BIOL1007 Lecture Notes - Lecture 7: Ultracentrifuge, Aldehyde, Leucine

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17 Aug 2018

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Then need to somehow convert a sequence of nucleotides (w 4 diff bases ) to a sequence of amino acids ( where there are 20) The need to have correct order of amino acds. The order of ac determines the fold and function of the protein. You need a precise start and stop for proteins. Extra ac may stuff uo the fold andalter function. The need to somehow convert a sequence of nucleotides (w 4 diff bases) to a sequence of amino acids (where there are 20 ) The combination of 3 bases whoch codes for an ac is called a codon. We refer to the codon as a triplet codon. The codon sequence is quoted 5" to 3" from the m rna. The codon will have the same sequence as the dna strand that wasn"t copied (non- template strand) except it will contain uracils instead of thymines.

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SEQUANCE:

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

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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?

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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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a) You also wish to synthesize a degenerate oligonucleotide primer that could be used eventually to clone DNA that encodes your mutant protein. Design this degenerate oligo, keeping in mind the rules we talked about in class. More information about designing degenerate primers and primers in general are located on the following page.

Write out the sequence of this oligo here

b) If you synthesized the above oligonucleotide primer, how many different oligos would be present in the mixture?

RULES TO FOLLOW for creating primers (Modified from the Thermo Fisher web site):

Designing degenerate primers:

Write out your amino acid sequence, label amino and carboxy termini

For each amino acid, use the genetic code to predict the various nucleic acid codons that can encode this amino acid. For all but Met, you should have redundancy present.

Now you need think about 5â and 3â considerations, and keep in mind that DNA within genes is double-stranded. It is helpful to keep in mind that the 5â end of a gene corresponds to the amino terminus of a protein. Using 5â and 3â labels write out a single strand DNA sequence corresponding to the your amino seq of interest. You may choose to start with only possible codon for each amino acid, or you can do all possibilities using the notation I showed you in class to account for the wobble position of codons.

Now make your DNA double stranded by filling in the opposite DNA strand; label the 5â and 3â ends of this second strand and make sure you have an anti-parallel arrangement of DNA strands.

If you havenât already, now consider the redundancy present on both strands, and make sure you have indicated sequences that account for all possible redundancies.

Realize that when a researched synthesizes a degenerate oligo that the final synthesis contains oligos with every substitution possible. The amount of degeneracy is defined by the number of different primer combinations in the mix. You can determine the degree of degeneracy by multiplying the number of changes present at each position together. If a position has no degeneracy then you multiple by 1 for that position.

Keep in mind that the trade-off between primer specificity and efficiency can be modified by altering the degeneracy of the primer. For example, the more degenerate the primers, the less specific annealing will be; however, decreased degeneracy will allow more potential to identify unknown variants. Anything over 1024 fold degeneracy is at the upper limit of potential usefulness.

Designing Primers:

Good primer design is essential for a successful PCR reaction. There are many factors to take into account when designing the optimal primers for your gene of interest. This information is taken from the Fisher Scientific website and are used when actually designing an experiment. Here are some tips to consider when designing primers.

In general, a length of 18 nucleotides is the minimum necessary for efficient primer binding.

Try to make the melting temperature (Tmm) of the primers between 65Â°C and 75Â°C, and within 5Â°C of each other.

If the Tmm of your primer is very low, try to find a sequence with more GC content, or extend the length of the primer a little.

Aim for the GC content to be between 40 and 60%, with the 3' of a primer ending in C or G to promote binding. This is called a GC clamp.

Typically, 3 to 4 nucleotides are added 5â of the restriction enzyme site in the primer to allow for efficient cutting. Restriction enzymes are ENDO nucleases, so they cut better when their recognition site is not on the end of DNA.

Try to avoid regions of secondary structure, and have a balanced distribution of GC-rich and AT-rich domains.

Avoid intra-primer homology (more than 3 bases that complement within the primer) or inter-primer homology (forward and reverse primers having complementary sequences). These circumstances can lead to self-dimers or primer-dimers instead of annealing to the desired DNA sequences.

BIOL1007 Lecture Notes - Lecture 7: Ultracentrifuge, Aldehyde, Leucine

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