Biology, asked by vineettyagi84192, 9 months ago

D
The given nucleotide sequence on mRNA is
Shown below!
as
5' AUGUCAUGGGAGUGA GUUGGGCUAAAA
UAG 3'
(@) How many amino acids will be inserted in a
polynucleotide chain under naumal conditions ?
(6) How many acide will be inserted in
polynucleotide chain in a mutated situation by
the deletion of 9th nucleotide in the cistron
part of the DNA ?​

Answers

Answered by Anonymous
0

Answer:

The given nucleotide sequence on mRNA is

Shown below!

as

5' AUGUCAUGGGAGUGA GUUGGGCUAAAA

UAG 3'

(@) How many amino acids will be inserted in a

polynucleotide chain under naumal conditions ?

(6) How many acide will be inserted in

polynucleotide chain in a mutated situation by

the deletion of 9th nucleotide in the cistron

part of the DNA ? ok dear this is right answer.....

Explanation:

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Answered by satyampandey1047
0

Answer:

. The template DNA strand, from which the mRNA is synthesized, is 5’ CAAACTACCCTGGGTTGCCAT 3’

(RNA synthesis proceeds in a 5’ à 3’ direction, so the template strand and the mRNA will be complementary to each other)

b. The coding DNA strand, which is complementary to the template strand, is 5’ ATGGCAACCCAGGGTAGTTTG 3’

c. The sequence of the mRNA is 5’ AUGGCAACCCAGGGUAGUUUG 3’

(the sequence of the mRNA is complementary to the template strand and identical to the coding strand with U substituted for T)

d. The third codon is 5’ ACC 3’. Therefore, the corresponding anti-codon is 5’ GGU 3’

2. Below is a table for the genetic code:

T

C

A

G

T

TTT Phe (F)

TTC "

TTA Leu (L)

TTG "

TCT Ser (S)

TCC "

TCA "

TCG "

TAT Tyr (Y)

TAC "

TAA Stop

TAG Stop

TGT Cys (C)

TGC "

TGA Stop

TGG Trp (W)

C

CTT Leu (L)

CTC "

CTA "

CTG "

CCT Pro (P)

CCC "

CCA "

CCG "

CAT His (H)

CAC "

CAA Gln (Q)

CAG "

CGT Arg (R)

CGC "

CGA "

CGG "

A

ATT Ile (I)

ATC "

ATA "

ATG Met (M)

ACT Thr (T)

ACC "

ACA "

ACG "

AAT Asn (N)

AAC "

AAA Lys (K)

AAG "

AGT Ser (S)

AGC "

AGA Arg (R)

AGG "

G

GTT Val (V)

GTC "

GTA "

GTG "

GCT Ala (A)

GCC "

GCA "

GCG "

GAT Asp (D)

GAC "

GAA Glu (E)

GAG "

GGT Gly (G)

GGC "

GGA "

GGG "

a. The following codons can be mutated by one base to produce an amber codon:

CAG Gln

AAG Lys

GAG Glu

TCG Ser

TTG Leu

TGG Trp

TAA Stop

TAT Tyr

TAC Tyr

b. From part a, CAG (Gln) and TGG (Trp) can become amber stop codons through EMS.

c. From part b, both of the resulting amber codons could be suppressed by amber nonsense suppressors generated by EMS.

3a. The codon is the three nucleotide sequence in the mRNA that indicates which amino acid should be incorporated in the growing polypeptide chain. The anticodon is the complementary three nucleotide sequence in the appropriate tRNA.

b. Template strand is the DNA strand off which the mRNA is synthesized. The coding, or non-template, strand is the DNA strand complementary to the template strand; it has the same sequence (except for T for U substitutions) as the mRNA.

c. The Pribnow box is a sequence of six nucleotides (TATAAT) positioned at -10 that signals where transcription initiation should begin in prokaryotic DNA. The Shine-Delgarno sequence is a short, purine-rich region in the mRNA that is complementary to the rRNA within the 16S ribosomal subunit. The sequence signals which AUG acts as the translation start in mRNA.

4a. False, a wobble allows the anticodon in the tRNA to hybridize with different codons in mRNA.

b. False, a frameshift mutation affects all the subsequent amino acids.

c. False, only one codon (AUG) encodes for the start of protein synthesis; three codons signal the end of protein synthesis.

d. False, the wobble is first base (5’ to 3’) in the anticodon.

e. True, RNA can be used as a template for DNA synthesis in a process known as reverse transcription.

f. True. For example, a single base substitution causing CAT to change to AAT would signal a termination.

g. False, the Wobble Hypothesis explains how alternate base pairing can occur with the first nucleotide (going from 5' to 3') in the anticodon.

Explanation:

.

TCC "

TCA "

TCG "

TAT Tyr (Y)

TAC "

TAA Stop

TAG Stop

TGT Cys (C)

TGC "

TGA Stop

TGG Trp (W)

C

CTT Leu (L)

CTC "

CTA "

CTG "

CCT Pro (P)

CCC "

CCA "

CCG "

CAT His (H)

CAC "

CAA Gln (Q)

CAG "

CGT Arg (R)

CGC "

CGA "

CGG "

A

ATT Ile (I)

ATC "

ATA "

ATG Met (M)

ACT Thr (T)

ACC "

ACA "

ACG "

AAT Asn (N)

AAC "

AAA Lys (K)

AAG "

AGT Ser (S)

AGC "

AGA Arg (R)

AGG "

G

GTT Val (V)

GTC "

GTA "

GTG "

GCT Ala (A)

GCC "

GCA "

GCG "

GAT Asp (D)

GAC "

GAA Glu (E)

GAG "

GGT Gly (G)

GGC "

GGA "

GGG "

a. The following codons can be mutated by one base to produce an amber codon:

CAG Gln

AAG Lys

GAG Glu

TCG Ser

TTG Leu

TGG Trp

TAA Stop

TAT Tyr

TAC Tyr

b. From part a, CAG (Gln) and TGG (Trp) can become amber stop codons through EMS.

c. From part b, both of the resulting amber codons could be suppressed by amber nonsense suppressors generated by EMS.

3a. The codon is the three nucleotide sequence in the mRNA that indicates which amino acid should be incorporated in the growing polypeptide chain. The anticodon is the complementary three nucleotide sequence in the appropriate tRNA.

b. Template strand is the DNA strand off which the mRNA is synthesized. The coding, or non-template, strand is the DNA strand complementary to the template strand; it has the same sequence (except for T for U substitutions) as the mRNA.

c. The Pribnow box is a sequence of six nucleotides (TATAAT) positioned at -10 that signals where transcription initiation should begin in prokaryotic DNA. The Shine-Delgarno sequence is a short, purine-rich region in the mRNA that is complementary to the rRNA within the 16S ribosomal subunit. The sequence signals which AUG acts as the translation start in mRNA.

4a. False, a wobble allows the anticodon in the tRNA to hybridize with different codons in mRNA.

b. False, a frameshift mutation affects all the subsequent amino acids.

c. False, only one codon (AUG) encodes for the start of protein synthesis; three codons signal the end of protein synthes

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