BMB 401 Lecture Notes - Lecture 5: Hydrolysis, Nucleophilic Substitution, Transcription Preinitiation Complex

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aquamarinemink463

16 Dec 2016

School

University of Miami

Department

Biochemistry and Molecular Biology

Course

BMB 401

Professor

Farooq

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Document Summary

4. 5a mrna has a code made up of 3 nucleotide bases called codons. Since there are 4 bases (acgu) that can be arranged in 3 different ways we get 4^3 = 64 possible variants. Of course there are only 20 aa so this shows how we have several codons for the same aa. These 64 codons make up the genetic code! Methionine always serves as the start codon aug! Aug is the only codon that codes for it whereas most of the others have several. The other 3 are nonsense 9stop) codons that do not specify an aa but the end of the polypeptide chain. In 2d form trna takes a cloverleaf shape. It is made up of 3 stem loops and a solo stem. The solo stem binds the corresponding aa through an ester linkage. One of the 3 stem lops has a codon complimentary to one of the mrna codons so we call it the anticodon.

Related Questions

Can someone please check and help me with these? Especially II and III?

I. Below is a DNA sequence that is a template strand and contains an Open reading frame for a gene:

3â TAA ATA CGT GAG GCA CTT CAA AGA CCG CAC ACT_ AAT TT 5â

5â AUU UAU GCA CUC CGU GAA GUU UCU GGC GUG UGA _ UUA AA 3â

a. Transcribe the DNA template into a strand of mRNA by typing the mRNA sequence below the template strand. Label the 5â and 3â end of the mRNA 5â AUU UAU GCA CUC CGU GAA GUU UCU GGC GUG UGA _ UUA AA 3â

b. Paste the mRNA below and do all 3 reading frames. Highlight all start codons green in all the reading frames. Highlight red the stop codons which are in frame with the start codons in all the 3 reading frames if they are present.

1) 5â AUU UAU GCA CUC CGU GAA GUU UCU GGC GUG UGA UUA AA 3â

2) 5â A UUU AUG CAC UCC GUG AAG UUU CUG GCG UGU GAU UAA A 3â

3) 5â AU UUA UGC ACU CCG UGA AGU UUC UGG CGU GUG AUU AAA 3â

c) In which reading frame is the ORF found? The second one. Since it has AUG.

d) Paste the mRNA showing the most likely ORF below and translate the codons. Label the 5â and 3â end of the mRNA and the N- and C-termini of the peptide.

5â A UUU AUG CAC UCC GUG AAG UUU CUG GCG UGU GAU UAA A 3â

MET HIS SER VAL LYS PHE LEU ARG CYS ASP STOP

N Terminus C Terminus

II. 3 pts. The base calling program made a mistake that you didnât catch and deleted the bolded G in the original template strand above. Paste the template DNA below and delete the G. Transcribe (show the codons) and translate the resulting sequence. Label the 5â and 3â ends of the DNA and mRNA and the N and C-termini of the peptide. Comment on the resulting peptide sequence.

5â AUUU AUG CAC UCC GUG AAG UUU CUG GCG UGU GAU UAA A â3

MET HIS SER VAL LYS PHE LEU ARG CYS ASP STOP

III. 3 pts. The base calling program made a mistake that you didnât catch and added an extra G after the bolded G in the original template strand above. Paste the template DNA below and insert the extra G. Transcribe (show the codons) and translate the resulting sequence. Label the 5â and 3â ends of the DNA and mRNA and the N and C-termini of the peptide. Comment on the resulting peptide sequence.

5â AUUU AUG CAC UCC GUG AAG UUU CUG GGG UGU GAU UAA A â3

MET HIS SER VAL LYS PHE LEU GLY CYS ASP STOP

thank you

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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BMB 401 Lecture Notes - Lecture 5: Hydrolysis, Nucleophilic Substitution, Transcription Preinitiation Complex

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