Question

Below is a template strand of DNA that contains a translation start codon. 5’ – CTTTACGGGATGGCATTG – 3’ a) Assuming this entire sequence is transcribed, write the sequence of the resulting mRNA transcript including appropriate end labels. b) Write the sequence of the polypeptide resulting from this transcript, Second mRNA base U C A G UUU UCU UAU UGU U Phe Tyr Cys UUC UCC UAC UGC C U Ser UUA UCA UAA Stop UGA Stop A Leu UUG UCG UAG Stop UGG Trp G CUU CCU CAU CGU U CUC CCC CAC CGC C C Leu Pro His Arg CUA CCA CAA CGA A Gln CUG CCG CAG CGG G AUU ACU AAU AGU U Asn Ser AUC Ile ACC AAC AGC C A Thr AUA ACA AAA AGA A Lys Arg AUG Met or start ACG AAG AGG G GUU GCU GAU GGU U Asp GUC GCC GAC GGC C G Val Ala Gly GUA GCA GAA GGA A Glu GUG GCG GAG GGG G First mRNA base (5’ end of codon) Third mRNA base (3’ end of codon)

          Below is a template strand of DNA that contains a translation start codon.
5’ – CTTTACGGGATGGCATTG – 3’
a) Assuming this entire sequence is transcribed, write the sequence of the resulting mRNA transcript including appropriate end labels.
b) Write the sequence of the polypeptide resulting from this transcript,
Second mRNA base
U C A G
UUU UCU UAU UGU U
Phe Tyr Cys
UUC UCC UAC UGC C
U Ser
UUA UCA UAA Stop UGA Stop A
Leu
UUG UCG UAG Stop UGG Trp G
CUU CCU CAU CGU U
CUC CCC CAC CGC C
C Leu Pro His Arg
CUA CCA CAA CGA A
Gln
CUG CCG CAG CGG G
AUU ACU AAU AGU U
Asn Ser
AUC Ile ACC AAC AGC C
A Thr
AUA ACA AAA AGA A
Lys Arg
AUG Met or start ACG AAG AGG G
GUU GCU GAU GGU U
Asp
GUC GCC GAC GGC C
G Val Ala Gly
GUA GCA GAA GGA A
Glu
GUG GCG GAG GGG G
First mRNA base (5’ end of codon)
Third mRNA base (3’ end of codon)

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Below is a template strand of DNA that contains a translation start codon.
5’ – CTTTACGGGATGGCATTG – 3’
a) Assuming this entire sequence is transcribed, write the sequence of the resulting mRNA transcript including appropriate end labels.
b) Write the sequence of the polypeptide resulting from this transcript,
Second mRNA base
U C A G
UUU UCU UAU UGU U
Phe Tyr Cys
UUC UCC UAC UGC C
U Ser
UUA UCA UAA Stop UGA Stop A
Leu
UUG UCG UAG Stop UGG Trp G
CUU CCU CAU CGU U
CUC CCC CAC CGC C
C Leu Pro His Arg
CUA CCA CAA CGA A
Gln
CUG CCG CAG CGG G
AUU ACU AAU AGU U
Asn Ser
AUC Ile ACC AAC AGC C
A Thr
AUA ACA AAA AGA A
Lys Arg
AUG Met or start ACG AAG AGG G
GUU GCU GAU GGU U
Asp
GUC GCC GAC GGC C
G Val Ala Gly
GUA GCA GAA GGA A
Glu
GUG GCG GAG GGG G
First mRNA base (5’ end of codon)
Third mRNA base (3’ end of codon)

Added by Matthew O.

Biology for AP Courses

Julianne Zedalis, John Eggebrecht

Instant Answer

Solved by Expert Mj A

08/04/2022

Step 1

Step 1: Transcribe the given DNA template into mRNA sequence. Show more…

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Below is a template strand of DNA that contains a translation start codon. 5’ – CTTTACGGGATGGCATTG – 3’ a) Assuming this entire sequence is transcribed, write the sequence of the resulting mRNA transcript including appropriate end labels. b) Write the sequence of the polypeptide resulting from this transcript,

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Transcript

00:01 Hello, so let's answer now the question.

00:02 Now here in the question, we are asked what will be the mrna sequence and also the sequence of the amino acids on the nucleotides.

00:13 So here we are given this dna template.

00:18 So the given dna template is we have 5 prime, c -t -t, t -a -c, g -g, g, a, t, g, g, g, c, a, and then we have t, t, t, g, g, 3 prime.

00:53 Okay, so this is the given dna.

00:57 And in the first question, we are asked, what will be the mrna? so this is our answer.

01:07 Okay.

01:07 So for the mrna, we will start with the 3 prime.

01:19 The pair of c, the pair of c is g, the pair of t.

01:30 Let me use green.

01:33 The pair of c is g.

01:34 The pair of t is a.

01:37 But since i remember also that in rna there is no t.

01:42 Okay.

01:43 So let us put a.

01:45 The pair of t is a.

01:47 Now the pair of a is t, but since there's no t and mrna, it should be replaced by eurasil.

01:54 So we have here a u.

01:57 Okay.

01:58 So the pair of a is u.

02:02 And then next the pair of g is c so the pair of g is c so we have or the pair of c is g rather so we have the pair of g is c the pair of g is c the pair of a is u the pair of t is a, the pair of g is c, c, the pair of g is g is g, the pair of a is u, pair of t is a, and the pair of g is c, 5 prime.

03:04 But since we have here, the mrna that starts with 3 prime, we need to reverse it so that we will put it in a 5 prime to 3 prime order...

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