Genetics Exam 1

If an organism contains 30% cytosine (C), the following is true:

It will contain 30% thymine

It will contain 20% adenine

It will contain 40% guanine

It will contain 60% guanine

It will contain 70% thymine

 It will contain 20% adenine

Griffith's experiment injecting a mixture of dead and live bacteria into mice demonstrated that

DNA is double-stranded

The genetic material of bacteria differs from the genetic material of mice

bacteria can recover from heat treatment if live helper cells are present

a factor was capable of transforming one bacterial cell type to another

Bacteria contain deoxyribonucleic acid

 a factor was capable of transforming one bacterial cell type to another

Which of the following statements is false

Genes are nucleotide sequences that encode functional RNAs

Genes are carried on chromosomes

Genes are expressed at the same level in all cell types and tissues of an organism

Gene products (RNAs and proteins) influence phenotypic traits by regulating biochemical processes

The nucleotide sequence of genes determines their function

 

Genes are expressed at the same level in all cell types and tissues of an organism

In the experiment of Avery, McLeod, and McCarty, the addition of RNAse and protease to the DNA extracts

prevented the conversion of type S bacteria into type R bacteria

allowed the conversion of type S bacteria into type R bacteria

prevented the conversion of type R bacteria into type S bacteria

allowed the conversion of type R bacteria into type S bacteria

killed all the bacteria

 

 

allowed the conversion of type R bacteria into type S bacteria

Okasaki fragments are a consequence of

the inability of the DNA polymerase to initiate a new DNA strand

the inability of the DNA polymerase to correct replication errors

random strand breakages resulting from supercoiling

mutations in the gene for DNA ligase

the inability of the DNA polymerase to polymerize in the 3' to 5' direction

 

the inability of the DNA polymerase to polymerize in the 3' to 5' direction

Which terms accurately reflect the nature of DNA replication in prokaryotes?

single ori, bidirectional, conservative

single ori, unidirectional, conservative

multiple ori, bidirectional, semiconservative

single ori, bidirectional, semiconservative

multiple ori, unidirectional, semiconservative

 

 

single ori, bidirectional, semiconservative

DNA polymerase III adds nucleotides

to the 3' end of the RNA primer

to the 5' end of the RNA primer

in the place of the primer RNA after it is removed

to both ends of the RNA primer

to internal sites in the DNA template

 

 

to the 3' end of the RNA primer

The two DNA strands in a DNA double helix are said to be complementary to one another. What does this mean?

One strand unequivocally determines the nucleotide (base) sequence of the other strand.

The two strands are interchangeable (have the same nucleotide or base sequence).

The two strands are interchangeable, but one strand is reversed with respect to the other.

The two strands are antiparallel (5'-end of one strand is where the 3'-end of the other strand is).

Separation of the two strands has no "cost" (does require no or only very little energy).

One strand unequivocally determines the nucleotide (base) sequence of the other strand.

DNA polymerases and RNA polymerases have the following in common:

A.	They use a 3' to 5' DNA strand as template
B.	They synthesize in the 5' to 3' direction
C.	They need a short primer to initiate synthesis
D.	A and B are correct
E.	All are correct

 

D.	A and B are correct

The poly(A) tail of mRNAs

A.	Is added to the 3' end of mRNAs.
B.	Is found on most mature eukaryotic mRNAs.
C.	is added in the nucleus.
D.	Helps prevent degradation of eukaryotic mRNAs.
E.	All of the above.

E.	All of the above.

The sigma subunit of bacterial RNA polymerase

A.	Binds to a bacterial gene's promoter
B.	Is composed of both polypeptide and RNA molecules
C.	Is required for RNA polymerization
D.	Is required for termination of transcription
E.	Is required for ribosomal binding

A.	Binds to a bacterial gene's promoter

Nucleosomes are composed of which class of molecules?

A.	histones
B.	glycoproteins
C.	lipids
D.	H1 histones
E.	nonhistone chromosomal proteins

A.

histones

Which of the following is (are) required for splicing to occur?

A.	two transesterification reactions
B.	intact, naturally occurring introns
C.	formation of a lariat like structure
D.	a branch-point A residue
E.	all of the above

E.	all of the above

The genetic code is _________, meaning that an amino acid may be coded by more than one codon

A.	Unambiguous
B.	Degenerate
C.	Comma-less
D.	Universal
E.	Overlapping

B.	Degenerate

Primary mRNAs undergo the following modifications in the eukaryotic cell nucleus

A.	3' polyadenylation
B.	5' capping
C.	Intron splicing
D.	A and B are correct
E.	A, B, and C are correct

E.	A, B, and C are correct

How would the artificial mRNA GUGUGUGUGU etc. be read by the translational machinery based on our knowledge of the genetic code?

A.	Two base, not overlapping
B.	Two base, overlapping
C.	Three base, not overlapping
D.	Three base, overlapping
E.	Four base, not overlapping

 

C.	Three base, not overlapping

Contrast the contributions made to an understanding of transformation by Griffith and by Avery and his colleagues

• Griffith was the first to observe and name the phenomenon of transformation, using an in vivo system (laboratory mice)

• Observed that a bacterial mixture containing heat-killed cells of a virulent strain of Diplococcus pneumonia and live cells of an avirulent strain killed the injected mice and led to the recovery of live cells of the virulent strain.

• Searched for the transforming principle originating from the heat-killed pathogenic strain and determined it to be DNA

When Avery and his colleagues had obtained what was concluded to be the transforming factor from the IIIS virulent cells, they treated the fraction with proteases, RNase, and DNase, followed in each case by the assay for retention or loss of transformation ability. What were the purpose and results of these experiments? What conclusions were drawn?

• Isolated the “transforming principle” as a bacterial extract that consisted of a mixture of macromolecules

• to identify which macromolecules the transforming principle, specific degradative enzymes selectively eliminated components of the extract

• The rationale was that if transformation were concomitantly eliminated, then the eliminated component would be the transforming principle

  • proteases

  • RNase

  • DNase

• Only DNase, which eliminated DNA, prevented transformation; therefore, DNA must be the transforming principle

Why were 32P and 35S chosen for use in the Hershey-Chase experiment? Discuss the rationale and conclusions of this experiment

• Nucleic acids contain large amounts of phosphorus and no sulfur, whereas proteins contain sulfur and no phosphorus

• substances injected into the bacterium is the substance responsible for producing the progeny phage and, therefore, must be the hereditary material

• The experiment demonstrated that most of the 23P-labeled material (DNA) was injected, whereas the 35S-labeled phage ghosts (protein coats) remained outside the bacterium - nucleic acid mist be the genetic material

Draw the chemical structure of the three components of a nucleotide, and then link the 3 together what atoms are removed from the structures when the linkages are formed

The structure of deoxyadenylic acid is given below and in the text. Linkages among the three components require the removal of water

Describe the various characteristics of the Watson-Crick double- helix model for DNA

• 2 polynucleotide chains, each formed by phosphodiester linkages between the 5-carbon sugars and the phosphates

• Bases are stacked 0.34 nm apart and in a platonic, antiparallel manner. There is one complete turn for each 3.4 nm, which constitutes 10 bases per turn.

• Hydrogen bonds hold the 2 polynucleotide chains together.

• There are 2 hydrogen bonds formed between the A to T pair

• 3 hydrogen bonds formed between the G to C pair

• Double helix exists as a twisted structure, approximately 2nm in diameter, with a topography of major and minor grooves

• Hydrophobic bases are located in the center of the molecule

• Hydrophilic sugar-phosphate backbone is on the outside

What might Watson and Crick have concluded had Chargaff’s data from a single source indicated the following?

%. A- 29 T-19 G-21 C-31

the data presented would have indicated a lack of pairing of these bases in favor of a single-stranded structure or some other nonhydrogen-bonded structure

List 3 main differences between DNA and RNA

DNA

• Thymine

• Deoxyribose

• most often occurs in a double-stranded form

RNA

• Uracil

• Ribose

• often occurs as both single- and partially double-stranded forms

What is the physical state of DNA after it is heated and denatured?

High temperatures disrupt the hydrogen bonds that hold together the complementary strands of DNA, causing double-stranded DNA to become single-stranded

Why is Tm related to base composition?

• G-C base pairs are formed w/ 3-Hydrogen bonds

• A-T base pairs are formed 2-Hydrogen bonds

• It takes more energy (higher temperatures) to separate G-C pairs

What did the Watson-Crick model suggest about the replication of DNA?

the model itself indicates the unwinding of the helix and separation of the double-stranded structure into 2 single strands immediately exposing the specific hydrogen bonds through which new bases are brought material.

Nature, they state:

“It has not escaped our notice that the specific pairing we have postulated immediately suggests a possible copying mechanism for the genetic material.”

One of the most common spontaneous lesions that occur in DNA under physiological conditions is the hydrolysis of the amino group of cytosine, converting the cytosine to uracil. What would be the effect of the DNA structure of a uracil group replacing cytosine?

The results would be a base substitution of G-C to A-T after 2 round of replication if not repaired

Compare conservative, semiconservative, and dispersive modes of DNA replication

The difference among the 3 models of DNA replication relates to the manner in which the new strands of DNA are oriented as daughter DNA molecules are produced

Conservative:

• The original double helix remains as a complete unit, and the new DNA double helix is produced as a single unit. The old DNA is completely conserved

Semiconservative:

• each daughter strand is composed of one old DNA strand and one new DNA strand. Breakage of hydrogen bonds is required

Dispersive:

• The original DNA strand is broken into pieces and the new DNA in the daughter strand is interspersed among the old pieces. Breakage of the individual covalent, phosphodiester bonds is required for this mode of replication

Describe the role of 15N in the Meselson-Stahl experiment

• labeling the pool of nitrogenous bases of the DNA of E. coli with the heavy isotope 15N, it would be possible to “follow” the “old” DNA - accomplished by growing the cells for many generations in a medium containing 15N

• Cells transferred to 14N medium so that “new” DNA could be detected

• Comparison of the density of DNA samples at various times in the experiment

• The initial 15N culture and subsequent cultures grown in the 14N medium showed that after one round of replication in the 14N medium, the DNA from bacteria grew only in the 15N medium.

• Sample taken after 2 rounds of replication in the 14N medium, half of the NDA was the intermediate density and the other half was as dense as DNA containing only 14N DNA

Predict the results of the experiment by Taylor, Woods, and Hughes if replication were (a) conservative and (b) dispersive

The cells that pass through the S phase in the presence of the 3H-thymidine are labeled and that each double helix (per chromatid) is “half-labeled.”

Conservative scheme

• both strands of the newly labeled DNA will go to one sister chromatid, and the other sister chromatid will consist entirely of the original

• Unlabeled DNA - contrasts, in a semiconservative scheme, the first replicative round would produce two labeled sister chromatids

  • each labeled on one strand of the double helix

Dispersive scheme

• all newly labeled DNA will be interspersed with unlabeled DNA — because metaphase chromosome preparations are highly coiled and condensed structures

• It’s impossible to detect the areas where labeled is not found. Rather, both sister chromatids would appear as evenly labeled structure- which would be similar to the results seen for semiconservative replication

• after a second round of replication (this time in an unlabeled medium) using a dispersive scheme, both sister chromatids will be labeled again. This is a contrast to the results in a semiconservative scheme, in which only one sister chromatid would be labeled and the other unlabeled

What are the requirements for in vitro synthesis of DNA under the direction of DNA polymerase I?

The in vitro replication requires a DNA template, a divalent cation (Mg2+),

- all 4 of the deoxyribonucleoside triphosphates

• dATP

• dCTP

• dTTP

• dGTP

Kornberg showed the nucleotides are added to the 3’ end of each growing DNA strand. In what way does an exposed 3’ -OH group participate in strand elongation?

exposed 3’-OH group is necessary for the attachment of the next nucleotide. Components of the 3’-OH group form a covalent bond with the 5’-phosphate of the added nucleotide and inorganic pyrophosphate is released

Distinguish between (a) unidirectional and bidirectional synthesis, and (b) continuous and discontinuous synthesis of DNA

Given a stretch of double-stranded DNA

One could initiate synthesis at a given point and replicate strands either in one direction only (unidirectional) or in both directions (bidirectional)

• synthesis of complementary strands occurs in a continuous 5’ to 3’ mode on the leading strand template in the direction of the replication fork

  • resulting in a single, long product stand

  • in a discontinuous 5’ to 3’ mode on the lagging strand template, resulting in shorter product strands called Okazaki fragments

Define and indicate the significance of (a) Okazaki fragments, (b) DNA ligase, and (c) prime RNA during DNA replication

Okazaki fragments

• relatively short 1000 to 2000 bases in bacteria

• DNA fragments that are synthesized in a discontinuous fashion on the lagging strand template during DNA replication - fragments appear to be necessary because template DNA is not available for 5’ to 3’ synthesis until some degree of continuous DNA synthesis occurs on the leading strand template in the direction of the replication fork - the isolation of such fragments provides support for the scheme of replication shown in the text

DNA ligase

• required to form phosphodiester linkages in nicks, which are generated when DNA polymerase I removes RNA primer and meets newly synthesized DNA ahead of it. Notice in the text that the discontinuous DNA strands are ligated together into a single continuous strand

Prime RNA

• formed by primase to serve as an initiation point for the production of DNA strands on a DNA template. None of the DNA polymerases are capable of initiating synthesis without a free 3’-hydroxyl group. The primer RNA provides that group and thus can be used by DNA polymerase III

Outline the current model for DNA synthesis

• Synthesis of DNA is thought to follow the pattern described in the text. The model involves opening and stabilizing the DNA helix, priming DNA synthesis with RNA primers, and moving replication forks in both directions

• Includes elongation of RNA primers in continuous and discontinuous 5’ to 3’ modes and their removal by the exonucleolytic activity of DNA polymerase I

• Okazaki fragments generated in the replicative process are joined together with DNA ligase. DNA gyrase relieves supercoils generated by DNA unwinding

Why is DNA synthesis expected to be more complex in eukaryotes than in bacteria? How is DNA synthesis similar in the 2 types of organisms?

• Eukaryotic DNA is replicated in a manner that is similar to that of E. coli

• synthesis is bidirectional, continuous on one template strand and discontinuous on the other, and the requirements of synthesis

  • 4 deoxyribonucleoside triphosphates, divalent cation, template, and primer) are the same

• Okazaki fragments of eukaryotes are about one-tenth the size of those in bacteria- because there is a much greater amount of DNA to be replicated and DNA replication is slower, there are multiple initiation sites for replication in eukaryotes (and increased DNA polymerase per cell)

• in contrast to the single replication origin in bacteria. Replication occurs at different sites during different intervals of the S phase. The proposed functions of four DNA polymerases are described in the text

Several temperature-sensitive mutant strains of E. coli display the following characteristics. Predict what enzyme of function is being affected by each mutation

a. Newly synthesized DNA contains many mismatched base pairs

b. Okazaki fragments accumulate, and DNA synthesis is never completed

c. No initiation occurs

d. Synthesis is very slow

e. Supercoiled strands remain after replication, which is never completed

a. no repair from DNA polymerase I and/or DNA polymerase III

b. no DNA Polymerase I or DNA ligase activity

c. no primase activity; other possibilities are no DnaA protein or faulty helicase

d. only DNA polymerase I activity (recall that DNA Pol III has high processivity, due to the sliding clamp subunit)

e. no DNA gyrase activity

Contrast the size of the single chromosome in bacteriophage and T2 and that of E. coli. How does this relate to the relative size and complexity of phages and bacteria?

• Bacteriophage has a linear, double-stranded DNA while in the phage coat and, upon infection, closes to form a circular chromosome

• It contains about 50kb. T2 phage also has a linear, double-stranded DNA chromosome; it is less than 200kb

• By contrast, E. coli has a circular, double-stranded DNA chromosome of about 4.2 X 10³ kb.

• Both intact phages are about 1/150 the size of E. coli.

• Since phages are obligate parasites of bacteria, they are dependent on their hosts for the manufacture of materials for their replication

• Bacteria contain all genetic information for metabolism, replication, and de novo synthesis of numerous life-supporting materials

• Phages, on the other hand, contain relatively few genes, namely, those needed to adsorb, inject, and produce progeny using primarily bacterial materials

Why might we predict that the organization of eukaryotic genetic material will be more complex than that of viruses or bacteria?

• Although greater DNA content per cell is associated with eukaryotes, one cannot universally equate genomic size with an increase in organismic complexity. There are numerous examples in which DNA content per cell varies considerably among closely related species. Because of the diverse cell types of multicellular eukaryotes, a variety of gene products are required, which may be related to the increase in DNA content per cell

• a much higher percentage of the genome of a bacterium is actually involved in phenotype production than in a eukaryote

• Eukaryotes have evolved the capacity to obtain and maintain what appear to be large amounts of “extra” DNA. This concept will be examined in subsequent chapters of the text. Bacteria, on the other hand, with their relatively short life cycles, are extremely efficient in their accumulation and use of their genome

• Given the larger amount of eukaryotic DNA per cell and the requirement that the DNA be partitioned in an orderly fashion to daughter cells during cell division, certain mechanisms and structures (mitosis, nucleosomes, centromeres, etc.) have evolved for packaging and distributing the DNA. In addition, the genome is divided into separate entities (chromosomes) to perhaps facilitate the partitioning process in mitosis and meiosis

Describe the molecular composition and arrangement of the components in the nucleosome.

Nucleosomes are octameric structures containing 2 molecules of each of the 4 core histones:

• H2A

• H2B

• H3

• H4

Each octamer consists of 2 tetramers

• (H2A)2

• (H2B)2

• (H3)2

• (H4)2

Between the nucleosomes and complexed with linker DNA is histone H1. A 147-base-pair sequence of DNA wraps around the nucleosome in a left-handed helix that completes about 1.7 turns

Describe the transitions that occur as nucleosomes are coiled and folded, ultimately forming a chromatid.

As chromosome condensation occurs, a 30nm fiber is formed. It appears to be composed of 5 or 6 nucleosomes coiled together. Such a structure is called a solenoid. These fibers form a series of loops that further condense into the chromatin fiber, which are then coiled into chromosome arms making up each chromatid

Provide a comprehensive definition of heterochromatin and list as many examples as you can.

Heterochromatin is chromosomal material that stains deeply and remains condensed when other parts of chromosomes, such as euchromatin, are otherwise pale and decondensed. Heterochromatic regions replicate late in the S phase and are relatively inactive in a genetic sense either because there are few genes present, or because the genes present are repressed. Telomeres and the areas adjacent to centromeres are composed of heterochromatin

Assuming the genetic code is a triplet, what effect would the addiction or loss of two nucleotides have on the reading frame? The addition or loss of 3, 6, or 9 nucleotides?

• Given a triplet code, the addition or loss of 2 nucleotides would change the reading frame, whereas the addition or deletion of three nucleotides (or nucleotides in any multiple of three: six or nine or twelve etc.) will have no effect on the reading frame. It will, however, result in the addition or loss of amino acids in the resulting protein

• To convince yourself of this, lay out a sequence such as THEBIGREDCAYATETHERAT and experiment with adding and deleting 1, 2, or 3 letters

In the triplet binding technique, radioactivity remains on the filter when the amino acid corresponding to the codon is labeled. Explain the rationale for this technique.

In the assay, a single population of charged tRNAs that carry the correct radioactively labeled amino acid is mixed with ribosomes and many copies of specific synthetic trinucleotide, and the mixture is applied to a filter. Trinucleotides and charged tRNAs alone are small and will pass through the filter, but ribosomes (and ribosome-tRNA complexes) will be retained due to their large size. If the trinucleotide provided is complementary to the anticodon of the labeled, charged tRNAs, they will remain on the filter. However, if the trinucleotide and the anticodon of the following tRNA are not complementary, the RNAs will pass through are not complementary, the RNAs will pass through the filter-no radioactivity will be retained

Predict the amino acid sequence produced during translation by the following short hypothetical mRNA sequences (note that the second sequence was formed from the first by a deletion of only one nucleotide):

Sequence 1: 5’-AUGCCGGAUUAUAGUUGA-3’

Sequence 2:5’-AUGCCGGAUUAAGUUGA-3’

What type of mutation gave rise to sequence 2?

Applying the coding dictionary, the following sequences are “decoded”

Sequence 1: Met-Pro-Asp-Tyr-Ser-(term)

Sequence 2: Met-Pro-Asp-(term)

The 12th base (a uracil) is deleted from Sequence 1, thereby causing a frameshift mutation, which introduced a terminating triplet UAA

Define the process of transcription. Where does this process fit into the central dogma of molecular biology (DNA makes RNA makes protein)?

Transcription describes the synthesis of RNA using a DNA template. The central dogma of molecular genetics and, to some extent, all of biology, states that DNA produces, through transcription, RNA which is “decoded” (during translation) to produce proteins.

Describe the structure of RNA polymerase in bacteria. What is the core enzyme? What is the role of the O subunit?

RNA polymerase from E. coli is a complex, large (almost 500,000 daltons) molecule composed of subunits in proportion to the holoenzyme. The B and B’ subunits provide catalytic function, whereas the sigma subunit is involved in the recognition of specific promoters. The core enzyme is the protein without the sigma subunit

Contrast the roles of tRNA and mRNA during translation, and list all enzymes that participate in the transcription and translation process

• Transfer RNAs are “adaptor” molecules in that they provide a way for amino acids to interact with sequences of bases in nucleic acids. Amino acids are specifically and individually attached to the 3’ end of tRNAs, which possess a three-based sequence (the anticodon) that can base-pair with three bases mRNA (codons). Messenger RNA, on the other hand, contains a copy of the triplet codes, which are stored in DNA. The sequences of bases in mRNA interact, three at a time, with the anticodons of tRNAs

• Enzymes involved in transcription include the following: RNA polymerase (E. coli) and RNA polymerase I, II, and III (eukaryotes). Those involved in translation include the following: aminoacyl tRNA synthetases, peptidyl transferase, and release factors

During translation, what molecule bears the codon? The anticodon?

The sequence of base triplets in mRNA constitutes the sequence of codons. A three-base portion of the tRNA constitutes the anticodon

The a chain of eukaryotic hemoglobin is composed of 141 amino acids. What is the minimum number of nucleotides in an mRNA coding for this polypeptide chain?

Since there are 3 nucleotides that code for each amino acid, there would be 423 code letters (nucleotides); 426 including a termination codon. This assumes that other features, such as the poly-A tail, the 5’ cap, noncoding leader sequences, and introns, are omitted

Summarize the steps involved in charging tRNAs with their appropriate amino acids

An amino acid in the presence of ATP, Mg2+, and a specific aminoacyl tRNA synthetase becomes activated as an amino acid-AMP enzyme complex (+PPi). This complex interacts with a specific tRNA and transfers the activated amino acid to this tRNA to produce the aminoacyl tRNA.

Histones have a ________ charge that helps their interaction with DNA, that has a __________ charge provided by the __________.

A. positive, negative, phosphates

B. positive, negative, bases

C. negative, positive, phosphates

D. negative, positive, bases

E. negative, positive, ribose

A. positive, negative, phosphates

Each molecule of tRNA is configured into a number of stems and loops. The anticodon loop matches up with a codon of mRNA. At least how many of the bases on this loop have to base pair exactly with the bases of the codon, and why?

A. one, because of the wobble position

B. two, because of the wobble position

C. three, because it pairs with a codon

D. one or two, depending on the codon

E. none of the above

B. two, because of the wobble position

What do DNA polymerases need to successfully synthesize DNA?

A. primer

B. template DNA

C. deoxynucleoside triphosphates (dNTPs)

D. all of the above

E. only A and C

D. all of the above

• primer

• template DNA

• deoxynucleoside triphosphates (dNTPs)

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

A. circular, double-stranded DNA.

B. linear, double-stranded DNA.

C. circular double-stranded DNA and histone proteins. D. circular, double-stranded RNA.

E. linear, double-stranded RNA.

A. circular, double-stranded DNA.

Why was the DNase treatment used by Avery, MacLeod, and McCarty an important step to elucidate the molecular nature of the “transforming principle”?

A. This allowed them to isolate pure DNA samples.

B. This allowed them to isolate pure protein samples.

C. This allowed them to demonstrate that removing the DNA prevents transformation.

D. This allowed them to demonstrate that mixing rough cells with DNA prevents transformation.

E. This allowed them to demonstrate that the acting factor was not RNA.

C. This allowed them to demonstrate that removing the DNA prevents transformation.

DNA isolated from a fungus has an adenine content of 25%. What would be the % (G+C) within the fungus DNA?

A. 0%

B. 12.5%

C. 25%

D. 50%

E. 75%

D. 50%

Introns

A) are spliced out by snRNPs including U1, U2, and U6. B) form a lariat structure during splicing

C) have specific nucleotide sequences at their 5’- and 3’-ends

D) are removed from the primary mRNA before translation

E) all of the above

E) all of the above

• are spliced out by snRNPs including U1, U2, and U6. B) form a lariat structure during splicing

• have specific nucleotide sequences at their 5’- and 3’-ends

• are removed from the primary mRNA before translation

What was the critical observation in Griffith’s 1928 experiments that demonstrated a genetic transformation of non-virulent into virulent bacteria (Streptococcus pneumoniae)?

A. Mice injected with a mix of dead S (virulent) bacteria and living R (non-virulent) bacteria died.

B. Virulent bacteria were transformed into non-virulent bacteria when they lost their polysaccharide capsule.

C. Uptake of a bacteriophage transformed non-virulent into virulent bacteria.

D. DNA released from virulent bacteria was taken up by non-virulent bacteria.

E. Virulent strains of bacteria could be isolated from dead mice that had been injected with a mix of dead S (virulent) bacteria and living R (non-virulent) bacteria.

E. Virulent strains of bacteria could be isolated from dead mice that had been injected with a mix of dead S (virulent) bacteria and living R (non-virulent) bacteria.

The compaction leading to a metaphase chromosome involves which of the following

A. the formation of nucleosomes

B. the formation of a 30-nm fiber

C. anchoring and further compaction of the radial loops

D. B and C

E. A, B, and C

E. A, B, and C

• the formation of nucleosomes

• the formation of a 30-nm fiber

• anchoring and further compaction of the radial loops

In the promoter region of a gene one would find

A. cis-acting sequences

B. consensus sequences for RNA polymerase binding

C. consensus sequences for transcription termination

D. A and B are correct

E. A and B and C are correct

D. A and B are correct

• cis-acting sequences

• consensus sequences for RNA polymerase binding

During DNA replication, once the replication bubble is formed, the following protein(s) are recruited to maintain the DNA template “open”

A. single-stranded binding proteins

B. DNA primase

C. Topoisomerases

D. DNA helicases

E. DNA polymerases

A. single-stranded binding proteins

Which of the following is not consistent with Erwin Chargaff’s findings about the ratios of bases in DNA?

A) (A + G) = (C + T)

B) (A + C) = (G + T)

C) (C + G) = (A + T)

D) A=T

E) (A + G) / (C + T) = 1

C). (C + G) = (A + T)

Which of the following is not true of tRNA?

A. contains modified bases

B. it is charged by peptidyl transferase

C. has a secondary structure

D. has anticodon loop

E. has amino acid accepting site

B. it is charged by peptidyl transferase

When examining the genetic code, it is apparent that

A. there can be more than one amino acid for a particular codon.

B. AUG is a terminating codon.

C. there can be more than one codon for a particular amino acid.

D. the code is ambiguous in that the same codon can code for two or more amino acids.

E. there are 44 non-coding triplets because there are only 20 amino acids.

C. there can be more than one codon for a particular amino acid.

The __________________________ is the cis-acting DNA sequence that serves as the binding site for RNA polymerase.

A) promoter

B) terminator

C) enhancer

D) regulator

E) Shine-Dalgarno

A) promoter

What is the major difference between the lagging and leading strands?

A. On the leading strand, DNA synthesis occurs from 5' to 3', while DNA synthesis occurs from 3' to 5' on the lagging strand.

B. DNA polymerase is able to continuously add new nucleotides on the leading strand while it must keep 'starting over' on the lagging strand.

C. The lagging strand requires only a single primer while the leading strand requires many.

D. Helicase opens the leading strand at a faster rate than the lagging strand.

E. The leading strand can not complete synthesis at telomeres.

B. DNA polymerase is able to continuously add new nucleotides on the leading strand while it must keep 'starting over' on the lagging strand.

The 40 S subunit od the ribosome is involved in:

A. scanning of the mRNA

B. association with first AUG.

C. peptidyl transferase activity.

D. A and B are correct

E. A, B, and C are correct

D. A and B are correct

• scanning of the mRNA

• association with first AUG.

During DNA replication, the enzymes below function in the following order:

F.DNA ligase, DNA polymerase, primase, DNA helicase

G. DNA helicase, primase, DNA polymerase, DNA ligase

H. DNA helicase, DNA polymerase, primase, DNA ligase

I. primase, DNA polymerase, DNA helicase, DNA ligase

J.DNA ligase, primase, DNA polymerase, DNA helicase

G. DNA helicase, primase, DNA polymerase, DNA ligase

Okasaki fragments are a consequence of

A. the inability of the DNA polymerase to initiate a new DNA strand.

B. the inability of the DNA polymerase to use RNA primers.

C. random strand breakages resulting from supercoiling.

D. mutations in the gene for DNA ligase.

E. the inability of the DNA polymerase to polymerize in the 3’ to 5’ direction.

the inability of the DNA polymerase to polymerize in the 3’ to 5’ direction.

The bacterial sigma factor allows

A. RNA polymerase to elongate the transcript.

B. RNA polymerase to terminate transcription.

C. RNA polymerase to identify and tightly bind promoter elements.

D. DNA polymerase III to bind at the ori and initiate DNA synthesis.

E. RNA polymerase to identify and tightly bind the TATA box sequence

C. RNA polymerase to identify and tightly bind promoter elements.

Ribosomes are composed of:

A. tRNA and protein.

B. mRNA and protein.

C. rRNA and protein.

D. proteins only

E. rRNA only

C. rRNA and protein.

The figure indicates a molecule of ________; the arrow indicates the location of______

A. RNA, 3’ hydroxyl

B. DNA, 3’ hydroxyl

C. RNA, 5’ phosphate

D. DNA, 5’ phosphate

E. RNA, 5’ oxygen

B. DNA, 3’ hydroxyl

Consider the human rRNA genes. What type of DNA would you expect them to be?

A. Nonrepetitive sequences.

B. Moderately repetitive sequences.

C. Highly repetitive sequences.

D. Retroelements.

E. None of these.

B. Moderately repetitive sequences.

When considering the initiation of transcription in prokaryotes, one finds consensus sequences located in the ______________ region that are important for binding the ___________that recruits RNA polymerase to the promoter.

A. -30 ; TBP

B. -10 and -35; sigma factor

C. Shine-Dalgarno; sigma factor

D. -35; RNA pol II

E. -30; sigma factor

B. -10 and -35; sigma factor

Telomerase

A. synthesizes telomeric repeats at the end of chromosomes

B. it is made of RNA and protein

C. it uses RNA that is part of the telomerase as template for DNA synthesis

D. binds to the 3’ overhang region of the telomere E. all of the above

E. all of the above

• synthesizes telomeric repeats at the end of chromosomes

• it is made of RNA and protein

• it uses RNA that is part of the telomerase as template for DNA synthesis

• binds to the 3’ overhang region of the telomere

During translation termination in both bacteria and eukaryotes,

A. A termination tRNA binds a stop codon at the A site

B. A termination tRNA associates with the Shine-Dalgarno sequence

C. The large ribosomal subunits spontaneously separates from the ribosome when a stop codon associates with the A site

D. A release factor binds to a stop codon at the E site

E. A release factor binds to a stop codon at the A site

E. A release factor binds to a stop codon at the A site

Splicing of transcripts normally occurs:

A) only in the cytoplasm of prokaryotes

B) only in the cytoplasm of eukaryotes

C) in both prokaryotes and eukaryotes

D) only in the nucleus of eukaryotes

E) in mitochondria, as they have their own DNA

D) only in the nucleus of eukaryotes

Occasionally, a mutation may occur in the telomerase enzyme in a cell that renders it nonfunctional. What is likely to be the result of this mutation on the DNA in the cell over the course of several rounds of mitosis?

A. Chromosome length will gradually increase.

B. Chromosome length will gradually decrease.

C. Chromosome length will stay constant.

D. Chromosomes will fail to dissociate after replication.

E. Chromosomes will not undergo recombination

B. Chromosome length will gradually decrease.

A short segment of an mRNA molecule is shown below. The polypeptide it codes for is also shown: 5'-AUGGUGAUGAAG : methionine-valine-methionine -lysine Assume that a mutation in the DNA occurs so that the fourth base (counting from the 5' end) of the messenger RNA is deleted. What sequence of amino acids will the mRNA now code for? (You do not need a copy of the genetic code to answer the question.)

A. methionine-valine-methionine-lysine

B. methionine-valine

C. methionine- valine-methionine

D. methionine

E. methionine-methionine

D. methionine

The spliceosome:

A. removes introns from primary mRNA

B. recognizes specific nucleotides within the introns

C. forms a lariat in the process of splicing

D. A and B are correct

E. A, B, and C are correct

E. A, B, and C are correct

• removes introns from primary mRNA

• recognizes specific nucleotides within the introns

• forms a lariat in the process of splicing

Based on their proposed structure of DNA published in the Nature 1953 paper, Watson and Crick suggested:

A. a mechanism that explained how mRNA is translated.

B. a copying mechanism for the genetic material.

C. a mechanism for how RNA polymerases can transcribe DNA.

D. a repair mechanism for DNA nicks.

E. a mechanism for genetic recombination.

B. a copying mechanism for the genetic material.

After being translated by the ribosome in a living cell, the following eukaryotic mRNA would result in a polypeptide containing:

5’-ACCGACAUGGUCGUAGCAACUUUUUGAUGACCC-3’

A. 9 amino acids

B. 11 amino acids

C. 5 amino acids

D. 6 amino acids

E. 10 amino acids

D. 6 amino acids

In the modification of eukaryotic mRNA, the following modifications are found at the 5’ end and 3’ end, respectively:

A. acetyl group in guanine; multiple cytosines

B. multiple guanines; methyl group

C. multiple thymines acetyl group

D. methyl group in guanine; multiple adenines

E. acetyl group in uracil; multiple adenines

D. methyl group in guanine; multiple adenines

Except for Met-tRNA used in translation initiation, all tRNAs pass through the A, E, and P sites of the ribosome in a specific order. What is that order?

A. A to E to P sites

B. P to A to E sites

C. A to P to E sites

D. P to E to A sites

E. E to P to A sites

C. A to P to E sites

The enzyme aminoacyl-tRNA synthetase ___________________, while the enzyme DNA ligase ______________________.

A. "recharges" tRNA molecules; catalizes glycosidic bonds between nucleotides

B. modifies the bases on the tRNA loops; catalizes glycosidic bonds between nucleotides

C. folds tRNA molecules in the proper conformation; catalizes phosphodiester bonds between nucleotides

D. "recharges" tRNA molecules; catalizes phosphodiester bonds between nucleotides

E. modifies the bases on the tRNA loops; synthesizes the complementary DNA strand

D. "recharges" tRNA molecules; catalizes phosphodiester bonds between nucleotides

What is Genetics?

The study of heredity and variation

Saccharomyces cerevisiae:

• S. cerevisiae is a nonpathogenic single-cell eukaryotic organism

• Can be easily grown

• it has a short generation time of around 2 hours and cycles between a haploid and diploid state which allows us for easy genetic analysis

• Amenable to genetic manipulation. easy to mutate, add, or knock out genes due to efficient homologous recombination

Life cycle Saccharomyces cerevisiae

G1

S

G2

M

What is a gene?

A gene is a sequence of nucleotides that encodes a functional RNA

Chromosome

is a single DNA (deoxyribonucleic acid) molecule which may contain many genes

• Long fiber of DNA

• genes are located on chromosomes

Promoter

RNA polymerase binding site; transcription starts here

Transcribed region

this DNA sequence is copied into RNA when the gene is expressed

Terminator

signal to stop transcription

“Gene expression”

• the process by which a gene’s genetic information (it's nucleotide sequence) is copied into a function RNA

• the function of a gene and its gene product(s) are determined by its nucleotide sequence

Many (but not all!) genes encode mRNAs, which are translated into protein

Gene products

RNAs and proteins influence phenotypic traits by regulating biochemical processes

ex: Human lactase (LCT) gene

• the LCT gene is expressed (transcribed and translated) in epithelial cells of the small intestine

• The LCT protein is an enzyme (lactase) that metabolizes lactose

“active”

Every cell has a complete copy of the genome, but different combinations of genes are “active” (expressed) in different cells

Intestinal epithelial cells:

• do express lactase

• do not express hemoglobin

Red blood cells:

• do not express lactase

• do express hemoglobin

Molecular nature of genes: summary

• genes are nucleotide sequences that encode functional RNAs, one type of which (mRNA) can be translated into protein

• Genes are carried on chromosomes

• Gene products (RNAs and proteins) influence phenotypic traits b regulating biochemical processes

• The nucleotide sequence of genes determines their function

• All of your cells carry the same genes, but different combination of genes are expressed in different cells

DNA is Reproduced by?

Semiconservative replication

• DNA strands of double helix serve as templates for synthesis of complementary strands

Daughter strands

the two newly-made strands