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20 Dec 2018

Q/ Plants have a complex immune system, which includes small peptides that are produced and secreted by cells, and served as signals to trigger cell death. You are a plant immunologist studying one such peptide. Youâve purified this small protein and sequence it, giving you the following amino acid sequence:

(N-terminal end) Met â Gln â Lys - Phe â Met â Asn (C-terminal end)

DNA codons for these amino acids are:

5â- ATG CAA/G AAA/G TTT/C ATG AAT/C -3â

(Note that Gln, Lys, Phe and Asn could be encode by more than one codon, for example Gln by codons CAA or CAG)

Describe a genetic engineering strategy you that would take to express and produce large amounts of this small protein to study its function in plant cells.

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Deanna HettingerLv2

23 Dec 2018

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Below you have the sequence of 33 nucleotides(part of a coding region of gene TTr34) from two different species of birds (Species A and Species B). For simplicity, the amino acids encoded by the codons of Species A is also indicated, and â.â indicates the same nucleotide as the one observed in the first sequence.

	Pro	Gly	Asn	Thr	His	Glu	Ala	Gln	Lys	Ala	Val
Species A	CCT	GGG	AAT	ACT	CAT	GAA	GCA	CAA	AAA	GCC	GTC
Species B	. . G	. . T	. . C	. . .	. . .	. C .	. . .	. C .	. . T	A . .	. . T

Use these two sequences to estimate the molecular distances.

a. How many nucleotide changes produce a change in amino acid (nonsynonymous changes)(use the genetic code)?

b. How many nucleotide changes are synonymous (do not cause a change in amino acid)(use the genetic code)?

c. Calculate the molecular distance for synonymous (Ds) and Nonsynonymous (Da) for this gene region without correction. To estimate the total number of synonymous and nonsynonymous sites, assume that 25% of the sites are synonymous and 75% of sites are nonsynonymous.

* Please show all work and explain. *

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

I need help with Part B. I believe that part A is correct.

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 based on this amino acid sequence (Met Ala Pro Met Val Ala Glu):

My answer (Is it correct?)

I got:

5â AUG - (GCU/C/A/G) - (CCU/C/A/G) - (AUG) - (GUU/C/A/G) - (GCU/C/A/G) - (GAA/G) â3

3â UAC - (CGA/G/U/C) - (GGA/G/U/C) - (UAC) - (CAA/G/U/C) - (CGA/G/U/C) - (CUU/C) 5â

______________________________________________________________________________

(What I actually need help with and have no idea where to begin. Please help me)

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

Follow these rules 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.