Genetics: Understanding the Base Sequence of Messenger RNA

[Soaring Crane]

Homework Statement

[/B]
I am reviewing an example on the basics of the genetic code; this example is listed at the bottom of the following webpage: https://www.atdbio.com/content/14/Transcription-Translation-and-Replication.I have produced the example below and added Roman numbers to better indicate the parts that I am questioning (V and VII).

One strand of genomic DNA (strand A, coding strand) contains the following sequence reading from 5′- to 3′-: I. TCGTCGACGATGATCATCGGCTACTCGAThis strand will form the following duplex: II. 5′-TCGTCGACGATGATCATCGGCTACTCGA-3'

III. 3′-AGCAGCTGCTACTAGTAGCCGATGAGCT-5'The sequence of bases in the other strand of DNA (strand B) written 5′- to 3′- is therefore IV. TCGAGTAGCCGATGATCATCGTCGACGAThe sequence of bases in the mRNA transcribed from strand A of DNA written 5′- to 3′- is V. UCGAGUAGCCGAUGAUCAUCGUCGACGAThe amino acid sequence coded by the above mRNA is VI. Ser-Ser-Ser-Arg-STOPHowever, if DNA strand B is the coding strand the mRNA sequence will be:VII. UCGUCGACGAUGAUCAUCGGCUACUCGAand the amino-acid sequence will be:VIII. Ser-Ser-Thr-Arg-Ser-Ser-Gly-Cys-Ser-

Homework Equations

Not applicable

(Well, I suppose Chargaff's rules apply for DNA.)

The Attempt at a Solution

[/B]

I understand that Strand A (5' -> 3') is this:5′-TCGTCGACGATGATCATCGGCTACTCGA-3'.However, if the above is (or designated) the coding strand, then why is the mRNA sequence (following transcription) not this (see below)?5'-UCGUCGACGAUGAUCAUCGGCUACUCGA-3' (Thought One)I thought that the defining trait of the DNA coding strand is that its base sequence is almost the same as the resulting mRNA sequence, with the exception that uracil (U) substitutes for thymine (T).I have the same question for Strand B (see IV):

5'-TCGAGTAGCCGATGATCATCGTCGACGA-3'.Why is the mRNA sequence of Strand B, if it is coding, not:

5'-UCGAGUAGCCGAUGAUCAUCGUCGACGA-3' (Thought 2)?(Although Strand B is the template strand in III, it clearly states later to assume that it is coding.)
As you may notice, my Thoughts 1 and 2 (for Strands A and B, respectively) are the switched answers for V and VII. That is, what I thought was the correct mRNA sequence corresponding to Strand A (see Thought 1) is actually the listed sequence that corresponds to Strand B (see VII).I really appreciate your help on where I went wrong.Thanks again.

 

[Ygggdrasil]

There are two ways of naming the two strands of a gene that is being transcribed. The first is from the perspective of the RNA polymerase enzyme that is synthesizing the RNA. RNA polymerase moves from the 3' to 5' of the template strand, producing an mRNA that is complementary to the template. The other strand of DNA that the polymerase does not read is called the non-template strand. Because the non-template strand is also complementary to the template strand, the mRNA ends up having the same sequence as the non-template strand (except that T is replaced with U). Therefore, we often refer to the non-template strand as the coding strand (or the sense strand). The template strand is therefore the non-coding strand or anti-sense strand.

So, in your example, if V is the mRNA, then I is the coding strand/non-template strand and III is the template strand/non-coding strand. sequence II is the template strand/non-coding strand and sequence III is the coding strand/non-template strand.

 

[Soaring Crane]

Dear Ygggdrasil,

If you look at the coding, or non-template, strand 5’-TCGTCGACGATGATCATCGGCTACTCGA-3’, how is this the same sequence as the mRNA strand in V (with exception of substituting T with U), 5’-UCGAGUAGCCGAUGAUCAUCGUCGACGA-3’?Thank you for any clarification.

 

[Ygggdrasil]

Given the template strand is 3′-AGCAGCTGCTACTAGTAGCCGATGAGCT-5'

If RNA polymerase uses the template strand as a template to synthesize mRNA, the mRNA will be the reverse complement of the template strand.

For the coding DNA strand to pair with the template DNA strand, the coding stand must also be the reverse complement of the template strand.

 

[Soaring Crane]

If I use 3′-AGCAGCTGCTACTAGTAGCCGATGAGCT-5' as the template strand and rules T = A and A = U,

then would not the mRNA strand be:
--------5'-UCGUCGACGAUGAUCAUCGGCUACUCGA-3'? [1]

From V. in the example, the above [1] does not match:
--------5’-UCGAGUAGCCGAUGAUCAUCGUCGACGA-3’.

I apologize for all these questions. Please let me know what I am doing incorrectly.

Thanks again.

 

[Ygggdrasil]

I previously did not look closely at the exact sequences, so I have corrected my post #2.

Based on a template strand of 3′-AGCAGCTGCTACTAGTAGCCGATGAGCT-5' (sequence III), sequence [1] (5'-UCGUCGACGAUGAUCAUCGGCUACUCGA-3') is indeed the correct sequence of the mRNA.

Sequence V (UCGAGUAGCCGAUGAUCAUCGUCGACGA) is the mRNA one would get when sequence II serves as the template strand.

 

[Soaring Crane]

Therefore, as indicated from my first post, mRNA sequence VII,
5'-UCGUCGACGAUGAUCAUCGGCUACUCGA-3',
results when sequence III
3′-AGCAGCTGCTACTAGTAGCCGATGAGCT-5'
is the template strand and sequence II (Strand A) is the coding, or non-template, strand

(5′-TCGTCGACGATGATCATCGGCTACTCGA-3').

Is this correct?

Thank you for your help.

 

[Ygggdrasil]

Therefore, as indicated from my first post, mRNA sequence VII,
5'-UCGUCGACGAUGAUCAUCGGCUACUCGA-3',
results when sequence III
3′-AGCAGCTGCTACTAGTAGCCGATGAGCT-5'
is the template strand and sequence II (Strand A) is the coding, or non-template, strand

(5′-TCGTCGACGATGATCATCGGCTACTCGA-3').

Is this correct?

Thank you for your help.

Yes that is correct. As you indicated, the quoted information in part 1 of post #1 seems to confuse the terms for coding strand and template strand.