Biology-Online • View topic - Calculations

## Calculations

Genetics as it applies to evolution, molecular biology, and medical aspects.

Moderator: BioTeam

### Calculations

Hi guys, I've got serious problems with calculation hope you guys can help me out.

Question 1.
A hypothetical organim has 2 distinct chromosomes (2n = 4) and fifty known genes, each with 2 alleles. If an individual is heterozygous at all known loci, how many types of gametes can be produced if all genes behave independently?

b) 4

My attempted answer is 2^15... which is option (d). Is it right?

Question 2.
How many types of gametes would you expect to be produced by individual described in question 1 if the genes are completely linked?
My answer is 2^2 = 4. Is it right?

Question 3.
Many restriction endonucleases used in genetic engineering recognise a sequence of 8 nucleotides in DNA where they cleave the phosphodiester bonds in the duplex backbone. How many such sites do you expect in the human genome if you assume a completely random occurrence of all 4 nucleotides throughout the genome?

For Question 3, i totally do not know where to start except that I know I need the information of genome size which is 6x10^9 bp.

Many thanks in advance. Really appreciate any help! Thanks!
simpleton
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Posts: 34
Joined: Thu Mar 18, 2010 3:48 am

1) why 2^15?
IMHO it's 2^50, what makes d correct answer
(BTW 2^15 is for me ~32 thousands)

2) yes

3) the stretch of 8 nts can have 4^8 possibilities, from which only one is recognized. That means, in average such ER cuts once per 65536 nts. Now you just have to divide it
http://www.biolib.cz/en/main/

Cis or trans? That's what matters.

JackBean
Inland Taipan

Posts: 5694
Joined: Mon Sep 14, 2009 7:12 pm

Opps typo! haha! wanted to type ^50 but typed ^15

Yeah I got them now! MANY MANY THANKS AGAIN, JACKBEAN!
simpleton
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Posts: 34
Joined: Thu Mar 18, 2010 3:48 am

Sorry Jackbean, for question 3, should I use genome size of haploid or diploid?

When I use (6 x 10^9) / 65536, I got ~100,000 random occurrence.

But the options given are just
a) ~12,000
b) ~24,000
c) ~120
d) ~1,000,000
e) ~45,500

And (e) is the answer that I will get if I were to use haploid genome size (3 x 10^9)/65536

simpleton
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Posts: 34
Joined: Thu Mar 18, 2010 3:48 am

### Re: Calculations

you see, you shouldn't thank before you know, it's correct

e.g. here they take as human genome 3.10^9.
I think it's because these two copies are basically the same...
http://www.biolib.cz/en/main/

Cis or trans? That's what matters.

JackBean
Inland Taipan

Posts: 5694
Joined: Mon Sep 14, 2009 7:12 pm

Opps!

Hmm, it's hard to determine when to use genome size for haploid or for diploid. Although the two homologue should theoretically be similar but then if we were to say through the genome.. there will be ~100,000 times this specific 8nt sequence will appear right?
simpleton
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Posts: 34
Joined: Thu Mar 18, 2010 3:48 am