BIOL 1011 Lecture Notes - Lecture 4: Start Codon, Cellular Respiration, Intron

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31 Aug 2016

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Regulatory coding- when dna reads certain code it knows to start making code. Introns- pieces of code that do not do anything, accidents of evolution, genetic baggage, (bacterial genes have no introns) Transcription- use a dna template to make a complementary rna. Dna template is 3" to 5" strand. Processing- removing of any-non coding(introns) turning strand into messenger. Translation- ribosome reads mrna from 5" end to 3" end trnas deliver the appropriate amino acids to the ribosome. Dna has a 4 letter alphabet- a, t, c, and g. At a gene that codes for a protien, you create 3 letter words. Example: ttt is the word for the amino acid lysine. Tac is the word for the amino acid methionine or start. In mrna, three letter words are codons- 64 possible codons. 10 to 15 minutes for mrna to transcribed. Each trna carries just one type of amino acid. Each trna has a 3 base anticodon sequence.

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5. Alpha amanitin is a drug that binds RNA polymerase andprevents it from functioning in the cell (this is the compound thatis found in poison mushrooms). The reason people die when theyconsume alpha amanitin is:

a) Alpha amanitin prevents gene expression from occurring in thecell

b) DNA replication is directly inhibited by alpha amantin

c) Alpha amanitin binds tightly to the messenger RNAs that aremade in the cell, preventing translation

d) Alpha amanitin binds to nuclear pores preventing thetransport of messenger RNA into the cytoplasm

6. In order for transcriptional initiation to begin in E. colithe following elements are needed:

a) an origin and RNA polymerase

b) initiator protein and a promoter

c) the sigma subunit of RNA polymerase and an RNA template

d) RNA polymerase and a promoter

e) helicase and primase

Questions 7 and 8. You are studying two species of closelyrelated snakes, one species lives in sandy desert area and has paletan skin and the species lives in a volcanic rock area and has darkbrown skin. You a sequence the DNA from each type of snake anddiscover a gene encoding a protein that transports pigment into theskin cells. The pigment transporter protein is responsible for skincolor. When you analyze the DNA coding sequence of the pigmenttransporter from each snake you find these two DNA sequences (hashmarks have been added to indicate the reading frame):

pale tan snake: GCG/TTG/GAC/TGC/ACC (DNA template strand)CGC/AAC/CTG/ACG/TGG (complementary strand) dark brown snake:ACG/TTG/GAC/TGC/ACC (DNA template strand) TGC/AAC/CTG/ACG/TGG(complementary strand)

7. From these sequences you conclude:

a) that the amino acid sequence of the pigment transporters forthe two proteins are almost identical

b) that the amino acid sequence of the pigment transporters forthe two proteins are very dissimilar

c) that the amino acid sequence of the pigment transporters forthe two proteins are identical

8. Since the skin color is different in each snake, you concludefrom the sequence data above that:

a) the genetic code in the skin cells in each of these snakes isdifferent

b) the DNA sequence encoding the entire gene for pigmenttransporters in these two snakes is different

c) that one of the codons in the gene for the pigmenttransporter in these two snakes is different

9. You are analyzing eukaryotic cells that have been exposed tothe radioactive uracil. In most cells the radioactive uracil isinitially localized in the nucleus and then moves from the nucleusto the cytoplasm. However, you observe that in some cells theradioactive molecules stay in the nucleus and never arrive in thecytoplasm. This is most likely a defect in:

a) the transcription of the DNA

b) the processing of the mRNA

c) the translation of the RNA

d) the labeling of the RNA

e) the replication of the DNA

10. Which of the following is true of both eukaryotic andbacterial transcription? (select all that are true)
a) Transcription factors must bind the promoter before RNApolymerase initiates transcription

b) Messenger RNAs must be spliced before translation begins inorder to produce the appropriate gene product

c) During transcription only one DNA strand is used as atemplate to make the messenger RNA

d) Termination of transcription occurs when the RNA polymerasetranscribes a hairpin structure in the RNA followed by a string ofU

erinporpoise384

SEQUANCE:

5'- GCU UGG GCU GCU GAU GCU CUU UAA CAU AGA GCU AAU AUU -3'

=======================================================================================

3- Convert your mRNA sequence into a DNA sequence, then make itdouble-stranded by writing the complementary sequence below.

4- Transcribe (write an RNA sequence) and translate (write anamino acid sequence) both the "top" and "bottom" strands startingwith the first nucleotide. You must use the genetic code in yourtextbook or in your lecture notes for this step. DO NOT use thetable above. If you encounter a stop codon, your protein sequenceis done at that point.

Use the single letter designation for amino acids as givenbelow.

Alanine (A)	Aspartic acid (D)	Asparagine (N)	Arginine (R)
Cysteine (C)	Glycine (G)	Glutamic acid (E)	Glutamine (Q)
Histidine (H)	Isoleucine (I)	Leucine (L)	Lysine (K)
Methionine (M)	Proline (P)	Phenylalanine (F)	Serine (S)
Threonine (T)	Tryptophan (W)	Tyrosine (Y)	Valine (V)

5- Use your original DNA sequence from #3. Add an A to the 5'end of the "top" strand, then write the complementary sequence onthe bottom. Add a GC to the 5' end of the

"bottom" strand, then write the complementary sequence on the top.Transcribe and translate both the "top" and "bottom" with this newsequence.

What kind of mutation(s) was introduced in the "top" strand?

What kind of mutation(s) was introduced in the "bottom"strand?

6- Use your original DNA sequence from #3. Starting at the 5'end on the "top" strand of your original sequence from part 3,delete the second A, the third G, and the fourth C along with thecomplementary nucleotides on the bottom strand. Transcribe andtranslate both the "top" and "bottom" with this new sequence.

What kind of mutation(s) was introduced in the "top" strand?

What kind of mutation(s) was introduced in the "bottom"strand?

7. Use your original DNA sequence from #3. Starting at the 5'end of the "top" strand, change the third G to an A, then changethe complementary nucleotide in the bottom. Starting at the 5' endof the "bottom" strand, change the third C to a T, then change thecomplementary nucleotide on the top. Transcribe and translate boththe "top" and "bottom" with this new sequence.

a. What kind of mutation(s) was introduced in the "top"strand?

b. Whatkind of mutation(s) was introduced in the "bottom" strand?

purplehippopotamus358

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

BIOL 1011 Lecture Notes - Lecture 4: Start Codon, Cellular Respiration, Intron

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