6 two different mutations are shown below along with the...

6. Two different mutations are shown below, along with the wild-type DNA sequence of this gene. Each is from the beginning of the coding region (exon) of the gene, and the top strand is the template strand. Wild type: 3'-CGGTACGGGATATCA-5' 5'-GCCATGCCCTATAGT-3' Mutant 1: 3'-CGGTACGGGATACCA-5' 5'-GCCATGCCCTATGGT-3' Mutant 2: 3'-CGGTACGGGGATATCA-5' 5'-GCCATGCCCCTATAGT-3' Use the genetic code to determine the effect on the amino acid sequence of the protein produced by the wild type and each mutant. Give the genetic code of the wild-type protein, along with the altered protein produced in each of the two mutants. Name the type of DNA mutation (point, indel) AND protein mutation (synonymous, missense, nonsense, frameshift) that has occurred in each mutant.

Transcript

00:02 Hello students, the concept given the question is about mutation and what is required for us is to discuss the type of mutation with respect to the given template sequence of dna.

00:25 Okay, so let us get into the explanation.

00:30 As you all know that the template dna runs from the direction of 3 prime to 4.

00:36 5 prime.

00:37 According to central dogma, the dna gets transcribed to mrna and mrna on translation forms the protein.

00:49 Okay, this is transcription, i've pulled out the short form and translation.

00:54 This entire sequence we call it as central dogma.

00:59 Correct? let us look at the template strand or the wild type of dna sequence given from 3 prime to 5 prime.

01:08 From 3 prime, it is cgg, i'm splitting off into frames from cure itself so that it will be easy for us to continue.

01:19 Okay, yes.

01:24 So here the frames of triplet codons have made it from the template dna itself so that it will be easy for us to proceed with this transcription and translation.

01:34 Now if this is being the template dna, this undergoes transcription, right, in order to form the 5 prime to 3 .3.

01:42 Prime mrna, correct? let us look at, so in mrna, the p will get replaced by u, okay? so here the sequence will be gcc, s, a, ug, ccc, and then uau, and we have ago.

02:06 This mrna on translation forms what kind of protein synthesis with the amino acid sequence gcc codes for alanin, aug for methionine, ccc for prulene, and uau for tyrosin, so a tryptophan, and ago for serene.

02:27 So this is the amino acid sequence of, you know, the codes which has been carried by the mrna from the message taken from the template d .a.

02:37 Now let us compare this with the two types of mutation.

02:40 The first mutation, what they have shown here, okay, the first musings.

02:43 If you see the sequence in the template dna, what does it become? it is cgg, the same thing, tac, g g g g, a, t, and it is cca.

03:03 It is c, c, a.

03:06 So this, you know, there is a replacement which has happened.

03:14 Okay if you see the c has replaced the t if you see it in the central dogma so c replacing the okay yes so accordingly after transcription okay memorand a sequence will be how it will be gcc a u c c c c u and it is g g g g you okay okay it is g -g -u.

03:49 I put here g -g -g -u.

03:50 You cannot see it and write it again g -g -u.

03:53 But this after translation, what does it form the protein? it will be alenine, correct, methenin, prolin, triptophen, and there is glycine getting formed here.

04:05 So the place of serene, which has to be formed, there is glycine taken place because of the substitution of one of the nucleotide.

04:13 When there is substitution of one of the nucleotide or single nucleotype, you call that mutation as point mutations.

04:24 So this is answering the first sub -question a under this...

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