
c. aminoacyl synthetase.
d. charged tRNA.
e. aminolinkage synthetase.

A

Identify the correct sequence of steps in protein synthesis.

A ‑ peptide bond formation

B ‑ binding of fmet‑tRNA to mRNA‑ribosome complex

C ‑ binding of mRNA to ribosome

D ‑ release of fmet‑tRNA and translocation

E ‑ binding of charged tRNA to A site

C, B, E, A, D

The correct sequence of events involving a given amino acid in protein synthesis is:

1 ‑ pairing of charged tRNA with mRNA

2 ‑ binds with aminoacyl synthetase

3 ‑ peptide bond formation

4 ‑ attached to tRNA

2, 4, 1, 3

How could you determine the number of amino acids in a polypeptide that was synthesized by a bacterial cell?
a. you would need to know how many bases are in the template strand; from this you could closely approximate the number of amino acids.
b. you would need to know how many amino acids are available in the bacteria.
c. you would need to know how many introns the gene has in order to determine polypeptide length.
d. you would need to know how many tRNA's are found in the species in order to determine the number of amino acids.
e. you would need to know how many genes were active at the same time as this would determine if sufficient ribosomes are available for translation.

A

A mad cartoonist has all the molecules necessary for protein synthesis in a test tube. He has inadvertently mixed up some of the components. Included in his test tube are:
-partner DNA from Fred Flintstone
-tRNA molecules from Bugs Bunny
-ribosomes from Tweedy Bird
-mRNA from George Jetson
-amino acids from Wiley Coyote
Assuming this concoction is functional and translation occurs, the first proteins generated will be the same proteins produced by which of the above mentioned critters?
a. George Jetson
b. Bugs Bunny
c. Fred Flintstone
d. Tweedy Bird
e. Wiley Coyote

A

A very odd life form is found in a cave in Kentucky. Compared to all other biological systems, this critter has a different number of amino acids and a different number of nucleotides. In research to determine the number of nucleotides necessary to specify a single amino acid, proflavin was used to induce frameshift mutations. When one, two or three nucleotides were added to a nucleotide sequence, the resultant polypeptide chains were all similar to the original polypeptide chain. What does this suggest about the number of nucleotides that specify a single amino acid?
a. 1 nucleotide specifies 1 amino acid.
b. 2 nucleotides specify 1 amino acid.
c. 3 nucleotides specify 1 amino acid.
d. 4 nucleotides specify 1 amino acid.
e. 5 nucleotides specify 1 amino acid.

A

In class we mentioned chromatin remodeling as means to regulate transcription in eukaryotes. Which of the following is the correct statement concerning gene regulation?
a. Regions of the DNA that are heterochromatic are not actively being transcribed
b. Euchromatin DNA regulation only occurs in prokaryotes
c. Heterochromatic DNA is being transcribed actively
d. When changes from heterochromatin to euchromatin turns transcription off
e. We have never talked about chromatin remodeling

A

The promoter within a gene is subdivided into regions based on their specific function. Which of the following statements is true regarding promoters?
a. The CAAT box is associated with template recognition in eukaryotes
b. The -10 box is associated with template binding in eukaryotes
c. Prokaryotic promoters almost always have a TATA box but not necessarily a CAAT box
d. The -35 box is associated with template recognition in eukaryotes
e. Almost all eukaryotic promoters have a GC box

A

Where are enhancers located?
a. Enhancers can be downstream, within, or upstream of the gene
b. Enhancers are located in the promoter
c. Enhancers are located in transcription region
d. Enhancers are located in the tRNA
e. Enhancers have no role in the cell

A

Describe the manner in which activators and repressors influence the rate of transcription initiation.
a. Activators and repressors change the relationship of RNAP II to the transcription complex.
b. Activators and repressors change the relationship of DNAP III to the chromatin remodeling complex.
c. Activators and repressors change the relationship of RNAP I to the transcription factors.
d. Activators and repressors modify histones and change the nucleosome composition.
e. Activators and repressors change the relationship of nucleosomes and the DNA strand.

A

Which statement is correct in regards to gene regulation?
a. Methylation patterns are tissue specific and heritable for all cells in that tissue
b. The methylation state of a gene cannot be determined by restriction enzymes
c. Transcriptional control in eukaryotes has a minor role
d. Prokaryotic mRNA is modified prior to translation
e. Posttranscriptional regulation is common in prokaryotes

A

Certain sequences have been found which enhance the transcription of genes, and are therefore called enhancer sequences. Which of the following is true with respect to these sequences?
a. sequences can be inverted without a loss of activity.
b. These sequences are part of promoters recognized by RNA polymerase II.
c. These sequences are never found upstream from the gene.
d. These sequences are found only in prokaryotes.
e. These sequences enhance your grade on Genetics exams.

A

The reason why the information stored in most genes is not constantly expressed is that:
a. there are regulatory mechanisms that turn genes off and on depending on the cells' need.
b. the lifespan of a gene is short.
c. the cells get tired.
d. the genetic information of a gene is expressed only once in the lifetime of the organism.
e. the information is destroyed in the long run

A

A structural gene may be defined as:
a. a sequence of DNA which codes for the amino acid sequence of a polypeptide.
b. a sequence of RNA which catalyzes the formation of peptide bonds.
c. a gene whose gene product is a component of the large subunit of the ribosome.
d. a sequence of DNA which is involved in promoter recognition by RNA polymerase.
e. a gene that is involved in the formation of the double helix structure of a DNA molecule.

A

In the functioning of the lac operon, an inducer substance binds to:
a. a repressor protein, which then dissociates from the operator.
b. the operator, causing an allosteric shift which initiates transcription.
c. lactose, which binds to DNA and completes transcription of the structural genes.
d. a repressor protein, causing it to bind tightly to the operator.
e. the promoter, which initiates transcription.

A

With regard to the lac operon, the repressor protein normally:
a. binds to the operator when lactose is rare.
b. binds to a structural gene when lactose is abundant.
c. binds to the promoter when lactose is abundant.
d. binds to the operator when lac is abundant.
e. binds to a lac molecule when lactose is rare

A

Which of the following represents the role of glucose in the catabolite repression system?
a. glucose interferes with adenyl cyclase preventing it from producing cAMP.
b. glucose interferes with the CAP site allowing transcription to occur.
c. glucose interferes with the CAP protein inhibiting transcription of the structural genes.
d. glucose interferes with the polymerase site of the promoter inhibiting RNA polymerase binding.
e. glucose has no role in catabolite repression

A

Which of the following is accurate with respect to catabolite repression?
a. it is under positive control.
b. it is an inducible system.
c. it is under negative control.
d. it is regulated by attenuation.
e. it is a system found in bacteria (technically correct but worth zilch)

A

The biosynthesis of tryptophan in bacteria is regulated by a repressor protein system and a secondary attenuation system. Which of the following is correct with respect to the trp attenuation system?
a. the availability of charged tRNA trp will determine whether or not transcription continues or is terminated.
b. the location of the ribosome relative to the polymerase determines whether transcription continues or is terminated.
c. termination will always be terminated when there is sufficient trp in the cell.
d. the rate of transcription does not decrease due to attenuation, but is actually increased.
e. there is no attenuation of the trp operon in bacteria.

A

The repressor protein in the trp operon normally:
a. binds to the operator when tryptophan is present.
b. binds to the operator when no tryptophan is present.
c. binds to a structural gene when tryptophan is present.
d. binds to a structural gene when no tryptophan is present.
e. binds to the promoter when tryptophan is present

A

The trp operon, lac operon, and catabolite repression systems are examples of what types of control?
a. negative; negative; positive.
b. positive; negative; positive.
c. negative; positive; negative.
d. positive; negative; negative.
e. negative; positive; positive.

A

In E. coli, the genetic material is composed of:

a. circular, double stranded DNA

b. linear, double stranded DNA

c. circular, single stranded DNA

d. linear, single stranded DNA

e. circular, double stranded RNA

A

Nucleosomes are structures composed of:
a. DNA and histone proteins
b. non-histone proteins
c. non-histone proteins and DNA
d. nucleotides that have been methylated
e. some nucleos

A

Histone proteins are involved in the packaging of the genetic material in eukaryotic organisms. Which histone is associated with linker DNA and is not a component of the nucleosome core particle?
a. H1
b. H2A
c. H2B
d. H3
e. H4

A

Solenoids are structures composed of:
a. 6 nucleosomes coiled in a loop.
b. nucleotides that have been methylated.
c. non-histone proteins.
d. non-histone proteins and DNA.
e. some nucleos.

A

Put the following DNA structural compaction events in the correct order:
1. nucleosome formation
2. solenoid formation
3. double helix formation
4. 700 nm chromatid diameter

3, 1, 2, 4

Albinism in humans is controlled by a recessive gene. Two normal parents produce an albino son. What are the chances that their next child will be a normal girl?
a. 3/8
b. 1/4
c. 3/4
d. 5/8
e. 1/2

C

If P produces yellow flowering soybean plants and ppproduces white flowering plants, what ratio from a testcross of a yellow flowering plant would indicate the yellow flowering plant was heterozygous?

a. all yellow

b. 3 yellow: 1 white

c. 1 yellow: 1 white

d. all white

e. 3 white: 1 yellow

C

In raccoons, the striped‑tail pattern is dependent on a dominant allele (P) and a plain‑tail pattern on the recessive allele (p). Long legs is controlled by a dominant allele (S) and short legs by the recessive allele (s).

A striped‑tail, long‑legged raccoon was mated to a plain‑tailed, long‑legged raccoon and produced the following progeny (over time, not all at once):

-28 striped‑tailed, long‑legged

-11 striped‑tail, short‑legged

-33 plain‑tailed, long‑legged

-8 plain‑tailed, short‑legged

What are the most probable genotypes of the parents?

a. ppss and PPSS

b. PpSS and ppSS

c. none of the above.

d. PpSs and ppSs

e. ppSs and ppss

D

In gators, green skin is dominant to orange, and long legs are recessive to short legs. A green, short‑legged gator and an orange, short‑legged gator have 16 baby gators with the following phenotypes: 6 green, short‑legged; 6 orange, short‑legged; 2 green, long‑legged, and 2 orange, long‑legged. What were the most probable genotypes of the parents? (Let G stand for the skin color gene and L for the leg gene)
A. GgLl X ggLl
B. ggLl X GGLl
c. ggLl X Ggll
d. GgLl X GgLl
e. Gg X ggll

A

In humans, cataracts in the eyes and excessive bone fragility are controlled by the separate dominant genes C and F. A man with cataracts and normal bones marries a woman with cataracts and fragile bones. His mother had normal eyes and normal bones. The woman's father had normal eyes and normal bones. What are the most probable genotypes of this couple?
a. the woman is CcFf and the man is Ccff.
b. the woman is CCFf and the man is ccFf.
c. none of the above.
d. the woman is Ccff and the man is CcFf.
e. the woman is CCFf and the man is Ccff.

A

A purebred long-tailed cat with long whiskers (TTww) is mated to a purebred short-tailed cat with short whiskers (ttWW). The kittens will be:
a. 50% long-tailed with short whiskers, 50% short-tailed with long whiskers.
b. 100% short-tailed with short whiskers.
c. 100% long-tailed with long whiskers.
d. 100% long-tailed with short whiskers.
e. 50% long-tailed with long whiskers, 50% short-tailed with short whiskers.

D

Mendel's postulates about independent assortment and segregation can be described as pertaining to ____and ____, respectively.
a. alleles at different loci, alleles at different loci.
b. none of the above.
c. alleles at the same locus, alleles at different loci.
d. alleles at the same locus, alleles at same locus
e. alleles at different loci, alleles at same locus

E

With complete dominance, what portion of the F2 from a trihybrid would be A_bbCc?

a. 1/32

b. 3/32

c. 1/16

d. 1/8

e. 1/4

B

Ima Mendel‑Tu, little known niece of Gregor, investigated flower colors in white clover. Unfortunately, Ima did not read her uncle's manuscript and made critical mistakes in her protocol. Eventually it was discovered that a white flower is dominant to a red flower. In one experiment, Ima crossed two white flowered plants and obtained 48 white flower progeny and 16 red flower progeny. Since we know so much more than Ima, how many of the progeny plants from this cross would you expect to be homozygous (true-breeding) for the white flower trait?
a. 16
b. 32
c. 48
d. 8
e. 40

A

In horses, the Palomino color is the result of incomplete dominance of the C locus. The homozygous dominant is chestnut and the homozygous recessive is albino. What phenotypic ratio would be expected from the mating of 2 Palomino types?
a. 1:1
b. 3:1
c. 1:2:1
d. all Palomino
e. 9:3:3:1

C

In the iris, leaf size is determined by 2 alleles. The genotype PP has large leaves; Pp produces medium leaves, and pp has small leaves. A cross was made between two plants each of which has medium leaves. Of the 80 progeny produced, how many had large leaves?
a. 20
b. 80
c. 40
d. 60
e. 0

A

In shorthorn cattle the heterozygous condition of the alleles for red coat color (R) and white coat color (r) is roan coat color. If two roan cattle are mated, what proportion of the progeny will resemble their parents in coat color?
a. 1/2
b. 1/4
c. all
d. 3/4
e. none

A

A mutant purple leaf corn plant is discovered by a corn breeder. The breeder crosses this mutant to a normal green leaf plant and obtains a purple F1. The F2 segregates 3 purple : 1 green. This mutant trait is:
a. dominant to green and controlled by a single gene.
b. dominant to green and controlled by two independent genes.
c. additional crosses are needed to determine the mode of inheritance of purple.
d. incompletely dominant to green and controlled by a single gene.
e. recessive to green and controlled by a single gene.

A

In a continuation of recent mishaps, the new born ward in a certain Tampa hospital has some problems. Patient records have been lost/misplaced to such a degree that one infant with O blood is being claimed by 5 sets of parents. Which set of parents are most probably the biological parents of this baby?
a. B, A
b. the situation is worse than we thought, as none of them could be the biological parents of the infant.
c. A, AB
d. AB, B
e. AB, O

A

Warthogs with small snouts, when crossed together, produce approximately twice as many small‑snouted progeny as large‑snouted progeny. This is a consistent trend when sufficient progeny are produced. What type of inheritance does this suggest?
a. normal inheritance with large snouts dominant to small snouts.
b. normal inheritance with small snouts incompletely dominant to large snouts.
c. small snouts dominant to large snouts, and large snouts lethal when homozygous.
d. normal inheritance with small snouts dominant to large snouts.
e. small snouts dominant to large snouts, and small snouts lethal when homozygous.

E

A dominant suppressor gene (S) is homozygous lethal, but in the heterozygous condition it suppresses another dominant gene (B) responsible for a black coat color (bb is white). What phenotypic ratio is expected in surviving progeny when double heterozygotes are mated?
a. 9 white: 3 black
b. 7 white: 3 black
c. 13 white: 3 black
d. 12 white: 0 black
e. 9 white: 7 black

A

In your job as a genetic counselor, you interview a man recently diagnosed with Huntington's disease. He asks you about the probability that his daughter has inherited this disease? What do you say?
a. tell him the probability is 1/2.
b. tell him the probability is 1/4.
c. tell him the probability is 3/4.
d. tell him the probability is 1.
e. tell him about holandric traits like this and that there is no chance that he transmitted the gene to her.

A