homework 2 - 1 Why do you expect to encounter a STOP codon about every 20 codons or so on average in a random sequence of DNA Hypothetically each of the

# homework 2 - 1 Why do you expect to encounter a STOP codon...

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1.Why do you expect to encounter a STOP codon about every 20 codons, or so, on average in a random sequence of DNA? 2. Human chromosome 1 contains 2.8 x 10 8 base pairs. At mitosis this chromosome is 10 m in length. Relative to its fully extended length, how compacted is the DNA molecule in chromosome 1 at mitosis? A DNA bp is 0.34 nm in length. The relative length of chromosome is = length X base pairs 2.8x 10^8 x 0.34nm= 95200000 nm = 95200µm compare with the chromosome at the stae of mitosis 10 µm means 952000µm=10µm gives a difference of 1000µm by this the relative length of chromosome at mitosis is compacted to 1000 times. 3. The human genome contains 6.4 x 10 9 base pairs and fits into a nucleus that is 6 m in diameter. What is the length of DNA in a human cell? If the diameter if the DNA helix is 2.4 nm, what fraction of the volume of the nucleus is occupied by DNA? Assume the nucleus is a sphere and the DNA is cylinder. Length of DNA in human cell = Base pair * inetrval at which base pair occurs Length = 6.4 x 10^9 bp * 34 x 10^-9 m/bp = 2.18 m radius of nucleus = 6 x 10^-6 / 2 = 3 x 10^-6 m V(nucleus) = 4/3 * pi * r^3 V(nucleus) = 4/3 * 3.14 * ( 3 x 10^-6)^3 = 1.13 x 10^-16 m^3 V(DNA) = pi * r^2 * h Radius of DNA = 2.4 / 2 = 1.2 nm = 1.2 x 10^-9 m V(DNA) = 3.14 * ( 1.2 x 10^-9)^2 * 2.18 = 9.86 x 10^-18 m^3 V(DNA) / V(nucleus) = 9.86 x 10^-18 / 1.13 x 10^-16 = 0.09 So DNA occupies 0.09 x 100 = 9% of volume of Nucleus  #### You've reached the end of your free preview.

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• Spring '14
• GaryL.Bowlin
• • • 