s******a
发帖数: 472 | 1 I have a DNA molecule which is 250 kb.
The question is,
how many molecules do I have in 100 ng sample?
It bothers me for some time. |
s******a
发帖数: 472 | 2 The result I got is,
720 billions,
which is too large to believe.
【在 s******a 的大作中提到】
: I have a DNA molecule which is 250 kb.
: The question is,
: how many molecules do I have in 100 ng sample?
: It bothers me for some time.
|
a*****x
发帖数: 901 | 3 rough estimation would be a few billions. you calculation is definitely
wrong |
s******a
发帖数: 472 | 4 Could you please briefly show me how to calculate?
【在 a*****x 的大作中提到】
: rough estimation would be a few billions. you calculation is definitely
: wrong
|
e*******e
发帖数: 1837 | 5 It's off by a factor of 2e3 based on my calculation. Did you take into
account that DNA is double-stranded? Also, make sure the nuclotide weight
conversion is based on Dalton, not kilo-Dalton.
【在 s******a 的大作中提到】
: The result I got is,
: 720 billions,
: which is too large to believe.
|
a*****x
发帖数: 901 | 6 The average MW is ~300, and therefore ~600 for a bp.
250kb molecular weight would be 600 x 250 x 1000 = 1.5E8;
100 ng therefore has 100E-9/1.5E8 = 6.6E-16 M
Therefore should be 6.23E23 * 6.6E-16 = 4E8 = 0.4 B |
s******s
发帖数: 13035 | 7 翻开gibson assembly protocol， 上书50ng 5kb dsDNA 0.015pmols
你这个乘以二除以50，大概6E-16x 6E23 = 360M？
【在 s******a 的大作中提到】
: I have a DNA molecule which is 250 kb.
: The question is,
: how many molecules do I have in 100 ng sample?
: It bothers me for some time.
|
z****g
发帖数: 972 | 8
~~~~~6.02?
My answer is 3.6E8
【在 a*****x 的大作中提到】
: The average MW is ~300, and therefore ~600 for a bp.
: 250kb molecular weight would be 600 x 250 x 1000 = 1.5E8;
: 100 ng therefore has 100E-9/1.5E8 = 6.6E-16 M
: Therefore should be 6.23E23 * 6.6E-16 = 4E8 = 0.4 B
|
s******a
发帖数: 472 | 9 参照 D.N.A. Box 3.3, J.M. Butler (2005) Forensic DNA Typing, 2nd Edition (
Elsevier Academic Press), p. 56
Molecular Weight of a DNA Base Pair = 618 g/mol
A = 313 g/mol;
T = 304 g/mol;
A-T base pairs = 617 g/mol
G = 329 g/mol;
C = 289 g/mol;
G-C base pairs = 618 g/mol
我用618和6.02E23的常数计算，结果是3.9E8, 最接近０.４B。
HOW:
MW of 1bp = 618 g/Mol
250 kb = 250000 bp = 250000/6.02E23 Mol= 4.15E-19 Mol
Mass of 250 kb BAC = 618*4.15E-19 g = 2.5647E-16 g = 2.5647E-7 ng
So, 100 ng of BAC contains: 100/2.5647E-7 = 3.899E8 molecules
【在 z****g 的大作中提到】
:
: ~~~~~6.02?
: My answer is 3.6E8
|
