Biology 2581B Lecture Notes - Lecture 11: Sequence Database, Single-Nucleotide Polymorphism, Sanger Sequencing

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maroonelk470

3 Mar 2013

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It has equal probability of being a, t, g, or c, therefore 4 permutations: chance that the base is a, chance of particular sequence = 1/sequence permutations. If the sequence is 2 bases long, what is the chance that is has the sequence gt? (1/4)^4 = 1/16. How many total positions are there of that length within. The e^-381 is the change that this sequence does not occur in the genome. There is a good chance that you wouldn"t find this sequence within a genome. The larger the sequence, the smaller the chance you will find it in the genome. Individual events so just multiply the probability of each event occurring (0. 37^2) Short sequences of 50 bases are not expected to occur at random in the sequence database. Sequencing the human genome 3 maps: high resolution genetic map, physical map, sequence map. Large scale efforts to find snps and ssrs to create a high resolution linkage map.

Related Questions

A transcription factor is a molecule that binds to a specificDNA sequence in order to control the expression of the nearbygenetic material. In this problem we will look at the process bywhich the factor finds the correct sequence to bind to. Assume thatfactor can bind loosely to a DNA sequence that does not match thetarget sequence. After every 10 ns the molecule will move 1 basepair with a 50/50 chance that the move is to the left or to theright.

(a) After 100 ns, what is the probability that the transcriptionfactor is 6 base pairs to the left of where it started? (hint:treat each step like the coin flip example from lecture)

(b) Where is the most likely place for the factor to be after100 ns? What is the probability that it ends up at the most likelyplace?

byzantiumcamel839

please help thank you :)

6) Guanosine consists of

a) A guanine base + a ribose sugar

b) A guanine base + a deoxyribose sugar

c) A guanine base + a ribose sugar + a phosphate group

d) A guanine base + a deoxyribose sugar + a phosphate group

e) A guanine base + a cytosine base

7) A deletion mutation in a viral DNA resulted in loss of 20 nm length of DNA. How many helical turns are missing?

a) 6 b) 20 c) 34 d) 60 e) 200

Question # 8 & 9.
Reverse transcriptase is an enzyme, which can utilize mRNA as a template and synthesize complementary DNA strand. If mRNA sequence is 5âAUGGUAUGCUCU 3â

8) What will be the sequence of complementary DNA strand?

a) 3â AGAGCATACCAT 5â

b) 3â ATGGTATGCTCT 5â

c) 3â TCTCGTATGGTA 5â

d) 3â TACCATACGAGA 5â

e) None of the above

9) How many codons are present in this mRNA?

a) 2

b) 4 c) 6 d) 8 e) 12

10) In gel electrophoresis, the migration of DNA molecules is unaffected by

a) Size of DNA

b) Shape of DNA

c) Strength of electric field

d) Strength of gel

e) None of the above

11) In biological macromolecules, the following two non-polar groups most commonly participates in the formation of hydrophobic interactions

a) N-H and O-H

b) N-H and C-H

c) C-H and C-C

d) C-H and O-H

e) C-C and O-P

1.) Explain how combinatorial control makes it possible to have far fewer transcription factors than the number of genes that are transcriptionally regulated.

2.) If a miRNA were 5 nucleotides long, what is the chance of finding a complementary sequence in a randomly selected 5 nucleotide sequence in mRNA? [Assume mRNA sequences to be random and that mRNAs have an equal amount of each nucleotide (A, U, G, and C)]. If an miRNA were 10 nucleotides long, what is the chance of finding a complementary sequence in a randomly selected 10 nucleotide sequence in mRNA? If you were designing a synthetic miRNA to specifically target a particular mRNA, would you want to design a 5 or 10 nucleotide long miRNA? Why?

3.) In the fungus Neurospora crassa, roughly 2 â 3% of cytosines are methylated. Researchers isolated Neurospora DNA sequences that contained 5-methyl cytosine and found that almost all methylated sequences were located in inactive copies of transposable elements. On the basis of these observations, propose a possible explanation for why Neurospora methylates its DNA and why DNA methylation in this species is associated with transposable elements.

4.) Some cases of bacterial gene regulation involve use of alternative transcriptional terminators. Propose a plausible way that a riboswitch could function in the use of alternative bacterial transcriptional terminators.

scarletporcupine910

Biology 2581B Lecture Notes - Lecture 11: Sequence Database, Single-Nucleotide Polymorphism, Sanger Sequencing

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