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BIOL239 Lecture Notes - Start Codon, Dna Replication, Thymine

39 views16 pages
pearsardine709
13 Sep 2014
School
University of Waterloo
Department
Biology
Course
BIOL239
Professor
Christine Dupont

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Document Summary

The process by which the polymerization of ribonucleotides guided by complementary base pairing produces an rna transcript of a gene. Like we saw from dna replication, it"s going to use a polymerase but its rna this time. 5" to 3" is a rule for polymerases. Nucleotides are added in the 5" to 3" direction. Uracil is incorporated in place of thymine in rna (both pair with adenine: transcription. Rna polymerase the enzyme that catalyzes transcription. Just like for translation, stop and start codon (methionine) There are places to start and stop transcription but they"re not codons because we"re not making protein, we"re making mrna. They"re the place where polymerase recognizes and comes in. Promoters dna sequences near the beginnings of genes that signal rna polymerase where to begin transcription. Makes the transcript and stops (not at the codon) but at the terminator (large sequence of dna again)

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Related Questions

1. Which of the following is the correct sequence of events (from start to finish) in cellular fatty acid oxidation?

A. Fatty acid activationà carnitine shuttleà beta-oxidation

B. Beta-oxidationà carnitine shuttleà Fatty acid activation

C. Fatty acid activationà carnitine shuttleà beta-oxidation

D. Carnitine shuttleà Fatty acid activationà beta-oxidation

2. What reaction is catalyzed by lipoprotein lipase, or lipase?

A. Hydrolysis of TAGs into fatty acids and glycerol

B. Hydrolysis of TAGs into amino acids and glycerol

C. Hydrolysis of cholesterol in to fatty acid and water

D. Hydrolysis of cholesterol into acetyl-CoA

3. How do fatty acids of different carbon lengths cross into the mitochondrial matrix for oxidation?

A. Small FAs of 10 carbons or fewer can freely diffuse through the membrane

B. Large FAs are cleaved into two smaller units that are actively transported across the membrane

C. A membrane-bound transferase catalyzes the conversion of long-chain FAs and FA-carnitine esters, which are transported across the membrane via a specific carrier

D. The breakdown of large FAs starts outside the matrix, allowing for the smaller particles to pass easily through the membrane

4. How can any RNA nucleotide be easily distinguished from a DNA nucleotide?

A. RNA nucleotides have a unique phosphodiester bond linkage

B. RNA nucleotides have uracil nitrogenous bases

C. RNA nucleotides have an -OH on the 2’ carbon of the ribose

D. RNA nucleotides cannot be easily distinguished from DNA nucleotides

5. Which of the following statement is true concerning the sequence pTACGTAACGTA?

A. This is an RNA sequence

B. In order to have a more rigid structure in a double-stranded molecule, the sequence given would need more adenosine

C. If the complementary strand were present, the number of purines would be equal to the number of pyrimidines

D. The amount of purines is equal to the amount to pyrimidines

6. Complementary base pairing is considered which level of structure?

A. Secondary

B. Tertiary

C. Quaternary

D. Primary

7. Which of the following is/are required for the polymerization reaction catalyzed by DNA polymerase III?

I. RNA primer

II. DNA primer

III. RNA template

IV. DNA template

V. NTPs

VI. dNTPs

VII. Magnesium

VIII. Calcium

A. II, IV, V, and VII

B. II, IV, VI, and VII

C. I, IV, V, and VIII

D. I, IV, VI, and VII

8. DNA polymerase I, from E. coli, has three different catalytic sites. Which of the following is NOT one of them?

A. RNA-dependent RNA polymerase

B. RNA-dependent DNA polymerase

C. DNA-dependent RNA polymerase

D. DNA-dependent DNA polymerase

9. Eukaryotes solve the problem of replicating the ends of their linear chromosomes by:

A. Using a protein as primer

B. Enzyme activity that uses an RNA template and reverse transcription in a series of repeat cycles

C. All of the listed solution

D. Terminal redundancy whereby each end of the chromosome is duplicated allowing recombination

Fill in the blank. Elongation during translation does NOT involve ____________.

Question 16 options:

the translation of codons according to the genetic code

the formation of bonds catalyzed by the ribosome

complementary base pairing between RNA molecules

amino acids being linked together in a polypeptide

reading the DNA template 3' to 5'

For a given gene, what establishes the reading frame for translation?

Question 17 options:

the location of the enhancer relative to the gene

the first three nucleotides at the 5' end of the mRNA

the first three nucleotides at the 3' end of the mRNA

the start codon in the mRNA

the location of the promoter relative to the gene

Which of the following is the LEAST likely direct consequence of a substitution mutation?

Question 18 options:

changing the length of a protein coded for by a gene

changing one amino acid in a protein

creating a stop codon

eliminating a start codon

changing the length of the DNA molecule containing a gene

Suppose that the pre-mRNA transcript from a eukaryotic gene is 30,000 nucleotides long, and the gene codes for a sequence of 300 amino acids. What is the best explanation for the relationship between these numbers?

Question 19 options:

only the first 900 nucleotides of the pre-mRNA transcript are translated

it takes 100 nucleotides to specify a single amino acid

300 of the nucleotides in the transcript are important, and the rest are "junk"

only the last 900 nucleotides of the pre-mRNA transcript are translated

large portions of pre-mRNA transcripts are cut out during RNA processing

Suppose an individual is born into a population with a novel mutation. Is the new mutation an evolutionary change, and why?

Question 20 options:

no, because it is not a big enough change to count

yes, because new mutations are always adaptive

yes, because the appearance of a new genetic variant is a genetic change in a population

no, because not enough individuals have the mutation for it to matter

no, because most mutations are not adaptive

If one is testing a goodness of fit to a 9:3:3:1 ratio, how many degrees of freedom would be associated with the Chi-square?

1

2

3

4

not enough information to determine

Which of the following choices does not describe a type of posttranslational modification of proteins?

The amino terminal methionine may be removed.
Glycoproteins are produced by the attachment of phosphates to newly synthesized proteins.
Precursor proteins are cleaved and trimmed by protease enzymes to produce a functional protein.
Molecular chaperones help proteins fold correctly.
Amino acids in the amino terminal of the protein can be acetylated.

A eukaryotic DNA sequence that affects transcription at distant promoters is called a(n)

long distance promotor
regulator
operon
enhancer

DNA methylation may be a significant mode of genetic regulation in eukaryotes. Meythylation refers to

altering RNA polymerase activity by adding methyl-groups
changes in DNA-DNA hydrogen bonding
altering translation activity via methylated tRNAs
adding methyl groups to certain cytosines in DNA

If a nucleotide lacks a 3’-OH group, what will happen?

it’s transformed into RNA
another nucleotide can bond at the 3’ site during replication
replication will halt
no effect

Which of the following describe posttranscriptional modifications commonly seen in mRNA of eukaryotes?

5’ capping, 3’-poly-A tail addition, splicing
3’ capping, 5’ poly-A tail addition, splicing
heteroduplex formation
5’ capping, 3’polyA tail addition, removal of exons

In birds, females are the heterogametic sex. This means that

females have chromosomes in pairs.
emales produce eggs.
females determine the gender of their offspring.

males have two identical sex chromosomes (ZZ)

An anticodon on a tRNA has the sequence 5'–GCA–3'. What would be the effect on translation if the G in the anticodon were mutated to a U?

The anticodon could pair to the serine codon 5'–UCA–3', and a cysteine would be incorporated at all points where a serine should have been incorporated.
The anticodon could pair to the threonine codon 5'–ACU–3', and a threonine would be incorporated at all points where a cysteine should have been incorporated.
The anticodon could pair to the serine codon 5'–AGU–3', and a serine would be incorporated at all points where a cysteine should have been incorporated.
The anticodon could pair to the stop codon 5'–UGA–3', and instead of stopping translation, a cysteine would be incorporated and translation could continue.