Answer the following questions: Note: Met (the first amino acid) should always b

Question

Answer the following questions: Note: Met (the first amino acid) should always be included in the total number of the amino acids of every polypeptie. 3- Which allele(s) produce(s) the shortest polypeptide? Just type the allele’s name(s). 4- How many amino acids do exist in the shortest polypeptide? Just type the number. ASSIGNMENT GUIDE: 1- Express the wild allele (M1, typed below, go through all of the steps of gene expression), and write down the normal product (P1). 2- Note that the promoter of the gene is on the right side of the gene! 3- Apply all of the changes (from # II to # VI, one at a time) on the original wild allele. Each change (mutation) results in generation of a new mutant allele (M2 to M6). 4- Express all of the new alleles (M2 to M6, one by one) to come up with a product for each one (P2 to P6 for the mutant products).

1

3’GCTATATAGGAAGATTAAATAATACAGTAAACGGGCGAGTCTACAAACCGTACTAAATAATGTTACAAAGTATATTTGGCCGATATATCC 5’

5’CGATATATCCT TCTAATTTATTATGTCAT TTGCCC GCTCAGATGTTTGGCATGATTTATTACAATGTTTCATATAAACCGGCTATATAGG 3’

Note: The promoter is at the right side, and the terminator is at the left side of the gene. I- Assume that the sequence above is the normal (wild) allele for gene M (let’s call it allele M1). The normal allele M1 encodes polypeptide P1. II- If base pair G/C is added between positions 35 and 36 of gene M (to allele M1), allele M2 is produced, which encodes polypeptide P2. G is added to the top strand. III- If base pair C/G is inserted between positions 60 and 61 of gene M (into allele M1), allele M3 is produced, which encodes polypeptide P3. C is added to the top strand. IV- If base pair number 53 (in allele M1) is changed from C/G to A/T, allele M4 is produced, which encodes polypeptide P4. V- If base pair A/T is inserted between nucleotides 40 and 41 (into allele M1), allele M5 is produced, which encodes polypeptide P5. A is added to the top strand. VI- If base pair number 64 (in allele M1) is changed from T/A to C/G, allele M6 is produced, which encodes polypeptide P6.

Explanation / Answer

The given DNA sequence is (M1):

3’GCTATATAGGAAGATTAAATAATACAGTAAACGGGCGAGTCTACAAACCGTACTAAATAATGTTACAAAGTATATTTGGCCGATATATCC 5’

Note that the 3’ to 5’ strand will be used as the template.

So, the mRNA sequence will be

5’-GCUAUAUAGGAAGAUUAAAUAAUACAGUAAACGGGCGAGUCUACAAACCGUACUAAAUAAUGUUACAAAGUAUAUUUGGCCGAUAUAUCC-3’

3’-CGAUAUAUCCUUCUAAUUUAUUAUGUCAUUUGCCCGCUCAGAUGUUUGGCAUGAUUUAUUACAAUGUUUCAUAUAAACCGGCUAUAUAGG-5’

The protein sequences will be:

Frame 1: AI.ED.IIQ.TGESTNRTK.CYKVYLADIS

Frame 2: LYRKIK.YSKRASLQTVLNNVTKYIWPIY

Frame 3: YIGRLNNTVNGRVYKPY.IMLQSIFGRYI

Frame 4: RYILLIYYVICPLRCLA.FITMFHINRLYR

Frame 5: DISF.FIMSFARSDVWHDLLQCFI.TGYI

Frame 6: IYPSNLLCHLPAQMFGMIYYNVSYKPAI.

Note that frame 1 and 2 will not produce protein, because methionine is lacking.

Let us consider frame 4 to be the correct protein. This is because it is the longest.

So, P1 (normal protein sequence) will be:

RYILLIYYVICPLRCLA.FITM

II

C/G is added to position 35 and 36. This means that two Cs are added at position 35 and 36 in the M1 (3’-5’) and two Gs are added in the (5’-3’) strand.

The resulting allele is called as M2.

M1 is:

5’-CGATATATCCTTCTAATTTATTATGTCATTTGCCCGCTCAGATGTTTGGCATGATTTATTACAATGTTTCATATAAACCGGCTATATAGG-3’

Structure of M2 will be:

CGATATATCCTTCTAATTTATTATGTCATTTGCCCGCTCAGATGTTTGGCATGATTCCTATTACAATGTTTCATATAAACCGGCTATATAGG

P2 is:

IYPSNLLCHLPAQMFGMIPITMFHINRLYR (polypeptide is from right to left and bolded).

M3 is:

5’-CGATATATCCTTCTAATTTATTATGTCATTTGCCCGCTCAGATGTTTGGCATGATTGGTATTACAATGTTTCATATAAACCGGCTATATAGG-3’

P3 will be:

IYPSNLLCHLPAQMFGMIPITMFHINRLYR (polypeptide is from right to left and bolded).

This means M2 and M3 will give the same product.

Similarly calculate:

P4 will be:

P5 will be:

P6 will be:   

**Note: This is a full assignment. I have answered the first 5 parts. Please post this again, for P4, P5 and P6 protein.

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