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Biology calculations

  1. Dec 20, 2014 #1
    I am not sure where to put this question in biology or maths forum.But i finally decided to ask here.Somatic cells have two "copies" of each of chromosomes, one inherited from each parent. Each sperm cell, on the other hand, only gets one "copy" of each chromosome (either the one from your mom or your dad). So there are something like 2^23 different possible sperm configurations.Similarly there are four different types of nitrogen bases ,so they can form 4^3 i.e 64 possible combinations of triplets,called codons.
    How these calculations are done,i am not getting ,if i will learn this it will make my life way easier. please help.How we can understand that where should we put ( ^)[to the power raised to].......?
     
  2. jcsd
  3. Dec 20, 2014 #2

    jedishrfu

    Staff: Mentor

    It looks like you're mixing and matching different things. On the one hand, you're thinking about chromosomes and all the possible that they can combine. On the other hand, you're thinking about the DNA strands that compose the chromosome and all the possible ways they can be coded.

    So what is it you're trying to figure out? Humans have largely the same DNA packaged into the same chromosomes. There is still a large variety of differences but not necessarily from codon to codon. We gain our traits mostly in how the chromosomes are paired from our mother's chromosomes and our fathers chromosomes. You might want to think more at the gene level than at the codon level.

    http://en.m.wikipedia.org/wiki/Chromosomes

    There's a table inside that shows the number of genes in each chromosome.
     
  4. Dec 20, 2014 #3
    I have given jut two examples,actually i want to understand how such( 4^3/ 2^23)calculation is done?
     
  5. Dec 20, 2014 #4

    Pythagorean

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    4^3 means multiply 4 three times. So it would be:

    4^3 = 4*4*4 = 16*4 = 64
     
  6. Dec 20, 2014 #5
    I know 4^3 = 4*4*4 = 16*4 = 64 .I just want to know how can i understand that here i have to calculate in this way.As in my examples ,there are four nitrogenous bases and i want to figure out how many different combinations i can make if i make pair of three of them i.e codon.If i will sit to write each and every combination it will be very tedious and tough task but in my textbook it is just calculated in this way 4^3 = 4*4*4 = 16*4 = 64 .64 different combinations.That's what i want to learn how to do such calculations.Four different bases .i want to make pair of three so how it comes out to do 4^3 why not 4 multiplied by 3?That's what i want to learn.
     
  7. Dec 20, 2014 #6

    Pythagorean

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    Think of it this way. If you have a switch with an on off state, that's one switch with two states, so: 2^1

    If you have two switches with two states each, there are four total states the whole two system switch can be in: 2^2=4

    If a widget has 4 states and you have three widgets, the total number of states is 4^3.

    You can also count them out. For two switches each with two states, all your states are:

    00
    01
    10
    11

    Four states, 2^2!
     
    Last edited: Dec 20, 2014
  8. Dec 20, 2014 #7
    I didn't understand this.Is it in this way
    one switch with two states.
     
  9. Dec 20, 2014 #8

    Pythagorean

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    I fixed that typo :) sorry, mobile keyboard.
     
  10. Dec 20, 2014 #9
    So basically what is general formula to do such calculations?Is it like number of things^their variable states(possible positions) right?
     
  11. Dec 20, 2014 #10

    Pythagorean

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    States^(number of things) right? Because 4^3 was three things with four states each. And bits are always 2^n.
     
  12. Dec 20, 2014 #11
    AA.............Four things with three states .There were four nitrogenous bases.
     
  13. Dec 20, 2014 #12
    Now i am getting confused.According to your examples States^(number of things),But according to mine number of things^their variable states.What next?
     
  14. Dec 20, 2014 #13

    Pythagorean

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    Well, let's go to 2^3 - that's 8. If you have three bitz with two states each, you have

    000
    001
    010
    011
    100
    101
    110
    111

    That's 8. Wheres 3^2 would be 9. So which formula is right?
     
  15. Dec 20, 2014 #14

    Pythagorean

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    Also, a codon is three base pairs and the pairs have four possible things they could be: A C G or T, so 4^3 is right!

    Edit: actually I think that T should be a U.
    edit 2: genes -> base pairs
     
    Last edited: Dec 20, 2014
  16. Dec 20, 2014 #15
    But i am taking it in this way there are 4 nitrogenous bases,and they are arranged in codon such that they can be given either of three positions in codon.So things^their variable states.I am taking number of things as number of bases you are taking number of things as number of genes in a codon.So ...........
     
  17. Dec 20, 2014 #16
    I am taking number of things as number of bases you are taking number of things as number of genes in a codon.So ...........What actually is number of things?
     
  18. Dec 20, 2014 #17

    Pythagorean

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    Sorry, gene was the wrong word. We are talking about the same thing - the units of genes (base pairs).
     
  19. Dec 20, 2014 #18
    So four base pair so four things. How three things^variable states?
     
  20. Dec 20, 2014 #19

    Pythagorean

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    No the value of the base pair is the state, the position is the thing, in this case. So three things (positions) with four possible states (four different things can go in each place).
     
  21. Dec 20, 2014 #20
    Now i understood.Thanks a lot.
     
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