anyone able to help?!

Answer:

1a. 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.

5a. Digestion of RNA with alkali will cleave the strand after each 3’ phosphate.  Therefore, the products remaining will consist of pppNp, Np, and N-OH

b. If RNA was synthesized in the 3’ to 5’ direction (i.e. by adding ribonucleotides to the 5’ end), then the pppNp and Np fragments should be labeled with tritium.

c. If RNA was synthesized in the 5’ to 3’ direction (i.e. by adding ribonucleotides to the 3’ end), then the Np and N-OH fragments should be labeled with tritium.

d. Since the N-OH fragments were labeled with tritium, RNA synthesis must occur in a 5' to 3' direction.

6. In a missense mutation, the new nucleotide alters the codon so as to produce an altered amino acid in the protein product.  With a nonsense mutation, the new nucleotide changes a codon that specified an amino acid to one of the stop codons (TAA, TAG, or TGA). Therefore, translation of the messenger RNA transcribed from this mutant gene will stop prematurely.

Explanation:

:)

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