# Cell and Molecular Revision Question Help!!!

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• #17549

Unsure if I posted in the right section so I posted twice, one for the two categories so forgive my double posting (Cell Biology and Molecular Biology)

Hello I would appreciate some assistance with the following review questions as my exam is coming up in a few weeks I’m just revising my content, however, there seems to be a limited range of practice questions and when it comes to these I seem to struggle, so would appreciate some explanations/walkthroughs as to how to do them so I may learn.

ESTIMATING GENE NUMBERS
Approximately 1% of the human genome consists of protein-coding sequences (exons). A diploid somatic human cell contains 5 picograms/pg of DNA and rapidly growing E.coli cells contain on average four genomes giving a total of 0.017pg of DNA. Assuming that the sequences within the E.coli DNA codes for protein and that the average size of proteins is similar in humans and E.coli, how many more proteins could theoretically be expressed by the haploid human genome than by the haploid E.coli genome?

LIGATION THEORY
A 100bp fragment is to be cloned into the EcoRI site of pBR328 (4.9 Kb) in order to detect successful insertions using blue/white selection. Given that there are 50 nanograms of EcoRI digested pBR328, on the basis of the optimal ratio of insert: vector for ligations, how much insert would be added to the 50ng of EcoRI digested pBR328 to ensure the best possible chance of successful ligation? (MW dsDNA = 660; MW ssDNA = 330)

Again would appreciate some feedback/walkthrough of these questions as I’m currently doing practice questions.

Cheers, Didact1

• #114429

What the "A diploid somatic human cell contains 5 picograms/pg of DNA"?
5 pg of what per pg of DNA?

ligation – just calculate the molecular weight of each fragment and then put them into 3:1 ration (by moles)

• #114434

I don’t know, that’s question presented to me. As for the Ligation theory, please elaborate? I still don’t get it…

• #114435

This is what I have..

Consider the Vector (Mass of Vector): pBR328 (4.9 x 103) x 660 = 2.94×106 Daltons

Given: Moles = Mass/MW

Moles = 50 x 10-9g / 2.94×106
Moles = 17 x10-15

Each vector has two ends therefore x2
= x2
Therefore there are 34 x10-15 moles of ends of vector

Consider the Insert:
Insert = 100 x 660 = 6.6 x 104 Daltons

Ideal molar ratio 3:1 therefore x3
= (34 x10-15) x 3 = 102 x 10-15 Moles of Insert

Given: Moles = Mass/MW … Therefore…

102 x 10-15 = Mass / 6.6 x 104
Therefore Mass = (6.6 x 104) x (102 x 10-15)
= 6.73 x 10-9
= 6.73ng
Each piece of insert has two ends therefore /2
Therefore Answer = 3.37ng (6.73 / 2)

Am I doing this right?

• #114444

except of the two ends, it seems OK. Both vector and insert have two ends so you do not need to include that 😉