BIOSC 0160 Lecture Notes - Lecture 6: Silent Mutation, Methionine, Protein Subunit

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25 Oct 2017

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Lecture 6: deciphering the genetic code: artificial mrna to direct protein synthesis, read the mrna to 5" 3", the complement would be found on the trna which is the anticodon. Reading frame: silent mutation: does not change our resulting protein. If ucu is mutated to uca it will be the same protein: point mutation: if mutation is in the third base, it generally is not an issue. Genetic code properties: redundant all amino acids are coded by more than one codon, the redundancy gives us that flexibility in terms of point mutation, unambiguous one codon never codes for more than one amino acid. Aug only codes for methionine and nothing else: conservative the first two bases are often identical when specifying the same amino acids. A mutation in the third base isn"t detrimental to the protein. Transcription: happens in the nucleus in eukaryotes, happens in prokaryotes in cytoplasm, happens through rna polymerase synthesizes dna into mrna.

Related Questions

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

QUESTION 1. Choose the true statement about prokaryotic and eukaryotic DNA replication.

A.Eukaryotic DNA replication involves less protein factors than does prokaryotic DNA replication.

B. Prokaryotic DNA polymerases replicate DNA 5' to 3', while eukaryotic DNA polymerases replicate DNA 3' to 5'.

C. Eukaryotic DNA replication begins at multiple origins of replication, while prokaryotic DNA replication begins at a single origin of replication.

D. Prokaryotic genomes take longer to replicate than eukaryotic genomes.

QUESTION 2. Bacterial cells use DNA replication to:

A. make back-up copies of their genome in case of a mutation.

B. manifest multiple phenotypes at once.

C. copy their genetic material prior to binary fission.

D. build functional proteins.

QUESTION 3. Choose the true statement about the genetic code.

A. A cell's mRNA code is read four nucleotides at a time to specify a single amino acid.

B. The genetic code encodes 64 amino acids, one for each of the 64 corresponding codons.

C. The genetic code is redundant, with multiple codons specifying a single amino acid.

D. The genetic code varies widely among organisms, with each having a different set of codons for a corresponding amino acid.

QUESTION 4. Of the following enzymes, which are correctly matched with their function? To be marked correct, you'll need to select all correctly matched statements, as there may be more than one correct answer.

A. Helicase, relieves torsion stress as DNA unwinds

B. Ligase, covalently bonds adjacent Okazaki fragments

C. Primase, removes RNA primers

D. DNA polymerase III, builds most of the DNA on the leading and lagging strand

QUESTION 5. Which of the following contributes to DNA replication accuracy? To be marked correct, you'll need to select all true statements, as there may be more than one correct answer.

A. The capacity to proofread improves DNA replication accuracy.

B. A slow rate of DNA replication contributes to accuracy.

C. The use of primase improves DNA replication accuracy.

D. DNA replication accuracy is improved thanks to complementary base pairing rules.

Question 9

Codominance is a form of inheritance in which two different allelesfor a gene are both expressed, and neither allele is dominant overthe other. One example of codominance is fur color in cattle; thealleles for red fur and white fur are codominant. If a homozygousred cow and a homozygous white bull mate, what would you expect tofind in their offspring?.912.l.16.2>

They will all have white fur.

Their fur will be a mixture of red and white hairs.

Their fur will be a lighter shade of red.

They will all have red fur.

Question 10

If a tRNA molecule has an anticodon which reads AUG, what will itmatch up with and what amino acid is it carrying?
.912.l.16.5>

	U		C		A		G
	UUU	Phenylalanine	UCU	Serine	UAU	Tyrosine	UGU	Cysteine	U
U	UUC	Phenylalanine	UCC	Serine	UAC	Tyrosine	UGC	Cysteine	C
	UUA	Leucine	UCA	Serine	UAA	Stop	UGA	Stop	A
	UUG	Leucine	UCG	Serine	UAG	Stop	UGG	Tryptophan	G
	CUU	Leucine	CCU	Proline	CAU	Hisitidine	CGU	Arginine	U
C	CUC	Leucine	CCC	Proline	CAC	Hisitidine	CGC	Arginine	C
	CUA	Leucine	CCA	Proline	CAA	Glutamine	CGA	Arginine	A
	CUG	Leucine	CCG	Proline	CAG	Glutamine	CGG	Arginine	G
	AUU	Isoleucine	ACU	Threonine	AAU	Asparagine	AGU	Serine	U
A	AUC	Isoleucine	ACC	Threonine	AAC	Asparagine	AGC	Serine	C
	AUA	Isoleucine	ACA	Threonine	AAA	Lysine	AGA	Arginine	A
	AUG	Methionine/Start	ACG	Threonine	AAG	Lysine	AGG	Arginine	G
	GUU	Valine	GCU	Alanine	GAU	Aspartate	GGU	Glycine	U
G	GUC	Valine	GCC	Alanine	GAC	Aspartate	GGC	Glycine	C
	GUA	Valine	GCA	Alanine	GAA	Glutamate	GGA	Glycine	A
	GUG	Valine	GCG	Alanine	GAG	Glutamate	GGG	Glycine	G

It is carrying arginine and will match with a CGU codon on themRNA.

It is carrying methionine and will match with a TAC codon on theDNA.

It is carrying methionine and will match with a UAC codon on themRNA.

It is carrying tyrosine and will match with a UAC codon on themRNA.

Question 11

While the DNA in a human skin cell was being replicated, a singlebase was miscopied. What will be the most likely result of this forthe cell in which it happened?.912.l.16.3>

All the proteins the cell creates from the miscopied strand willdo different jobs than the old ones.

If the new sequence codes for the same amino acid as theoriginal cell, it will function normally.

Both new DNA strands will end up together in a new cell, and theinaccurate one will be discarded.

Any miscopied DNA will be replaced with an accurate DNA copyonce the cell divides.

Question 12

A ferret's haploid number of chromosomes is 20. How would thenumber of chromosomes in the ferret's body cells compare to thenumber of chromosomes in its gametes?.912.l.16.17>

Its body cells would have 20 chromosomes, and its gametes wouldhave 40 chromosomes.

Its body cells and gametes would both have 40 chromosomes.

Its body cells and gametes would both have 20 chromosomes.

Its body cells would have 40 chromosomes, and its gametes wouldhave 20 chromosomes.

Question 13

A body cell is in the longest stage of its life cycle. The cellgrows, synthesizing proteins and increasing in size. Eventually,the cell will grow too large to carry out normal activities. Whichof the following is the best conclusion you can make about the lifecycle of this cell?.912.l.16.14>

The cell is close to the end of its life cycle, and a chemicalsignal will initiate cell death.

The cell is ready to undergo mitosis, and a chemical signal willsend the cell to prophase.

The cell is in the G1 phase of interphase, and a chemical signalwill send the cell to the S phase.

The cell is in the S phase of interphase, and a chemical signalwill move the cell to the G2 phase.

Question 14

While mRNA strands are being created a sequence is sometimesmiscopied. What is the best possible outcome for the cell shouldthis take place?.912.l.16.5>

The miscopied sequence codes for the same amino acids as theoriginal sequence.

The new sequence creates a protein that serves a differentfunction from the original.

The ribosomes will correct the mistake before the tRNA matchesan amino acid to it.

The mRNA will only be used to create non-critical proteins forthe cell.

Question 15

Over the last several decades, the scientific community hasgathered a large amount of information regarding genetics andgenetic variation. What are two main sources that lead toincreased genetic variation?.912.l.15.15>

Selective breeding

Gamete mutations

Recombination

Genetic drift

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BIOSC 0160 Lecture Notes - Lecture 6: Silent Mutation, Methionine, Protein Subunit

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