23- Given that there are three bases.
If there are two bases in one codon then the number of codons will be 3*3 = 9
If there are three bases in one codon then the number of codons will be 3*3*3 = 27
If there are four bases in one codon then the number of codons will be 3*3*3*3 = 81
If there are five bases in one codon then the number of codons will be 3*3*3*3*3 = 343
Given that only 26 amino acids are needed for the Arth organism. So minimum codon could be 26 + 1 (Stop codon)
So the answer is three bases per codon.
Option B is the correct answer.
24 -
As shown in the figure open reading Start codon and ends before stop codon, So in the given sequence open reading frame starts from 57th base and ends at 65th base.
So answer is option B - 57-65.
25- Protein coding sequence is AUG GUC ACA. It encodes for N-terminal MET VAL THR- C-terminal.
So correct answer is option D - N-Met Val Thr -C
5 pts Question 23 Life on Earth uses 3 nucleotide long code words, with 4 letters...
5 pts Question 33 The following prokaryotic DNA (same as above) is transcribed into RNA and the MRNA transcript is translated into protein. 31 21 11 1 ATGGGTTACT ATGAGGAGTT GACACACAAG AGGAGGTAGC 71 61 51 41 AGTATGGGTA TAATCTAATG CGTAATTGAG GAGGTAGTTG 101 111 91 81 ACGTATGCAT AGATAACGTA CGGGGGGGAA ACCCCCCCTT 141 131 121 TTTTTTTTTC GAGCAATAAA AGGGTTACAG Useful sequences: -35: 5' TTGACAT 3', -10: 5' TATAAT 3 (Pribnow box), Shine Dalgarno sequence: 5' AGGAGGU 3 THE GENETIC CODE THE GENETIC CODE SECOND LETTER A...
I. Use the DNA sequence below, which encodes a prokaryotic gene to answer the following questions. 11 21 31 41 51 71 81 61 91 CGTAATTGAG GAGGTAGTTG ACGTATGAAT AGTTAACGTA CGGGGGGGAA 101 111 121 131 141 ACCCCCCCTT TTTTTTTTTC GAGCAATAAA AGGGTTACAG ATTGCATGCT a) Write down the corresponding sequences, find them in the sequence above and label them: -35 Consensus sequence: _(label as -35) -10 Consensus sequence (Pribnow box): _ __ (label as Pribnow) Shine-Dalgarno sequence in corresponding mRNA: __ (label as SD)...
Question 10 (15 points) Given the following sequence for a template strand of DNA 3 - ATACTTTGTCGAGACCCGCTTCTTGCAGACTGGG A. Provide the mRNA sequence following transcription (include polarity) B. Provide the amino acid sequence using either the one letter or three letter abbreviations. Include polarity (N-or C-terminus) and be careful to start in the correct place: C. What if the "C" underlined above was changed to a T. What is the new codon? How does that affect the amino acid sequence? What...
A template strand of DNA in a gene reads 3’ CCA AGC TCT 5’.
Using the codon chart provided, answer the following
questions:
-What is the sequence of amino acids that is produced when this
gene is translated?
-If a mutation causes a substitution (an A instead of a T) 3’
CCA AGC ACT 5’, what effect will it have on the mRNA transcript AND
on the protein?
-What do we call this type of mutation?
Second letter U С...
If the sequence of an mRNA molecule is: 5' AUG GGA UUU CGA 3' (a) Give the sequence of the template strand of DNA. (Please indicate where the 5' and 3" ends are located) (1.5 marks) (b) Give the sequence of the non-template strand of DNA. (Please indicate where the 5' and 3" ends are located) (1.5 marks) (c) Give the amino acid sequence. (1.5 marks) You will require the following genetic code to answer this question. Second letter с...
Follow the instructions below to answer questions about
Replication, Transcription & Translation.
3’- T A C A
C C G G
T C A G
G T G A T C
-5’
A. Imagine that the sequence shown represents one strand of a
gene sequence. What would be the sequence of the complementary
strand of DNA? Write out your answer, indicating correct
polarity (5' and 3' ends) on your new strand. (1.5
points)
B. Now imagine that the new strand...
Some amino acids are post-translationally removed from the
C-terminal end of the beta-lactamase enzyme from B.
imaginarium (i.e. - after it is translated and
released from the ribosome, a protease chews off a some amino
acids). The wild-type enzyme, which has had the amino
acids removed from the C’-terminus, is 246 amino acids in length
and the C-terminal amino acids are shown below aligned with the
C-terminal amino acids of a frameshift mutant, which – due to a
frameshift mutation -...