ALL 9,11,12,13

According to Figure 13.19, explain why the ribosome translocates along the mRNA in a 5' to 3' direction rather than a 3' to 5' direction.

...

In the experiment of Figure 13.4, what would be the predicted amounts of amino acids incorporated into polypeptides if the RNA was a random polymer containing 50% C and 50% G?

...

With regard to the experiment described in Figure 13.11, answer the following questions: A. Why was polyUG mRNA template used? B. Would you radiolabel the cysteine with the isotope 14C or 35S? Explain your choice. C. What would be the expected results if the experiment was followed in the same way except that a polyGC template was used? Note: A polyGC template could contain two different, alanine condons (GCC and GCG), but it could not contain any cysterine codons.

...

RNA polymerase-Proaryotes

1 polymerase

Which of the following modifications occur to a eukaryotic pre-mRNA?

5' cap added, removal of introns, 3' poly A tail added.

DNA and RNA directionality

5' to 3'

DNA polymerase III does not have the following activity:

5' to 3' exonuclease activity

If a tRNA molecule carries a glutamic acid, what are the two possible anticodon sequences that it could contain? Be specific about the 5' and 3' ends.

3'-CUU-5' or 3'-CUC-5'

The figure shows an intron flanked by two exons. Splicing requires three sites to be recognized by the spliceosome, the 5' splice site, the 3' splice site, and the branch site. Label these sequences on the figure.

5', branch 3'

What is the complementarity rule that governs the synthesis of an RNA molecule during transcription? An RNA transcript has the following sequence: 5'-GGCAUGCAUUACGGCAUCACUAGGGAUC-3' What is the sequence of the template and coding strands of the DNA that encodes this RNA? On which side (5' or 3') of the template strand is the promoter located?

DNA-G/RNA-C DNA-C/RNA-G DNA-A/RNA-U DNA-T/RNA-A The template strand is the 3'-CCGTACGTAATGCCGTAGTGATCCCTAG-5' and the coding strand is 5'-GGCATGCATTACGGCATCACACTAGGGATC-3'. The promoter would be to the left (in the 3' direction) of the template strand.

In the experiment by Avery, Macleod, and McCarty, ______ destroyed the transforming factor

DNAase

An organism that has a G + C content of 64% in its DNA. What are the percentages of A, T, G, and C?

G = 32%, C = 32%, A = 18%, T = 18%

What is the consensus sequence of the following six DNA sequences? GGCATTGACT GCCATTGTCA CGCATAGTCA GGAAATGGGA GGCTTTGTCA GGCATAGTCA

GGCATTGTCA

The enzyme that unwinds a segment of the DNA molecule is

Helicase

The initiator tRNA is special because it is the only tRNA that can initially bind at this site in the ribosome.

P site

Phenylketonuria occurs when the gene for the enzyme phenylalanine hydroxylase is mutated, leading to an undesirable build-up of the compound usually acted upon by phenylalanine hydroxylase. Based on what you know about the metabolic pathway involving this enzyme, what dietary changes do you presume phenylketonurics must make?

Phenylketonurics should eat reduced amounts of phenylalanine and increased amounts of tyrosine.

The RNA transcript being produced by the RNA polymerase is complementary to the template strand of the DNA.

T

The length of the poly A tail affects the stability of the mRNA

T

Which eukaryotic transcription factor(s) shown in Figure 12.14 plays an equivalent role to o- factor found in bacterial cells?

TFIID and TFIIB would play equivalent roles to sigma factors. Sigma factor does two tihngs: it recognizes the promoter (as does TFIID), and it recruits RNA polymerase to the promoter (as does TFIIB).

Place the sentences in the correct order for formation of the open complex at eukaryotic promoters TFIIID binds to TATA box TFIIB binds to TFIID TFIIB promotes binding of RNA polymerase II TFIIF binds to RNA pol II TFIIH acts as helicase to form open complex TFIIE and TFIIH bind to RNA pol II to form closed complex

TFIIID binds to TATA box TFIIB binds to TFIID TFIIB promotes binding of RNA polymerase II TFIIF binds to RNA pol II TFIIE and TFIIH bind to RNA pol II to form closed complex TFIIH acts as helicase to form open complex

Look at the sequences in the figure. They all represent different bacterial promoter regions. Which of the sequences listed below is the best consensus sequence for the -35 region based on these seven promoters?

TTGACA

Initiation of transcription includes

The RNA polymerase holoenzyme binds to the promoter.

A eukaryotic structural gene contains two introns and three exons: exon 1-intron 1-exon 2-intron 2-exon 3. The 5' splice site at the boundary between exon 2 and intron 2 has been eliminated by a small deletion in the gene. Describe how the pre-mRNA encoded by this mutant gene would be spliced. Indicate which introns and exons would be found in the mRNA after splicing occurs.

Only the first intron would be spliced out. The mature RNA would be: exon 1-exon 2-intron 2-exon 3.

Describe how bases interact with each other in a double helix. This discussion should address the issues of complementarity hydrogen bonding, and base stacking.

The bases conform to the AT/GC rule of complementarity. There are two hydrogen bases between the A and T bases and three hydrogen bonds between the G and C bases. The planar structures of the bases stack on top of each other within the helical structure to provide even more stability.

Look at the promoter for the gene shown to the left. Where does transcription start?

Transcription starts at the +1 site.

Describe what happens to the chemical bonding interactions when transcription termination occurs. Be specific about the type of chemical bonding.

Transcriptional termination occurs when the hydrogen bonding is broken between the DNA and the part of the newly made RNA transcript that is located in the open complex.

Look up the meaning of the word transformation in a dictionary and explain whether it is an appropriate word to describe the transfer of genetic material from one organism to another.

Transformation means changing from one form to another. IN bacterial genetics, transformation involves the uptake of DNA into another bacterium. This may change the form (i.e., phenotype) of the bacterium. For example, transformation may change a rough bacterial strain into a smooth bacterial strain. The form or phenotype of the strain has been changed.

List the components required for translation. Describe the relative size of these different components. In other words, which components are small molecules, macromolecules, or assemblies of macromolecules?

Translation requires mRNA, tRNAs, ribosomes, proteins such as initiation, elongation, and termination factors, and many small molecules. ATP and GTP are small molecules that contain high-energy bonds. The mRNA, tRNAs, and proteins are macromolecules. The ribosomes are a large complex of macromolecules.

The IP3 receptor, which functions as a calcium channel in the endoplasmic reticulum

Transport

RNA editing can result in different ___ as well as new ___.

amino acids, stop codons

Telomerase is unique because it contains

an RNA molecule.

protein coat that surrounds the genetic material of the T2 phage

capsid

structure of dADP

diphosphate, deoxyribose, adenine

5. Removes RNA primers and fills in the region with DNA.

dna polymerase

6. Extends DNA from the RNA primer, synthesizes the leading strand and the lagging strand.

dna polymerase

Enzymes that are processive

does not dissociate from the growing strand once it has joined two nucleotides.

capping-Proaryotes

does not occur

structure of purine

double ring

5. Initiation requires a 7-methylguanosine cap on the mRNA.

eukaryotes

The DNA probe must be from an___.

exon

Which of the following occurs first in transcription?

formation of a holoenzyme

rRNA genes in some protozoa have_______ introns.

group I

The proofreading function of DNA polymerase reduces the error rate from about one in a million basepairs to about one in a ________ basepairs.

hundred million

2 DNA strands are connected through

hydrogen bonds

Where does splicing occur in the cell?

in the nucleus

Which of the following is correct regarding sigma factor?

it recognizes the promoter region

1. Covalently attaches adjacent Okazaki fragments.

ligase

If the control had been performed and the results were 339 cpm, then the conclusion would be that the RNA the probe could bind to is in the___ but not ___

liver, spleen and brain.

1. The first tRNA used is charged with a formyl methionine.

prokaryotes

3. Requires a Shine-Dalgarno sequence to be present on the mRNA.

prokaryotes

4. Translation occurs cotranscriptionally

prokaryotes

In group I and group II introns, the RNAs act as_________, which means they are RNAs with enzymatic activity.

ribozymes

Since the control is missing, the results of the _____ cannot be interpreted as positive or negative.

spleen and brain

A_________, composed of many proteins and RNAs is required for pre-mRNA splicing to occur.

spliceosome

A single strand of RNA with sequence 5'-UGAAUUGCAUCGGCAAUUGG-3' has the potential to form which structure?

stem loop

The code is considered____ although there are some exceptions to the rules.

universal

The third base in a codon is called the___ base.

wobble

RNA polymerase II

transcribes mRNA

RNA polymerase III

transcribes most rRNA

Following base removal, DNA polymerase can add nucleotides in the 5' to 3' direction.

T

Improper base-pairing during DNA replication causes a pause in chain elongation.

T

Rna polymerases demonstrate the statement function follows form

T

Telomeres consist of direct repeat sequences.

T

The DNA replication machinery is assembled at the replication fork.

T

If one DNA strand is 5'-GGCATTACACTAGGCCT-3', what is the sequence of the complementary strand?

3'-CCGTAATGTGATCCGGA-5'

Which of the following DNA sequences is complementary to 5' TAGAC 3'?

5' GTCTA 3'

Select all the anticodons that could bind to the codon for serine. Choose all that apply.

****

Promoter-Proaryotes

-10 TATA

Promoter-Eukaryotes

-25 TATA for mRNA

5' - 3' -10 sequence -35 sequence Rho terminator cDNA of insulin

-35 sequence -10 sequence cDNA of insulin Rho terminator

List and briefly describe the three types of sequences within bacterial origins of replication that are functionally important.

1. DnA boxes-Binding sites for the DnaA protein 2. Methylation sites- Sites of adenine methylation that are important for regulating DNA replication 3. AT-rich region- Site where the DNA initially separates to form an opening that is sometimes called a replication bubble

For each of the following sequences, rank them in order (from best to worst) as sequences that could be used to initiate translation according to Kozak's rules. GACGCCAUGG GCCUCCAUGC GCCAUCAAGG GCCACCAUGG

1. GCCACCAUGG 2. GACGCCAUGG 3. GCCUCCAUGC 4. GCCAUCAAGG The last one does not have a start codon, so it would not work. The third one may be translated, but very poorly.

An interesting trait that some bacteria exhibit is resistance to killing by antibiotics. For example, certain strains of bacteria are resistant to tetracycline, whereas other strains are sensitive to tetracycline. Describe an experiment you would carry out to demonstrate that tetracycline resistance is an inherited trait encoded by the DNA of the resistant strain.

1. Isolate and purify DNA from resistant bacteria. 2. In three separate tubes, add DNase, RNase, or protease. 3. Add sensitive bacteria to each tube. A small percentage may be transformed. 4. Plate on petri plates containing tetracycline. Expected results: Tetracycline-resistant colonies should only grow when the DNA has been exposed to RNase and protease, but not to DNase.

Describe the three most important functions of the DnA protein.

1. It recognizes the origin of replication. 2. It initiates the formation of a replication bubble. 3. It recites helicase to the region.

DNA polymerases use their ________ activity to remove a mismatched basepair.

3' -> 5' exonuclease

bases and length of turn

10 bases, 34 angstroms/ 3.4 nm

B DNA

10 bp per turn, centrally located bases, occurs normally, right hand helix

Z DNA

12 bp per turn, left hand helix, occurs in GC rich sequences

At a single origin of replication, how many leading strand primers are required for the complete replication of the DNA strand?

2

hydrogen bonds between AT

2

Arrange the following proteins in the proper order in which they participate in DNA replication. 1. Primase 2. Helicase 3. Single-stranded binding proteins 4. DNA polymerase

2, 3, 1, 4

Arrange the following in the proper sequence in which they occur during RNA splicing. 1. Lariat is formed 2. U2 binds to branch site 3. 3' splice site is cut

2,1,3

How many RNA polymerases are there in eukaryotes?

3

hydrogen bonds between CG

3

RNA polymerase-Eukaryotes

3 polymerases

distance between bases

3.4 A / .34 nm

A piece of double-stranded DNA has 100 base pairs. What is the length of the piece of DNA? How many complete turns can this double helix make?

34 nm; 10

initiation-Eukaryotes

5 general factors

Here are two strands of DNA. ---------------------------DNA polymerase-> -------------------------------------------------------------- The one on the bottom is a template strand, and the one on top is being synthesized by DNA polymerase in the direction shown by the arrow. Label the 5' and 3' ends of the top and bottom strands.

5'-----------DNA polymerase------------>3' 3'---------------------------------------------------------------------------------------------5'

There are__ possible codons.

64

What is the meaning of the term consensus sequence? Give an example. Describe the locations of consensus sequence within bacterial promotors. What are their functions?

A consensus sequence is the most common nucleotide sequence that is found within a group of related sequences. An example is the -35 and -10 consensus sequences found in bacterial promotors. At -35, it is TTGACA, but it can differ by one or tow nucleotides and still function efficiently as a promotor. In the consensus sequences within bacterial promotors, the -35 site is primarily for recognition by sigma factor. The -10 site, also known as the Pribnow box, is the site where the double-stranded DNA will begin to unwind to allow transcription to occur.

How and when does formylmethionine become attached to the initiator tRNA in bacteria?

A formyl group is covalently attached to methionine after the methionine has been attached to the tRNA containing a UAC anticodon.

What does it mean to say that gene expression is colinear?

A gene is colinear when the sequence of bases in the coding strand of the DNA (i.e., the DNA strand that is complementary to the template strand for RNA synthesis) corresponds to the sequence of bases in mRNA. Most prokaryotic genes and many eukaryotic genes are colinear. Therefore, you can look at the gene sequence in the DNA and predict the amino acid sequence in the polypeptide. Many eukaryotic genes, however, are not colinear. They contain introns that are spliced out of the pre-mRNA.

What chemical group (phosphate group, hydroxyl group, or a nitrogenous base) is found at the 3' end of a DNA strand? What group is found at the 5' end?

A hydroxyl groups is at the 3' end and a phosphate group is at the 5' end.

What is a polysome?

A polysome is an mRNA molecule with many ribosomes attached to it.

A short genetic sequence, which may be recognized by DNA primase, is repeated many times throughout the E. coli chromosome. Researchers have hypothesized that DNA primase may recognize this site as a site to begin the synthesis of an RNA primer for DNA replication. The E. coli chromosome is roughly 4.6 million bp in length. How many copies of the DNA primase recognition sequence would be necessary to replicate the entire E. coli chromosome?

A primer is needed to make each Okazaki fragment. The average length of an Okazaki fragment is 1000 to 2000 bp. If we use an average value of 1500 bp for each Okazaki fragment, then there needs to be approximately 4,600,000/1500 = 3,067 copies

The term subunit can be used in a variety of ways. Compare the use of the term subunit in proteins verses ribosomal subunit.

A protein subunit is a polypeptide. A ribosomal subunit is a much larger complex that is composed of RNA and many proteins. A ribosomal subunit is a much larger structure compared to a protein subunit.

During the elongation stage of translation, new tRNAs enter at which site?

A site

Genetic material acts as a blueprint for the organism's traits. Explain how the experiments of Griffith indicate that genetic material was being transferred to the type R bacteria.

A trait of pneumococci is the ability to synthesize a capsule. There needs to be a blueprint for this ability. The blueprint for capsule formation was being transferred from the type S to the type R bacteria. (Note: At the molecular level, the blueprint is a group of genes that encode enzymes that can synthesize a capsule.)

polyA addition - Defn

A's added to 3' end

Elongation factors dissociate from RNA Pol2

Allosteric Model

RNA Pol 2 is destabalized

Allosteric Model

Answer the following questions that pertain to the experiment of Figure 11.3. A. What would be the expected results if the Meselson and Stahl experiment were carried out for four or five generations? B. What would be expected results of the Meselson and Stahl experiment after three generations if the mechanism of DNA replication was dispersive? C. As shown in the data, explain why three different bands (i.e., light, half-heavy, and heavy) can be observed in the CsCl gradient.

A. Four generations: 7/8 light, 1/8 half-heavy; five generations: 15/16 light, 1/16 half-heavy B. All of the DNA double helices would be 1/8 heavy. C. The CsCl gradient separates molecules according to their densities. 14N-containing compounds have a lighter density compared to 15N-containing compounds. The bases of DNA contain nitrogen. If the bases contain only 15N, the DNA will be heavy; it will sediment at a higher density. If the bases contain only 14N, the DNA will be light; it will sediment at a lower density. If the bases in one DNA strand contain 14N and the bases in the bases in the opposite strand contain 15N, the DNA will be half-heavy; it will sediment at an intermediate density.

Termination factors may bind to RNA pol 2 complex

Allosteric Model

Mutations that occur at the end of a gene may alter the sequence of the gene and prevent transcriptional termination. A. What types of mutations would prevent p-independent termination? B. What types of mutations would prevent p-dependent termination? C. If a mutation prevented transcriptional termination at the end of a gene, where would gene transcription end? Or would it end?

A. Mutations that alter the uracil-rich region by introducing Gs and Cs, and mutations that prevent the formation of the stem-loop structure. B. Mutations that alter the termination sequence, and mutations that alter the p recognition site. C. Eventually, somewhere downstream from the gene, another transcriptional termination sequence would be found, and transcription would terminate there. This second termination sequence might be found randomly or it might be at the end of an adjacent gene.

Do the following events during bacterial translation occur primarily within the 30S subunit, within the 50S subunits, or at the interface between these two ribosomal subunits? A. mRNA-tRNA recognition B. Peptidyl transfer reaction C. Exit of the polypeptide chain from the ribosome D. Binding of initiation factors IF1, IF2, and IF3

A. On the surface of the 30S subunit and at the interface between the two subunit B. Within the 50S subunit C. From the 50S subunit D. To the 30S subunit

For each of the following transcription factors, how would eukaryotic transcriptional initiation be affected if it were missing? A. TFIIB B. TFIID C. TFIIH

A. RNA polymerase would not be bound to the core promoter. B. TFID contains the TATA-binding protein. If it were missing, RNA polymerase would not bind to the TATA box. C. The formation of the open complex would not take place.

If the following RNA polymerases were missing from a eukaryotic cell, what types of genes would not be transcribed? A. RNA polymerase I B. RNA polymerase II C. RNA polymerase III

A. Ribosomal RNA (5.8S, 18S, and 28S) B. All of the mRNA and certain snRNA genes C. All of the tRNA genes and the 5S rRNA genes

In which of the ribosomal sites, the A site, P site, and/or E site, could the following be found? A. A tRNA without an amino acid attached B. A tRNA with a polypeptide attached C. A tRNA with a single amino acid attached

A. The E site and P sites. (Note: A tRNA without an amino acid attached is only briefly found in the P site, just before translocation occurs). B. P site and A site (Note: A tRNA with a polypeptide chain attached is only briefly found in the A site, just before translocaiton occurs). C. Usually the A site, except the initiator tRNA, which can be found in P site.

For each of the following initiation factors, how would eukaryotic initiation of translation be affected if it were missing? A. eIF2 B. eIF4 C. eIF5

A. The initiation tRNA would not bind to the 40S ribosomal subunit. B. The mRNA would not bind to the 40S ribosomal subunit, and/or the start codon would not be recognized because the secondary structure in the mRNA was not unwound. C. The 60S ribosomal subunit would not assemble after the start codon had been identified.

According to the adaptor hypothesis, are the following statements true or false? A. The sequence of anticodons in tRNA directly recognizes codon sequences in mRNA, with some room for wobble. B. The amino acid attached to the tRNA directly recognizes codon sequences in mRNA. C. The amion acid attached to the tRNA affects the binding of the tRNA to a codon sequence in mRNA.

A. True B. False C. False

The ability of DNA polymerase to digest a DNA strands of end is called its exonuclease activity. Exonuclease activity is used to digest RNA primers and also to proofread a newly made DNA strand. Note: DNA polymerase I does not change direction while it is removing an RNA primer and synthesizing new DNA. It does change direction during proofreading. A. In which direction, 5' to 3' or 3' to 5', is the exonuclease activity occurring during the removal of RNA primers and during the proofreading and removal of mistakes following DNA replication? B. Figure 11.6 shows a drawing of the 3' exonuclease site. Do you think this site would be used by DNA polymerase I to remove RNA primers? Why or why not?

A. The removal of RNA primers occurs in the 5' to 3' direction, while the proofreading function occurs in the 3' to 5' direction. B. No. The removal of RNA primers occurs from the 5' end of the strand.

With regard to the experiment described in Figure 9.2, answer the following: A. List several possible reasons why only a small percentage of the type R bacteria was converted to type S. B. Explain why an antibody must be used to remove the bacterial that are not transformed. What would the results look like, in all five cases, if the antibody/centrifugations step had not been included in the experimental procedure?

A. There are different possible reasons why most of the cells were not transformed. 1. Most of the cells did not take up any of the type S DNA. 2. The type S DNA was usually degraded after it entered the type R bacteria. 3. The type S DNA was usually not expressed in the type R bacteria. B. The antibody/centrifugation steps were used to remove the bacteria that that had not been transformed. It enabled the researchers to determine the phenotype of the bacteria that had been transformed. If this step was omitted, there would have been so many colonies on the plate it would have been difficult to identify any transformed bacterial colonies, because they would have represented a very small proportion of the total number of bacterial colonies.

With regard to the experiment of Figure 9.5, answer the following: A. Provide possible explanations why some of the DNA is in the supernatant. B. Plot the results if the radioactivity in the pellet, rather than in the supernatant, had been measured. C. Why were 32P and 35S chosen as radioisotopes to label the phages? D. List possible reasons why less than 100% of the phage protein was removed from the bacterial cells during the shearing process.

A. There are several possible explanations why about 35% of the DNA is in the supernatant. One possibility is that not all of the DNA was injected into the bacterial cells. Alternatively, some of the cells may have been broken during the shearing procedure, thereby releasing the DNA. If the radioactivity in the pellet had been counted instead of the supernatant, the following figure would be produced. [Figure is graph of % of total isotope (in the pellet)/time in blender (min) where there are curved lines with 35S at 20% and 32S at 65%] C. 32P and 35S were chosen as radioisotopes to label the phages because phosphorus is found in nucleic acids, while sulfur is found only in proteins. D. There are multiple reasons why less than 100% of the phage protein is removed from the bacterial cells during the shearing process. For example, perhaps the shearing just is not strong enough to remove all of the phages, or perhaps the tail fibers remain embedded in the bacterium and only the head region is sheared off.

In bacteria, researchers have isolated strains that carry mutations within tRNA genes. These mutations can change the sequence of the anticodon. For example, a normal tRNAtrp gene would encode a tRNA with the anticodon 3'-ACC-5'. A mutation could change this sequence to 3'-CCC-5'. When this mutation occurs, the tRNA still carries a tryptophan at its 3' acceptor stem, even though the anticodon sequence has been altered. A. How would this mutation affect the synthesis of polypeptides within the bacterium? B. What does this mutation tell you about the recognition between tryptophanyl-tRNA synthetase and tRNAtrp? Does the enzyme primarily recognize the anticodon or not?

A. This mutant tRNA would recognize glycine condons in the mRNA but would put in tryptophan amino acids where glycine amino acids are supposed to be in the polypeptide chain. B. This mutation tells us that the aminoacyl-tRNA synthetase is primarily recognizing other regions of the tRNA molecule besides the anticodon region. In other words, tryptophanyl-tRNA synthetase (the aminoacyl-tRNA synthetase that attaches tryptophan) primarily recognizes other regions of the tRNAtrp sequence aminoacyl-tRNA sythetases recognized only the anticodon region, we would expect glycyl-tRNA sythetase to recognize this mutant tRNA and attach glycine. This is not what happens.

As shown in Figure 11.5, five DnA boxes are found within the origin of replication in E. coli. Take a look at these five sequences carefully. A. Are the sequences of the five DnA boxes very similar to each other? (Hint: Remember that DNA is double-stranded; think about these sequences in the forward and reverse direction.) B. What is the most common sequence for the DnA box? In other words, what is the most common base in the first position, second position, and so on until the ninth position? The most common sequence is called the consensus sequence. C. The E. coli chromosome is about 4.6 million bp long. Based on a random chance, is it likely that the consenus sequence for a DnA box occurs elsewhere in the E. coli chromosome? If so, why aren't there multiple origins of replication in E. coli?

A. When looking at Figure 11.5, the first, second, and fourth DnaA boxes are running in the same direction, and the third and fifth are running in the opposite direction. Once you realize that, you can see that the sequences are very similar to each other. B. According to the direction of the first DnaA box, the consensus sequence is TGTGGATAA ACACCTATT C. This sequence is nine nucleotides long. Because there are four kinds of nucleotides (i.e., A, T, G, and C), the chance of this sequence occurring by random chance is 4^(-9), which equals once every 262,144 nucleotides. Because the E. coli chromosome is more than 10 times longer than this, it is fairly likely that this consensus sequence occurs elsewhere. The reason why there are not multiple origins, however, is that the origin has five copies of the consensus sequence very close together. The chance of having five copies of this consensus sequence occurring close together (as a matter of random chance) is very small.

An absentminded researcher follows the steps of Figure 11.3, and when the gradient is viewed under UV light, the researcher does not see any bands at all. Which of the following mistakes could account for this observation? Explain how. A. The researcher forgot to add 14N-containing compounds. B. The researcher forgot to add lysozyme. C. The researcher forgot to turn on the UV lamp.

A. You would probably still see a band of DNA, but you would only see a heavy band. B. You would probably not see a band because the DNA would not be released from the bacteria. The bacteria would sediment to the bottom of the tube. C. You would not see a band. UV light is needed to see the DNA, which absorbs light in the UV region.

Genes may be structural genes that encode polypeptides, or they may be nonstructural genes. A. Describe three examples of genes that are not structural genes. B. For structural genes, one DNA strand is called the template strand, and the complementary strand is called the coding strand. Are these two terms appropriate for nonstructural genes? Explain. C. Do nonstructural genes have a promotor and terminator?

A. tRNA genes encodes tRNA molecules, and rRNA genes encode the rRNAs found in ribosomes. There are also nonstructural genes for the RNAs found in snRNPs and so on. B. The term template strand is still appropriate because on eof the DNA strands is used as a template to make the RNA. The term coding strand is not appropriate because the RNA made from nonstructural genes does not code for a polypeptide sequence. C. Yes.

Beadle and Tatum used Neurospora to identify mutants affecting various biosynthetic pathways. Their research led them to develop the one gene, one enzyme hypothesis. We now know this hypothesis is too narrow in its description of genes. In what way(s) was their hypothesis is too narrow? (Check all that apply) Some genes make RNAs that are not translated. Some genes make RNAs that are not translated. Many proteins in the cell are not enzymes. Many proteins in the cell are not enzymes. They were unable to isolate all of the enzymes involved in methionine biosynthesis. They were unable to isolate all of the enzymes involved in methionine biosynthesis. Many proteins are composed of two or more different polypeptides; so more than one gene is involved in making the functional protein.

All Apply

According to Chargaff's rule, which of the following statements about double-stranded DNA is true? amount of A = amount of T; amount of C = amount of G A + G = C + T The total number of purines = the total number of pyrimidines

All of these responses are correct.

Telomerase replicates the ends of linear chromosomes. Which of the following accurately describes telomerase and telomeres? The telomerase enzyme is made up of both protein and RNA. Telomerase uses reverse transcriptase activity to add the telomere repeats. Telomere sequences vary among organisms. Telomeres prevent the loss of genetic material caused by telomere shortening.

All of these responses are true.

Describe the structural features that all tRNA molecules have in common.

All tRNA molecules have some basic features in common. They all have a cloverleaf structure with three stem-loop the antiocodon sequence that recognizes the codon sequence in mRNA. At the 3' end, there is an acceptor stem, with the sequence CCA, that serves as an attachment site for an amino acid. Most tRNAs also have base modifications that occur within their nucleotide sequences.

Describe the anticodon of a single tRNA that could recognize the codons 5'-AAC-3' and 5'-AAU-3'. How would this tRNA need to be modified for it to also recognize 5'-AAA-3'?

An anticodon that was 3'-UUG-5' would recognize the two codons. To recognize 5'-AAA-3', it would have to be modified to 3'-UUI-5'.

The covalent attachment of an amino acid to a tRNA is an endergonic reaction. In order words, it requires an input of energy for the reaction to proceed. Where does the energy come from to attach amino acids to tRNA molecules?

As shown in Figure 13.13, the energy comes from ATP. It is the energy conversion that explains the term charged tRNA.

_______ of the DNA double helix generates major and minor grooves.

Asymmetrical spacing of the backbones

Discuss the significance of modified bases within tRNA molecules.

Bases that have been chemically modified can occur at various locations throughout the tRNA molecule. The significance of all of these modifications is not entirely known. However, within the anticodon region, base modification may alter base pairing to allow the anticodon to recognize two or more different bases within the codon.

With regard to DNA replication, define the term bidirectional replication.

Bidirectional replication refers to DNA replication in both directions starting one origin.

If Nirenberg and Matthaei performed their cell-free translation experiment using an mRNA composed of 60% C and 40% A what radiolabeled amino acids would be incorporated into the precipitated polypeptides?

CCC, CCA, CAA, CAC, AAA, ACC, AAC, ACA

MreB, a bacterial protein important for helping maintain the rod-shape of certain bacteria

Cell shape and organization

RNA Editing - Defn

Change of base in sequence

You are doing research in bacteria on the effects of mutating the Shine-Dalgarno sequence. You change some nucleotides in the Shine-Dalgarno sequence and find that translation is greatly reduced. What other element in the bacteria could you change that would have a good chance of restoring translation in the bacterial strain with the altered Shine-Dalgarno sequence?

Change the sequence of the 16S rRNA

splicing - Defn

Cleavage and joining of RNA

Processing - Defn

Cleavage into smaller RNAs

What provides the energy for the formation of the bond between an incoming dNTP and the growing DNA chain?

Cleavage of the dNTP into dNMP and PPi

5'-capping- Presence in eukaryotes

Common

Base modification - Presence in eukaryotes

Common

Base modification - Presence in prokaryotes

Common

Processing- Presence in eukaryotes

Common

Processing- Presence in prokaryotes

Common

RNA editing- Presence in eukaryotes

Common

polyA addition- Presence in eukaryotes

Common

splicing- Presence in eukaryotes

Common

Base modification - Defn

Covalent modification of base

In the experiment by Hershey and Chase, the bacteriophage injected ____ into the E. coli cells.

DNA

part of the phage that enters the bacterial cell following infection

DNA

Which of the following best describes transcription?

DNA -> RNA

In E. coli, which DNA polymerase carries out the majority of the DNA synthesis?

DNA Pol 3

The enzyme that travels along the leading strand assembling new nucleotides on a growing new strand of DNA is

DNA Pol 3

Elongation

DNA and polymerase in open complex

In Chapter 11, we discussed the function of DNA helicase, which is involved in DNA replication. The structure and function of DNA helicase and p protein are rather similar to each other. Explain how the functions of these two proteins are similar and how they are different.

DNA helicase and p protein bind to a nucleic acid strand and travel in the 5' to 3' direction. When they encounter a double-stranded region, they break the hydrogen bonds between complementary strands. The p protein is different from DNA helicase in that it moves along an RNA strand, while DNA helicase moves along a DNA strand. The purpose of DNA helicase funciton is to promote DNA replication, while the purpose of p protein function is to promote transcriptional termination.

Which enzyme replaces the RNA primer with DNA?

DNA pol 1

A DNA strand has the following sequence: 5'-GATCCCGATCCGCATACATTTACCAGATCACCACC-3' In which direction would DNA polymerase slide along th8is strand (from left to right or from right to left)? If this strand was used as a template by DNA polymerase, what would be the sequence of the newly made strand? indicate the 5' and 3' ends of the newly made strand.

DNA polymerase would slide from right to left. The new strand would be 3'-CTAGGGCTAGGCGTATGTAAATGGTCTAGTGGTGG-5'

Which eukaryotic DNA polymerase interacts with the primase?

DNA polymerase α

What activity does the replisome has the the primosome lacks?

DNA synthesis ability

Bacterial cells need to tightly regulate DNA replication so that replication does not initiate multiple times before cell division occurs. The cells use several methods to regulate the initiation of DNA replication. In order for replication to initiate which of the following needs to occur?

DnaA is bound to OriC and the GATC sites on both strands are methylated

5' capping- Presence in prokaryotes

Does not occur

RNA Editing- Presence in prokaryotes

Does not occur

Draw the structure of deoxyribose and number the carbon atoms. Describe the numbering of the carbon atoms in deoxyribose with regard to the directionality of a DNA strand. In a DNA double helix, what does the term antiparallel mean?

Double-stranded RNA is more likely A DNA than B DNA. See the text for a discussion of A-DNA structure.

Ubiquitin ligase, which attaches ubiquitin molecules to a protein, targeting it for destruction

Enzymes

What sequence elements are found within the core promoter of structural genes in eukaryotes? Describe their locations and specific functions.

Eukaryotic promoters are somewhat variable with regard to the pattern of sequence elements that may be found. In the case of structural genes that are transcribed by RNA polymerase II, it is common to have a TATA box, which is about 25 bp upstream from a transcriptional start site and the assembly of RNA polymerase and various transcription factors. The transcriptional start site defines where transcription actually begins.

2. Lagging strand synthesis occurs in the same direction as the replication fork.

F

2. Prokaryotic cells have multiple origins of replication.

F

2. The wrong amino acid is used to charge a tRNA about 1 out of every 1000 charging events.

F

4. Eukaryotic cells use polymerase I to remove the RNA primers.

F

4. Okazaki fragments remain in the DNA. This means that one strand has a series of breaks in the sequence.

F

4. There are 64 different tRNAs in the cell.

F

5. In bacteria, DNA polymerase lays down the RNA primer first and then continues replication.

F

6. The 3' nucleotide of the anticodon on the tRNA is involved in wobble.

F

A eukaryotic promoter will be functional in prokaryotic cells

F

A precursor tRNA molecule must be cut with exonucleases to produce a mature tRNA

F

A prokaryotic promoter will be functional in eukaryotic cells

F

All RNAs undergo RNA editing

F

All transcribed RNAs are translated

F

As helicase unwinds the DNA molecule, the separated strands are kept apart by DNA polymerase.

F

Before DNA replication can occur, the OH bonds between the strands must be broken.

F

DNA replication can begin at any site along the DNA molecule.

F

During DNA replication, the lagging strand is synthesized continuously, while the leading strand is synthesized discontinuously.

F

Eukaryotic pre-mRNA can be used for protein expression in prokaryotes

F

Polyadenylation primarily occurs in the cytoplasm.

F

Pre-mRNA would be useful for an in vivo translation assay

F

The 3' splice site contains a GU, whereas the 5' splice site contains an AG.

F

The newly-synthesized strand of DNA has exactly the same base sequence as that of its template strand.

F

The terminator of a gene is located upstream of the coding region.

F

Unwinding of the DNA during transcription is the result of the activity of a helicase enzyme downstream of the RNA polymerase.

F

3. Eukaryotic ribosomes are assembled in the cytosol.

False

4. Ribosomes are composed solely of rRNAs.

False

In bacteria, DNA polymerase I removes RNA primers. What enzyme plays this role in eukaryotic cells?

Flap endonuclease

The technique of dideoxy sequencing of DNA is described in Chapter 18. The technique relies on the use of dideoxyribnucleotides (shown in Figures 18.18 and 18.19). A dideoxyribonucleotide has a hydrogen atom attached to the 3'-carbon atom instead of an -OH group. When a dideoxyribonucleotide is incorporated into a newly made strand, the strand cannot grow any longer. Explain why.

For a DNA strand to grow, a phosphoester bond is formed between the 3'-OH group on one nucleotide and the innermost 5' phosphate group on the incoming nucleotide (see Figure 11.12). If the -OH group is missing, a phosphoester bond cannot form.

RNA polymerase I

Forms phosphodiester bonds

The following sentences describe the Meselson and Stahl experiment. Place them in order so they explain the steps of the experiment. Grow E. coli in the presence of heavy nitrogen (¹⁵N) for several generations. Transfer the E. coli to ¹⁴N media. Remove samples of the E. coli at multiple time points, about every thirty minutes. Isolate the DNA from the E. coli and load it onto a cesium chloride density gradient. Determine where the DNA is located in the cesium chloride density gradient.

Grow E. coli in the presence of heavy nitrogen (¹⁵N) for several generations. Transfer the E. coli to ¹⁴N media. Remove samples of the E. coli at multiple time points, about every thirty minutes. Isolate the DNA from the E. coli and load it onto a cesium chloride density gradient. Determine where the DNA is located in the cesium chloride density gradient.

Initiation

Holoenzyme binds to promoter

The initiation phase of eukaryotic transcription via RNA polymerase II is considered an assembly and disassembly process. Which types of biochemical interactions-hydrogen bonding, ionic bonding, covalent bonding, and/or hydrophobic interactions-would you expect to drive the assembly and disassembly process? How would temperature and salt concentration affect assembly and disassembly?

Hydorgen bonding is usually the predominant type of interaction when proteins and DNA follow an assembly and disassembly process. In addition, ionic bonding and hydrophobic interactions could occur. Covalent interactions would not occur. High temperature and high salt concentrations tend to break hydrogen bonds. Therefore, high temperature and high salt would inhibit assembly and stimulate disassembly.

The chromosome of E. coli contains 4.6 million bp. How long will it take to replicate its DNA? Assuming DNA polymerase III is the primary enzyme involved and this enzyme can actively proofread during DNA synthesis, how many base pair mistakes will be made in one round of DNA replication in a bacterial population containing 1,000 bacteria?

If we assume there are 4,600,000 bp of DNA, and that DNA replication is bidirectional at the rate of 750 nucleotides per second: If there were just a single replication fork: 4,600,000/750 =6,133 seconds, or 102.2/2 minutes Because replication is bidirectional: 102.2/2 = 51.1 minutes Actually, this is an average value based on a variety of growth conditions. Under optimal growth conditions, replication can occur substantially faster. With regards to errors, if we assume an error rate of one mistake per 100,000,000 nucleotides: 4,600,000 x 1,000 bacteria =4,600,000,000 nucleotides of replication DNA 4,600,000,000/100,000,000 = 46 mistakes When you think about it, this is pretty amazing. In this population, DNA polymerase would cause only 46 single mistakes in a total of 1,000 bacteria, each containing 4.6 million bp of DNA.

The experiment of Figure 11.19 described a method for determining the amount of DNA made during replication. Let's suppose that you can purify all of the proteins required for DNA replication. You then want to "reconstitute" DNA synthesis by mixing together all of the purified components necessary to synthesize a complementary strand of DNA. If you started with single-stranded DNA as a template, what additional proteins and molecules would you have to add for DNA synthesis to occur? What additional proteins would be necessary if you started with a double-stranded DNA molecule?

If you started with single-stranded DNA, you would need to add a primer (or primase), dNTPs, and DNA polymerase. If you started with double-stranded DNA, you would also need helicase. Adding single-strand binding protein and topoisomerase may also help.

What is alternative splicing? What is its biological significance?

In alternative splicing, variation occurs in the patter of splicing so that the resulting mRNAs contain alternative combinations of exons. The biological significance is that two or more different proteins can be produced from a single gene. This is a more efficient use of the genetic material. In multicellular organisms, alternative splicing is often used in a cell-specific manner.

In eukaryotes, what types of modification occur to pre-mRNA?

In eukaryotes, pre-mRNA can be capped, spliced, edited, and then exported out of the nucleus.

If a tRNA has an anticodon sequence 3'-CCL-5', what codon(s) can it recognize?

It can recognize 5'-GGU-3', 5'-GGU-3', and 5'-GGA-3'. All of these specify glycine.

In Trypanosomes, RNA editing involves the addition and deletion of uracil nucleotides.

T

Discuss the differences between p-dependent and p-independent termination.

In p-dependent termination, the p protein binds to the rute site in the RNA transcript after the p site has been transcribed. Eventually, an RNA sequence forms a stem-loop structure that will cause RNA polymerase to pause in the transcription process. As it is pausing, the p protein, which functions as a helicase, will catch up to RNA polymerase and break the hydrogen bonds between the DNA and RNA within the open complex. When this occurs, the completed RNA strand is separated from the DNA along with RNA polymerase. In p-independent transcription, there is no p protein. The RNA forms a stem-loop complex that causes RNA polymerase to pause. However, when it pauses, a uracil-rich region is bound to the DNA template strand in the open complex. Because this is holding on by fewer hydrogen bonds, it is rather unstable. Therefore, it tends to dissociate from the complex and thereby end transcription.

What is an activated amino acid?

In the context of translation, an activated amino acid has had AMP attached to it. This provides necessary energy so that the amino acid can be attached to the correct tRNA.

Suppose it was possible to change the anticodon on a tRNA from 3'-CCG-5' to 3'-UGG-5' after the amino acid was already loaded. Predict the result in a translated polypeptide. What if you changed the sequence prior to the amino acid being loaded onto the tRNA?

In the first case glycine would be found in the polypeptide where threonine was supposed to be. In the second case the changed tRNA would behave like a tRNA for threonine.

Select the stages of translation (check all that apply).

Initiation Elongation Termination

Which of the following best describes the function of telomerase at the telomere?

It adds new DNA to the longer strand of the telomere overhang.

Does the experiment of Figure 9.5 rule out the possibility that RNA is the genetic material of T2 phage? Explain your answer. If it does not, could you modify the approach of Hershey and Chase to show that it is DNA and not RNA that is the genetic material of T2 bacteriophage? Note: It is possible to specifically label DNA or RNA by providing bacteria with radiolabeled thymine or uracil, respectively.

It does not rule out the possibility that RNA is the genetic material, because RNA and DNA both contain phosphorus. One way to distinguish RNA and DNA is to provide bacteria with radiolabeled uracil in order to label RNA or provide bacterial with radiolabeled thymine to label DNA. (Note: Uracil is found only in RNA and thymine is found only in DNA.) IF they had a propagated T2 phages contain DNA and not RNA, because radiolabeled uracil would not be label the genetic material of T2 bacteriophage.

What key structural features of the DNA molecule underlie its ability to be faithfully replicated?

It is a double-stranded structure that follows the AT/GC rule.

Which statement best describes synthesis of the lagging strand?

It proceeds 5'-3' away from the replication fork

Which is a correct statement about Rho independent termination?

It requires a stem loop to form in the RNA.

Compare the structural features of a double-stranded RNA structure with those of a DNA double helix.

Its nucleotide base sequence.

In the absence of telomerase activity, chromosomes are shortened slightly after every round of replication.

T

An mRNA has the following sequence: 5'-CCCAUAUGCUGGGAUAUCGGUGAC-3'. What is the first codon translated? What is the last codon translated? How many amino acids are in the polypeptide?

Methionine; Arginine; 5

5' capping - Defn

Methylated guanine added to 5' end

Describe the sequence in bacterial mRNA that promotes recognition by the 30S subunit.

Most bacterial mRNAs contain a Shine-Dalgarno sequence, which is necessary for the binding of the mRNA to the small ribosomal subunit. This sequence, UUAGGAGGU, is complementary to a sequences will hydrogen bond to each other during the initiation stage of translation.

Actin assembles and diassembles, allowing cells to migrate

Movement

Mutations in bacterial promotors may increase or decrease the level of gene transcription. Promoter mutations that increase transcription are termed up-promoter mutations, and those that decrease transcription are termed down-promoter mutations. As shown in Figure 12.5, the sequence of the -10 region of the promoter for the lac operon is TATGTT. Would you expect the following mutations to be up-promoter or down-promoter mutations? A. TATGTT to TATATT B. TATGTT to TTTGTT C. TATGTT to TATGAT

Mutations that make a sequence more like the consensus sequence are likely to be up promoter mutations, while mutations that cause the promoter to deviate from the consensus sequence are likely to be down promoter mutations. Also, in the -10 region, AT paris are favored over GC pairs, because the role of this region is to form the open complex. AT pairs are more easily separated, because they form only two hydrogen bonds compared to GC pairs, which form three hydrogen bonds. A. Up promoter B. Down promoter C. Up promoter

Is the promoter sequence part of the DNA that is transcribed?

Neither the -35 nor the -10 sequences are present on the new transcript.

Is it necessary for a cell to make 61 different tRNA molecules, corresponding to the 61 codons for amino acids? Explain your answer.

No, it is not. Due to the wobble rules, the 5' base in the anticodon of a tRNA can recognize two or more bases in the third (3') position of mRNA. Therefore, any given cell type synthesizes far fewer than 61 types of tRNAs.

The genetic material found within some viruses is single-stranded DNA. Would this genetic material contain equal amounts of A and T and equal amounts of G and C?

Not necessarily. The AT/GC rule is required only of double-stranded DNA molecules.

polyA addition- Presence in prokaryotes

Occasional

The short DNA segments formed on the discontinuously replicated strand are called

Okazaki fragments.

Termination

Polymerase stops synthesizing copy of DNA template

Elongation

Polymerase synthesizes copy of DNA template

Prokaryotic chromosomes do not have telomeres because

Prokaryotic chromosomes are circular.

It is possible to get more than one protein product from a eukaryotic pre-mRNA.

T

Single-stranded binding proteins keep the two parental strands of DNA separated from each other until DNA polymerase has an opportunity to replicate the strands. Suggest how single-stranded binding proteins keep the strands separated and yet do not impede the ability of DNA polymerase to replicate the strands.

Primase and DNA polymerase are able to knock single-stranded binding proteins off the template DNA.

The most important control that is missing are the results of a sample with no____ added.

RNA

Discuss the similarities and differences between RNA polymerase (described in this chapter) and DNA polymerase (described in Chapter 11).

RNA and DNA polymerase are similar in the following ways: 1. They both use a template strand. 2. They both synthesize in the 5' to 3' direction. 3. The chemistry of synthesis is very similar in that they use incoming nucleoside triphosphates and make a phosphoester bond between the previous nucleotide and incoming nucleotide. 4. They are both processive enzymes that slide along a template strand of DNA. RNA and DNA polymerase are different in the following ways: 1. RNA polymerase does not need a primer. 2. RNA polymerase does not have a proofreading function.

Termination

RNA forms stem loop structure for this to occur

what is the subunit composition of bacterial RNA polymerase holoenzyme? What are the functional roles of the different subunits?

RNA polymerase holoenzyme consists of sigma factor plus the core enzyme, which is composed of five subunits, a2BB'w. The role of sigma factor (o-) is to recognize the promoter sequence. The a subunits are necessary for the assembly of the core enzyme and for loose DNA binding. The B and B' subunits are the portion that catalyzes the covalent linkages between adjacent ribonucleotides. The w subunit is important for the proper assembly of the core enzyme.

Describe the movement of the open complex along the DNA.

RNA polymerase slides down the DNA and forms an open complex as it goes. The open complex is a DNA bubble of about 17 bp. Within the open complex, the DNA strand running in the 3' to 5' direction is used as a template for RNA synthesis. This occurs as individual nucleotides hydrogen bond to the DNA template strand according to the rules described in question C12. As the RNA polymerase slides along, the DNA behind the open complex rewinds back into a double helix.

The enzyme that accomplishes transcription is termed

RNA polymerase.

Before the lagging strand can begin assembling new DNA nucleotides, which of the following must occur?

RNA primase constructs a short RNA primer.

Why is it important for living organisms to regulate DNA replication?

Regulating DNA replication is necessary to have the correct number of chromosomes per cell. If DNA replication is too slow, daughter cells may not receive any chromosomes. If it is too fast, they may have too many chromosomes.

Mature mRNA from a eukaryotic cell will not be identical to the primary transcript

T

Termination-Proaryotes

Rho dependent and independent models

Mature mRNA from a prokaryotic cell will be identical to the genomic sequence

T

What is meant by the term self-splicing? What types of introns are self-splicing?

Self-splicing means that an RNA molecule can splice itself without the aid of a protein. Group I and II introns can be self-splicing, although proteins can also enhance the rate of processing.

5. DNA replication in eukaryotes takes a much longer time than DNA replication in prokaryotes.

T

At the molecular level, describe how o- factor recognizes bacterial promotors. Be specific about the structure of o- factor and the type of chemical bonding.

Sigma factor can slide along the major groove of the DNA. In this way, it is able to recognize base sequences that are exposed in the groove. When it encounters a promoter sequence, hydrogen bonding between the bases and the sigma factor protein can promote a tight and specific interaction.

initiation-Proaryotes

Sigma factor needed

5. There are 20 different aminoacyl-tRNA synthetases in the cell.

T

6. DNA polymerase III is a holoenzyme.

T

7. Binds to single stranded DNA to keep it from reforming double stranded DNA.

Single strand binding protein

If DNA adenine methyltransferase (Dam) had been discovered via isolation of temperature sensitive mutants defective in DNA replication would it have been a rapid-stop mutant or a slow-stop mutant?

Slow-stop mutant

A 7-methylguanosine cap is necessary for an mRNA to leave the nucleus.

T

Which of the following statements is not true? Explain why. A. A DNA strand can serve as a template strand on many occasions. B. Following semiconservative DNA replication, one strand is a newly made daughter strand and the other strand is a parental strand. C. A DNA double helix may contain two strands of DNA that were made at the same time. D. A DNA double helix obeys the AT/GC rule. E. A DNA double helix could contain one strand that is 10 generations older than its complementary strand.

Statement C is not true. A new strand is always made from a preexisting template strand. Therefore, a double helix always contains one strand that is older than the other.

1. All tRNAs have the sequence CCA at their 3' end.

T

1. During DNA replication new nucleotides are added at the 3' end of the molecule.

T

1. Eukaryotic cells have multiple DNA polymerases.

T

3. Eukaryotic cells must replicate the mitochondrial DNA with polymerase gamma.

T

3. Leading strand synthesis occurs continuously.

T

3. The codon in the mRNA and the anticodon on the tRNA bind in an antiparallel fashion.

T

Alternative splicing is more important in yeast than humans.

T

Explain the funcitonal roles of the A, P, and E sites during translation.

The A (aminoacyl) site is the location where a tRNA carrying a single amino acid initially binds. The only exception is the initiator tRNA, which binds to the P (peptidyl) site. The growing polypeptide chain is removed from the tRNA in the P site and transferred to the amino acid attached to the tRNA in the A site. The ribosome translocates in the 3' direction, with the result that the two tRNAs in the P and A sites are moved to the E (exit) and P sites, and the uncharged tRNA in the E site is released.

What is meant by the term DNA sequence?

The DNA sequence refers to the sequence of nucleotide bases.

An mRNA has the following sequence: 5'-AUG UAC UAU GGG GCG UAA-3' Describe the amino acid sequence of the polypeptide that would be encoded by this mRNA. Be specific about the amino and carboxyl terminal ends.

The amino acid sequence is methionine-tyrosine-tyrosine-glycine-alanine. Methionine is at the amino terminal end, alanine at the carboxyl terminal end. The peptide bonds should be drawn as shown in Figure 13.6b.

The wobble rules for tRNA-mRNA pairing are shown in Figure 13.14. If we assume that the tRNAs do not contain modified bases, what is the minimum number of tRNAs needed to efficiently recognize the codons for the following types of amino acids? A. Leucine B. Methionine C. Serine

The answer is three. There are six leucine codons: UUA, UUG, CUU, CUC, CUA, and CUG. The anticodon AAU would recognize UUA and UUG. You would need two other tRNAs to efficiently recognize the other four leucine codons. These could be GAG and GAU or GAA and GAU. B. The is answer is one. There is only one codon, AUG, so you need only one tRNA with the anticodon UAC. C. The answer is three. There are six serine codons: AGU, AGC, UCU, UCC, UCA, and UCG. You would need only one tRNA to recognize AGU and AGC. This tRNA could have the anticodon UCG or UCA. You would need two tRNAs to efficiently recognize the other four tRNAs. These could be AGG and AGU or AGA and AGU.

Describe the components of eukaryotic ribosomal subunits and where the assembly of the subunits occurs within living cells.

The assembly process is very complex at the molecular level. In eukaryotes, 33 proteins and 1 rRNA assemble to form a 40S subunit, and 49 proteins and 3 rRNAs assemble to form a 60S subunit. This assembly occurs within the nucleolus.

What are the building blocks of a nucleotide? With regard to the 5' and 3' positions on a sugar molecule, how are nucleotides linked together to form a strand of DNA?

The building blocks of a nucleotide are a sugar (ribose or deoxyribose), a nitrogenous base, and a phosphate group. In a nucleotide, the phosphate is already linked to the 5' position on the sugar. When two nucleotides are hooked together, a phosphate on one nucleotide forms a covalent bond with the 3' hydroxyl group on another nucleotide.

A tRNA has an anticodon sequences 3'-GGU-5'. What amino acid does it carry?

The codon is 5'-CCA-3', which specifies proline.

You have engineered additional ter sequences into the bacterial chromosome. They are spaced throughout the chromosome. What effect do you predict this will have on DNA replication in this cell?

The entire chromosome will not be replicated.

What enzymatic features of DNA polymerase prevent it from replicating one of the DNA strands at the ends of linear chromosomes? Compared with DNA polymerase, how is telomerase different in its ability to synthesize a DNA strand? What does telomerase use as its template for the synthesis of a DNA strand? What does telomerase use as its template for the synthesis of a DNA strand? How does the use of this template result in a telomere sequence that is tandemly repetitive?

The inability to synthesize DNA in the 3' to 5' direction and the need for a primer prevent replication at the 3' end of the DNA strands. Telomerase is different than DNA polymerase in that it uses a short RNA sequence, which is part of its structure, as a template for DNA synthesis. Because it uses this sequence many times in row, it produces a tandemly repeated sequence in the telomere at the 3' ends of linear chromosomes.

Which steps during the translation of bacterial mRNA involve an interaction between complementary strands of RNA?

The initiation phase involves the binding of the Shine-Dalgarno sequence to the rRNA in the 30S subunit. The elongation phase involves the binding of anticodons in the tRNA to codons in mRNA.

Discuss the similarities and differences in the synthesis of DNA in the lagging and leading strands. What is the advantage of a primosome and a replisome as opposed to having all replication enzymes functioning independently of each other?

The leading strand is primed once, at the origin, and then DNA polymerase III synthesizes DNA continuously in the direction of the replication fork. In the lagging strand, many short pieces of DNA (Okazaki fragments) are made. This requires many RNA primers. The primers are removed by DNA polymerase I, which then fills in the gaps with DNA. DNA ligase then covalently connects the Okazaki fragments together. Having the enzymes within a complex such as a primosome or replisome provides coordination among the different steps in the replication process and thereby allows it to proceed faster and more efficiently.

According to the examples shown in Figure 12.5, which positions of the -35 sequence (i.e., first, second, third, fourth, fifth, or sixth) are more tolerant of changes? Do you think that these positions play a more or less important role in the binding of o- factor? Explain why.

The most highly conserved positions are the first, second, and sixth. In general, when promoter sequences are conserved, they are more likely to be important for binding. That explains why changes are not found at these positions; if a mutation altered a conserved position, the promoter wold probably not work very well. By comparison, changes are occasionally tolerated at the fourth position and frequently at the third and fifth positions. The positions that tolerate changes are less important for binding by sigma factor.

What is the function of the nucleolus?

The nucleolus is a region inside the eukaryotic nucleus where the assembly of ribosomal subunits occurs.

What parts of a nucleotide (namely, phosphate, sugar and/or bases) occupy the major and minor grooves of double-stranded DNA, and what parts are formed in the DNA backbone? If a DNA-binding protein does not recognize a specific nucleotide sequence, do you expect that it recognizes the major grooves, the minor grooves, or the DNA backbone? Explain.

The nucleotide bases occupy the major and minor grooves. Phosphate and sugar are found in the backbone. If a DNA-binding protein does not recognize a nucleotide sequence, it probably is not a binding in the grooves, but instead is binding to the DNA backbone (i.e., sugar-phosphate sequence). DNA-binding proteins that recognize a base sequence must bind into a major or minor groove of the DNA, which is where the bases would be accessible to a DNA-binding protein. Most DNA-binding proteins that recognize a base sequence fit into the major groove. By comparison, other DNA-binding proteins such as histones, which do not recognize a base sequence, bind to the DNA backbone.

As shown in Figure 11.26, telomerase attaches additional DNA, six nucleotides at a time, to the ends of eukaryotic chromosomes. However, it works in only one DNA strand. Describe how the opposite strand is replicated.

The opposite strand is made in the conventional way by DNA polymerase using the strand made via telomerase as a template.

In bacteria, what event marks the end of the initiation stage of transcription?

The release of sigma factor marks the transition to the elongation stage of transcription.

How does a eukaryotic ribosome select its start codon? Describe the sequences in eukaryotic mRNA that provide optimal context for a start codon.

The ribosome binds at the 5' end of the mRNA and then scans in the 3' direction in search of an AUG start codon. If it finds one that reasonably obeys Kozak's rules, it will begin translation at the site. Aside from an AUG start codon, two other important features are a guanosine at the +4 position and a purine at the -3 position.

What is the role of aminoacyl-RNA synthetase? The ability of the amnioacyl-tRNA synthetase to recognize tRNAs has sometimes been called the "second genetic code". Why has the function of this type of enzyme been described this way?

The role of the aminoacyl-tRNA synthetase is to specifically recognize tRNA molecules and attach the correct amino acid to them. This ability is sometimes described as the second genetic code because the specificity of the attachment is critical step in deciphering the genetic code. For example, if a tRNA has a 3'-GGG-5' anticodon, it will recognize a 5'-CCC-3' codon, which should specify proline. It is essential that the aminoacyl-tRNA synthetase known as prolyl-tRNA-synthetase recognize this tRNA and attach proline to the 3' end. The other aminoacyl-tRNA synthetases should not recognize this tRNA.

Write out a sequence of an RNA molecule that could form a stem-loop with 24 nucleotides in the stem and 16 nucleotides in the loop.

The sequence in part A would be more difficult to separate because it has a higher percentage of GC base pairs compared to the one in part B. GC base pairs have three hydrogen bonds compared with AT base pairs, which only have two hydrogen bonds.

An RNA has the following sequence: 5'GGCGAUGGGCAAUAAACCGGGCCAGUAAGC-3' Identify the start codon and determine the complete amino acid sequence that would be translated from the mRNA.

The start codon begins at the fifth nucleotide. The amino acid sequence would be Met Gly Asn Lys Pro Gly Gln STOP.

Draw the structure of a phosphodiester linkage.

The structure is a phosphate group connecting two sugars at the 3' and 5' positions, as shown in Figure 9.10.

Draw the structure of guanine, guanosine, and deoxyguanosine triphosphate.

The structures can be deduced from Figures 9.7 and 9.9. Guanine is the base by itself. Guanosine is the base attached to a ribose sugar. Deoxyguanosine triphosphate is a base attached to a deoxyribose sugar with three phosphates.

List the structural differences between DNA and RNA.

The sugar in DNA is deoxyribose; in RNA it is ribose. DNA containing the base thymine, while RNA has uracil. DNA is a double helical structure. RNA is single stranded, alothough parts of it may form double-stranded regions.

What is the meaning of the term genetic material?

The term genetic material refers to the actual substance that contains genetic information. It is usually DNA, but in some viruses it can be RNA.

What are the three stages of translation? Discuss the main events that occur during these three stages.

The three stages are: 1. Initiation: The mRNA, initiator tRNA, and initiation factors associate with the small ribosomal subunit; then the large subunit associates. 2. Elongation: The ribosome moves one codon at a time down the mRNA, adding one amino acid at a time to the growing polypeptide chain. Three sites on the ribosome, the A, P, and E sites, are important in the process. The A site is where the tRNA (execpt for the initiator tRNA) binds to the ribosome and recognizes the codon in the mRNA. The growing polypeptide chain is then transferred to the amino acid attached to this tRNA. The ribosome then translocates so that this tRNA is now moved to the P site. The empty tRNA that was in the P site is moved into the E site. This empty tRNA in the E site is then expelled and the next charged tRNA can bind to the A site. 3. Termination: A stop codon is reached and a termination factor binds to the A site. The hydrolysis of GTP initiates a series of events that leads to the disassembly of the ribosomal subunits and the release of the completed polypeptide chain.

After the DNA from type S bacteria is exposed to type R bacteria, list all of the steps that you think must occur for the bacterial to start making a capsule.

The transformation process is described in Chapter 7. 1. A fragment of DNA binds to the cell surface. 2. It penetrates the cell wall/cell membrane. 3. It enters the cytoplasm. 4. It recombines with the chromosome. 5. The genes within the DNA are expressed (i.e., transcription and translation). 6. The gene products create a capsule. That is, they are enzymes that synthesize a capsule using cellular molecules as building blocks.

How many different sequences of mRNA could encode a peptide with the sequence proline-glycine-methionine-serine?

There are four proline codons, four glycine codons, one methionine codon, and six serine codons. We apply the product rule to solve this problem. 4 x 4 x 1 x 6 = 96

As the minor and major grooves of the DNA wind around a DNA double helix, do they ever intersect each other, or do they always run parallel to each other?

They always run parallel.

In the tertiary structure of tRNA, where is the anticodon region relative to the attachment site for the amino acid? Are they adjacent to each other?

They are very far apart, at opposite ends of the molecule.

Which of the following is not true about ARS (autonomously replicating sequence) elements?

They have the same sequence as oriC.

A double stranded DNA molecule contains about 560 nucleotides. How many complete turns would be found in this double helix?

This DNA molecule contains 280 bp. There are 10 base pairs per turn, so there are 28 complete turns.

IN this chapter, we considered two experiments-one by Avery, MacLeoa, and McCarthy and the second by Hershey and Chase- that indicated DNA is the genetic material. Discuss the strengths and weaknesses of the two approaches. Which experimental approach did you find the most convincing? Why?

This is really a matter of opinion. The Avery, MacLeod, and McCarty experiment seems to indicate directly that DNA is the genetic material, because DNase prevented transformation and RNase and protease did not. However, one could argue that the DNA is required for the rough bacteria to take up some other contaminant in the DNA preparation. It would seem that the other contaminant would not be RNA or protein. The Hershey and Chase experiments indicate that DNA is being injected into bacteria, although quantitatively the results are not entirely convincing. Some 35S-labeled protein was not sheared off, so the results do not entirely rule out the possibility that protein could be the genetic material. But the results do indicate that DNA is more likely the candidate.

What is the function of the enzyme primase during DNA replication?

To add a RNA primer to enable replication of DNA

Termination-Eukaryotes

Torpedo and allosteric models

Exonuclease degrades transcript in 5' to 3' direction

Torpedo model

Exonuclease stops RNA Pol

Torpedo model

RNA Pol 2 is physically removed from DNA

Torpedo model

RNA downstream from polyA sequence is cleaved

Torpedo model

1. Bacteria have only one kind of ribosome.

True

2. Ribosomes have large and small subunits.

True

5. In eukaryotes, ribosomes found in mitochondria and chloroplasts have a different composition than that of cytosolic ribosomes.

True

The genetic code is a triplet code, meaning that three nucleotides are necessary to code for one amino acid. If one or two nucleotides are inserted into a protein coding region it can cause a frame shift which would change the amino acid sequence. Assume instead, that the genetic code is made up of codons consisting of six nucleotides. The statements below describe nucleotide insertions into a protein coding region. Which would not cause a frame shift?

Twelve nucleotides are inserted.

What does it mean when we say that the genetic code is degenerate? Discuss the universality of the genetic code?

When we say the genetic code is degenerate, it means that more than one codon can specify the same amino acid. For example, GGG, GGC, GGA, and GGU all specify glycine. In general, the genetic code is nearly universal, because it is used in the same way by viruses, prokaryotes, fungi, plants, and animals. As discussed in Table 13.3, there are a few exceptions, which occur primarily in protists and yeast and mammalian mitochondria.

You are studying a protein that contains the amino acid selenocysteine. To understand the protein better, you place the gene for this protein into a plasmid, so that you can control the expression of the gene in cells. When you look at the protein expressed from the plasmid, however, you are surprised to find that the protein encoded from the plasmid is shorter than the protein encoded from the genome. Why?

You forgot to include the SECIS, and therefore UGA functioned as a stop codon rather than a codon for selenocysteine.

Figure 11.4b shows an autoradiograph of a replicating bacterial chromosome. If you analyzed many replicating chromosomes, what types of information could you learn about the mechanism of DNA replication?

You might be able to determine the number of replication forks and their approximate locations. For chromosomes with a single origin, you could determine that replication is bidirectional. However, you would not get any molecular information about the DNA replication process.

The base composition of an RNA virus was analyzed and found to be 14.1% A, 14.0% U, 36.2% G, and 35.7% C. Would you conclude that the viral genetic material is single-stranded RNA or double-stranded RNA?

You would conclude that it is probably double-stranded RNA because the amount of A equals U and the amount of G equals C. Therefore, this molecule could be double stranded and obey the AU/GC rule. However, it is also possible that it is merely a coincidence that A happens to equal U and G happens to equal C, and the genetic material is really single stranded.

Using the reconstitution strategy described in experimental question E4, what components would you have to add to measure the ability of telomerase to synthesize DNA? Be specific about type of template DNA that you would add to your mixture.

You would need to add a DNA template with a 3' overhang that was complementary to the telomere sequence. You would also have to add telomerase and dNTPs.

For a protein encoding gene, what marks the start and end of the DNA region that will be transcribed?

a promoter and a terminator

purines

adenine and guanine

Mature T2 phage particles are released

after lysis of bacterial host cell

Transcription factors are proteins that can increase the rate of transcription. can decrease the rate of transcription. bind to DNA.

all of these are correct

enhancer

cis-acting

promoter

cis-acting

silencer

cis-acting

Tailing-Eukaryotes

common for stability

Splicing-Eukaryotes

common in mRNA genes

capping-Eukaryotes

common on mRNA

RNA editing occurs when bases are____.

deaminated

Multiple codons can code for the same amino acid so, we say the genetic code is___.

degenerate

Phage T2 infection of a bacterial cell is accompanied by

degradation of host cell DNA

X-ray crystallography of DNA suggested

helical structure, not single stranded, 10 base pairs per complete turn

3. Unwinds the two strands of DNA to create a single stranded region.

helicase

Genes are used to make RNA and some RNA is used to make proteins. Not all RNAs in the cell are destined to be translated. What RNAs will be translated?

mRNAs

RNA editing occurs in____ trypanosomal mitochrondrial mRNAs.

many

While group I and II introns can self splice, in cells the process is often helped by proteins called_________.

maturases

Living R type + heat killed s type

mouse died

Living S Type injected

mouse died

Living R type injected

mouse survived

levels of DNA complexity in order from simplest to most complex

nucleotides, DNA strand, double helix, 3D structure

structure of pyrimidine

one ring

triplex DNA

only T and C

labeled with radioactive phosphorus

phage DNA

labeled with radioactive sulfur

phage protein

correct structure of DNA components can be presented as

phosphate - sugar - base

structure of CMP

phosphate, ribose, cytosine

Introns in _______ are not self splicing.

pre-mRNAs

2. Synthesizes an RNA primer.

primase

2. Translation occurs in the cytoplasm.

prokaryotes and eukaryotes

6. Requires a release factor to terminate translation.

prokaryotes and eukaryotes

Tailing-Proaryotes

promotes degredation

Spacer regions removed

rRNA

Transcribed mainly by RNA pol 1

rRNA

splicing- Presence in prokaryotes

rare

Splicing-Proaryotes

rare; self splicing

difference between ribose and deoxyribose

ribose has OH group on 2' carbon of pentose

Proofreading by DNA polymerase involves the removal of

several bases on the newly-synthesized strand of DNA.

A research group has sequenced the cDNA and genomic DNA from a particular gene. Here are the DNA sequences. cDNA: 5-ATTGCATCCAGCGTATACTATCTCGGGCCCAATTAATGCCAGCGGCCAGACTATCACCCAACTCGGTTACTAGTATATCCCATATACTAGCATATATTTTACCCATAATTTGTGTGTGGGTATACAGTATAATCATATA-3' Genomic DNA (contains one intron): 5'-ATTGCATCCAGCGTATACTATCTCGGGCCCAATTAATGCCAGCGGCCAGACTATCACCCAACTCGGCCCACCCCCAGGTTTACACAGTCATACCATACATACAAAAATCGCAGTTACTTATCCCAAAAAAACCTAGATACCCCACATACTATTAACTCTTTCTTTCTAGGTTACCTACTAGTATACCCATATACTAGCATATATTTTACCCATAATTTGTGTGTGGGTATACAGTATAATCATATA-3' Indicate where the intron is located. Does the intron contain the normal consensus splice site sequences based on those described in Figure 12.21? Underline the splice site sequences, and indicate whether or not they fit the consensus sequence.

s

An mRNA encodes a polypeptides that is 312 amino acids in length. The 53th codon in this polypeptide is a tryptophan codon. A mutation in the gene that encodes this polypeptides changes this tryptophan codon into a stop codon. How many amino acids would be in the resulting polypeptide: 52, 53, 259, or 260?

s

Explain what is meant by coupling of transcription and translation in bacteria. Does coupling occur in bacterial and/or eukaryotic cells? Explain.

s

If a gene contains three introns, draw what it would look like in an R loop experiment.

s

The lactose permease of E. coli is a protein composed of a single polypeptide that is 417 amino acids in length. By convention, the amino acids within a polypeptide are numbered from the amino-terminal end to the carboxyl-terminal end. Are the following questions about the lactose permease true or false? A. Because the 64th amino acid is glycine and the 68th amino acid is aspartic acid, the codon for glycine, 64, is closer to the 3' end of the mRNA than the codon for aspartic acid, 68. B. The mRNA that encodes the lactose permease must be greater than 1241 nucleotides in length.

s

What is an R loop? In an R loop experiment, to which strand of DNA does the mRNA bind, the coding strand or the template strand?

s

Group I and group II introns are ______

self-splicing

DNA replication is said to be

semiconservative

RNA editing is the process where the base______ of a transcribed RNA can be changed.

sequence

The snRNPs are also called

snurps

Codons that specify the same amino acid are termed____.

synonymous

Extensive base modification

tRNA

Processed by RNaseD

tRNA

Processed by RNaseP

tRNA

Processed by endonuclease

tRNA

Processed by exonuclease

tRNA

Transcribed by RNA pol 3

tRNA

The Rho protein is involved in the _____ stage of transcription.

termination

DNA polymerases cannot replicate

the 3' ends of linear DNA strands.

pyrimidines

thymine, uracil, cytosine

4. Removes positive supercoiling ahead of the replication fork.

topoisomerase

RNA polymerase

trans-acting

TFIID

trans-acting

transcription factors

trans-acting

RNA polymerase III

transcribes 5s rRNA and tRNA

A male cat and a female cat produce a litter of kittens. Which criteria of genetic material is represented?

transmission

Splicing joins together

two exons.

Deamination of cytosine results in___ and with adenine it results in____.

uracil, inosine

Griffith

used Pneumococcus strains and mice to demonstrate there is a "transforming factor" that could change R bacteria into S bacteria

Avery, McLeod, and McCarthy

used Pneumococcus strains and protease, DNase, and RNase to demonstrate DNA was the transforming factor

Hershey and Chase

used bacteriophage and E. coli to demonstrate DNA was the genetic material injected by the bacteriophage into E. coli and used to code for new phage

RNA polymerase III

uses DNA as a template

RNA polymerase II

uses ribonucleotides