1.How long would it take for the E. coli RNA polymerase to synthesi...

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1.How long would it take for the E. coli RNA polymerase to synthesize the primary transcript for E. colirRNAs (6500 bases), given that the rate of RNA chain growth is 50 nucleotides per second?
2.What is the maximum rate of initiation at a promoter, assuming that the diameter of RNA polymerase is about 204 Angstroms and the rate of RNA chain growth is 50 nucleotides per second?
3.The following DNA fragment:
5’... AGCAG ... 3’
3’ ... TCGTC ... 5’
is allowed to duplicate in an environment with radioactive nucleotides that are symbolized as A*, T*, C*, and G*. Write the daughter molecules that result from the first and the second duplication. Justify your answer.
4.A protein contains 1,998 peptide bonds. If the gene responsible for its synthesis contains 4,000 adenines, find the number of hydrogen bonds in the gene.
5.The adenines in a gene from a prokaryotic cell are 20% of its total nucleotides. The gene contains 11,700 hydrogen bonds and is transcribed to mRNA. Find:
a.the total number of nucleotides in the mRNA,
b.the number of amino acids of the protein.
6.A eukaryotic gene contains 20,000 base pairs and encodes a 3000 amino acid-long polypeptide. During processing of the mRNA, ten phosphodiester bonds were hydrolyzed. The 5’- and 3’-untranslated regions of the mRNA contain 447 nucleotides.
a.How many water molecules were consumed during splicing of the exons?
b.How many nucleotides are there in the introns (assume all introns have the same length)?
c.How many phosphodiester bonds are there in the mRNA?
d.What is the molecular weight of the pre-mRNA if the average molecular weight of each nucleotide is 300?
7.We isolated two RNA molecules from a frog cell. The first molecule resulted from the transcription of a protein-coding gene, while the second one was transcribed from a rDNA gene:
1st molecule:
2nd molecule:
a.Which of the two molecules is the product of the protein-coding gene?
b.The two RNA molecules are partially complementary because of a 6-nucleotide long region. Locate the regions responsible for complementarity in the two molecules and explain their biological significance.
8.The first six amino acids of a bacterial polypeptide chain are:
With the help of the genetic code, find the sequence of the gene fragment that encodes these six amino acids, given that, in the fragment A/G=1 and the antisense strand contains four thymines.
9.By looking at the chart for the genetic code:
a.Determine which codons can be mutated by a single base change to produce an amber codon (TAG). What amino acids are encoded by these codons?
b.The mutagen ethyl methane sulfanate (EMS) causes G to A or C to T mutations. Which of the possibilities in part a could arise with EMS?
c.Of the possibilities in part b, which could be suppressed by amber nonsense suppressor tRNAs generated by EMS?
10.The direction of RNA synthesis was determined by growing bacteria for a short period of time with trinucleotides labeled with tritium (3H) on their bases. The structure of RNA looks like:5' pppNpNpNpNp...N-OH 3' (N = nucleosides; p = phosphates). Cleavage of RNA by alkali occurs after the 3' phosphate.
a.What products are produced if RNA is digested with alkali?
b.The products obtained in Part a can be distinguished by thin layer chromatography. Which of the products would you expect to be differentially labeled if new nucleotides are added at the 5' end (i.e., synthesis is in the 3' to 5' direction)?
c.Which products would be differentially labeled if RNA is synthesized from the 5' to 3' direction?
d.When the experiment was done the tritium label was found in the spot containing N-OH. In which direction is RNA synthesized?

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1. Ans:
Given, the rate of RNA chain growth = 50nucleotides per second.
The length of rRNA transcript = 6500.
Hence, the time taken by E.coli RNA polymerase = 6500/50 = 130 sec.
2. Ans:
Since the typical nucleotide helix has an advancement of 3.4Å per base pair, the rate of RNA chain growth in Å per seconds is 50 x 3.4.
Since, the diameter of RNA polymerase is given as 204Å, hence the time taken for initiation would be 204/(50x3.4) = 1.2 sec...

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