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Default The base sequence of codons 85-86 of troponin I gene is...

The base sequence of codons 85-86 of troponin I gene is CGCGAC. The mRNA produced upon transcription of this gene will contain the sequence:

A) CGCGAC
B) GCGCTG
C) GCGCUG
D) GUCGCG
E) GTCGCG

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D
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3' GCGCTG 5' - Coding Strand
5' CGCGAC 3' - Template Strand
3' GCGCUG 5' - mRNA


answer has to be provided in 5'-3' direction -> D) GUCGCG
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Can it be A? Cause mRNA corresponds to the codon sequence, and U replaces T?
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Quote:
Originally Posted by beka-CTS View Post
3' GCGCTG 5' - Coding Strand
5' CGCGAC 3' - Template Strand
3' GCGCUG 5' - mRNA


answer has to be provided in 5'-3' direction -> D) GUCGCG
Man these questions are my achilles heal...would somebody be kind enough to go through explaining this thing through and through....i didn't like kaplans explanations
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Quote:
Originally Posted by DefeatTheBeast View Post
Man these questions are my achilles heal...would somebody be kind enough to go through explaining this thing through and through....i didn't like kaplans explanations
actually it's not a question to be afraid of.

the main things i remember are:

1. every base sequence given on the test or in a literature is always written in 5' -> 3' direction

2. RNA polymerase II reads the template strand only in 3' -> 5' direction i.e. the new molecule of mRNA is always synthesized in 5' -> 3' direction.

in our example: given gene is the always a TEMPLATE strand, unless question stem says that it coding. So we have the TEMPLATE strand 5' CGCGAC 3', for translation the RNA plymerase II will attach to 3' site and will begin creating mRNA in 5'->3' direction. so the firt nucelotide corresponds to C -> G , the second corresponds to A -> U ( hense not G -> C ! ) and at the end we'll get mRNA containing the base sequence: 5' GUCGCG 3'. remember in the answer choices we have to choose the sequence written in 5' -> 3' direction.

3' GCGCTG 5' - Coding Strand
5' CGCGAC 3' - Template Strand <- given in the Question
3' GCGCUG 5' - mRNA <- answer should be written in oposite direction i.e. GUCGCG
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Quote:
Originally Posted by DefeatTheBeast View Post
Man these questions are my achilles heal...would somebody be kind enough to go through explaining this thing through and through....i didn't like kaplans explanations
Just think of this question like this -

They tell you that "troponin I gene is CGCGAC." And they're asking for mRNA strand. So now, follow these steps -

1) They didn't tell you the 3' & 5" directions. So you ASSUME its 5'-CGCGAC-3' because that's the direction you read it in.
2) mRNA is synthesized from 5' to 3' so the 5' end of mRNA would match with 3' end on the DNA. This gives you your starting point -
READ...........5'-CGCGAC-3'
Transcribe: 3'- ------ -5'

4) Since you are transcribing mRNA, just substitute U for all the T's. So this is what the final product/answer should look like:
READ...........5'-CGCGAC-3'
Transcribe: 3'-GCGCUG-5'
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Quote:
Originally Posted by beka-CTS View Post
actually it's not a question to be afraid of.

the main things i remember are:

1. every base sequence given on the test or in a literature is always written in 5' -> 3' direction

2. RNA polymerase II reads the template strand only in 3' -> 5' direction i.e. the new molecule of mRNA is always synthesized in 5' -> 3' direction.

in our example: given gene is the always a TEMPLATE strand, unless question stem says that it coding. So we have the TEMPLATE strand 5' CGCGAC 3', for translation the RNA plymerase II will attach to 3' site and will begin creating mRNA in 5'->3' direction. so the firt nucelotide corresponds to C -> G , the second corresponds to A -> U ( hense not G -> C ! ) and at the end we'll get mRNA containing the base sequence: 5' GUCGCG 3'. remember in the answer choices we have to choose the sequence written in 5' -> 3' direction.

3' GCGCTG 5' - Coding Strand
5' CGCGAC 3' - Template Strand <- given in the Question
3' GCGCUG 5' - mRNA <- answer should be written in oposite direction i.e. GUCGCG
Got it...thanks man, cleared up doubts i had...appreciate it!
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Originally Posted by DefeatTheBeast View Post
Got it...thanks man, cleared up doubts i had...appreciate it!
you're welcome i wish i could understand every concept like i get this one
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Default correct answer

Quote:
Originally Posted by beka-CTS View Post
actually it's not a question to be afraid of.

the main things i remember are:

1. every base sequence given on the test or in a literature is always written in 5' -> 3' direction

2. RNA polymerase II reads the template strand only in 3' -> 5' direction i.e. the new molecule of mRNA is always synthesized in 5' -> 3' direction.

in our example: given gene is the always a TEMPLATE strand, unless question stem says that it coding. So we have the TEMPLATE strand 5' CGCGAC 3', for translation the RNA plymerase II will attach to 3' site and will begin creating mRNA in 5'->3' direction. so the firt nucelotide corresponds to C -> G , the second corresponds to A -> U ( hense not G -> C ! ) and at the end we'll get mRNA containing the base sequence: 5' GUCGCG 3'. remember in the answer choices we have to choose the sequence written in 5' -> 3' direction.

3' GCGCTG 5' - Coding Strand
5' CGCGAC 3' - Template Strand <- given in the Question
3' GCGCUG 5' - mRNA <- answer should be written in oposite direction i.e. GUCGCG
The logic and explanations are correct EXCEPT for the crucial first step SO WATCH OUT!!!!

In the stem they say it's a "sequence of codons" - so IT'S A CODING strand, NOT a template. That was actually the point of me posting this question.

So we have 5' CGCGAC 3' CODON STRAND which is basically the same as mRNA (except for T that will change into U).

Why?

1) if not stated otherwise, we assume that sequence they present to us is 5' to 3' - just like mentioned in previous posts
so codon DNA is 5' CGCGAC 3', hence template DNA (from which mRNA will be transcribed) is 3'GCGCTC 5'.
2) so we take this template 3'GCGCTC 5 and we trascribe it into
mRNA which will be: 5' CGCGAC 3' (so basically the same as codon sequence).
As stated in 1) we present sequences from 5' to 3' end so mRNA sequence is: 5' CGCGAC 3'

A) CGCGAC is the correct answer
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Can it be A? Cause mRNA corresponds to the codon sequence, and U replaces T?
exactly!
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Quote:
Originally Posted by Casandra View Post
"sequence of codons" - so IT'S A CODING strand, NOT a template.


does the words sequence of codons always mean the Coding strand?
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Originally Posted by beka-CTS View Post


does the words sequence of codons always mean the Coding strand?
I have done 3 questions where they mentioned that term and it always meant coding strand... Kaplan, old Pretest and from some other online resource... can't remember the name now...

btw this question (a bit modified I admit is from Kaplan
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does the words sequence of codons always mean the Coding strand?
the other explanation to this question could be: codons are on mRNA (anti-codon on tRNA) so in the stem they actually gave us the answer but they are trying to confuse us BIG TIME!
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Finally, I get something right!! Woohoo!

Quote:
Originally Posted by Casandra View Post
exactly!
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CORRECT ANSWER IS A BECAUSE
CODING STRAND IS IDENTICAL TO mRNA JUST T---TO U,TEMPLATE STRAND IS COMPLEMENTARY,ANTIPARALEL TO mRNA
FOR EXAMPLE
CODING STRAND 5'-ATGGGGCTCAGCGAC-3'
TEMPLATE STRAND 3'-TACCCCGAGTCGCTG-5'
mRNA 5'-AUGGGGCUCAGCGAC-3'

BUT TO BE A PALINDROME BOTH STRAND MUST HAVE SAME SEQUENCE WHEN READ IN 5'---3' DIRECTION
FOR EXAMPLE
AAGGAA
AAGAAG
AAGTTC
AAGCTT
AAGAGA
PALINDROME IS AAGCTT BECAUSE
5'-AAGCTT-3'
3'-TTCGAA-5'
IF READ 5'--3' WILL BE
AAGCTT
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