| Q. | When does combinatory diversification take place in lymphocytes? | Related Search:
Medicine | | | When does the lymphocyte undergo the diversification (recombination of DNA with sequences of variable, constant, joining and other parts od heavy and light chain of imunoglobulin) ? Or is it even a lymphocyte or is it just some stem cell or lymphoblast who diversificate? And does it occur before or during gaining the imunocompetency?
Thank you for the answer.
| | A. | Finally, a question that made me think. Your B cell, in this case, will start out as a pleuripotent stem cell in the bone marrow, of which it will go through several phases of early B cell development before it finally turns into a fully mature B cell. I will list the stages and then explain them:
Stem cell-->early pro-B cell-->late pro-B cell-->large pre-B cell-->small pre-B cell-->immature B cell--> Mature B cell
Early Pro B: in this phase the D-J rearrangement of the heavy chain (these two always begin first)
Late Pro B: you get V-DJ rearrangement
Large pre B: the VDJ of the heavy chain is now completely rearranged, and you now have a u (mu) chain pre-B cell receptor which is mostly intracellular at this point
Small pre: you now have VJ rearranging on the light chain (remember there is no D region on the light chain).
Immature B cell: VJ of light chain is now also rearranged, and you have IgM expressed on the cell surface.
Mature B cell: the B cell has co-expression of IgM and IgD which are produced from alternatively-spliced H-chain transcripts.
The B cell is not immunocompetent until it is mature. | | | |
| Q. | Does someone knows where I can get or has Discrete & Combinatory Math from R.P. Grimaldi with solved exercises | Related Search:
Mathematics | | | Looking for solved exercises from chapters 1 - 7.
| | A. | Search for it at [Link] 
Go to the public library and ask how much they would charge to find this book through InterLibrary Loan. Most libraries absorb the cost, but you want to make sure first. If you can afford it, then submit a request and see what happens. | | | |
| Q. | Combinatory exercice (induction)? | Related Search:
Mathematics | | | Demostrate that what follows works with all n in N.
a) (n choose 0)+(n choose 1)+...+(n choose n)=2^n.
b) (n choose 0)-(n choose 1)+...+(-1)^n(n choose n)=0.
I tried hardly on the a section, without an answer. I tried by induction. I looked for a formula that says " (m choose n-1)+(m choose n)=(m+1 choose n) but I realized it doesn't help. Please help me and explain when you think it's necessary to. Thanks a lot!
| | A. | (a) Seeing these symmetric expressions makes us think of the coefficients in the binomial expansion of the expression (x+y)^n. In fact, the coefficients of the binomial expansion match the terms in the sum exactly. If we let x = y = 1, then the expansion of (x+y)^n becomes the left hand side of the idenity we are trying to prove. Of course (1+1)^n = 2^n and we are done.
(b) This is very similar to the result in part (a) so we wonder if we can employ a similar tactic to derive this result and indeed we can. This time let x = 1 and y = -1. When we expand the binomial expression we get the left hand side of the identity. This should be equal to (1-1)^n = 0 and we are done.
As an aside there is a nice counting argument that validates (a) as well. See if you can find it. Hint: Think about what both sides of the expression count and convince yourself that they are the same thing.
You can prove (a) by induction as well. There's a nice proof of that which does make use of Pascal's identity. (I think you can prove (b) by induction as well). This does require some ingenuity so it's more challenging the other methods. See if you can find it. | | | |
| Q. | statistics question, combinatory analysis? | Related Search:
Mathematics | | | a pack of 40 cards, 4 suits ..suit a, suit b, suit c and suit d ...each suit has ten unique cards .
9 are chosen from the deck. without replacement.
how many possible combinations are there...
i did 40*39*38*37*36*35*34*33*32*31 all divided by 9 factorial
then . how many of these combinations will have exactly 3 cards of suit a.
i did 10*9*8*30*29*28*27*26*25 all divided by 9 factorial
how many will have less than 2 number 7s
i did 4*36*35*34*33*32*31*30*29 all divided by 9 factorial
and added it to 36 *35*34*33*32*31*30*29*28 all divided by 9 factorial
... then how many will have at least 2 number 8s.
i did total number of combinations minues the answr from above...
please tell me exactly how wrong i am ..
| | A. | Your results are expressed in a way which is not exactly easy to follow (but at least you have shown that you have tried ! Too many people think that they can fool us by saying "I have done them and just want to check my answers". As if !)
Your first answer is almost correct.
In short form, it should be : 40C9 = 40! / (31! x 9!)
or 40 x 39 x . . . . . . . . . 32 / 9 x 8 x . . . . . 1
(You have got one factor too many on the top. The number of factors on the top should match the factorial on the bottom (the 9!). So you should have 9 factors on the top, you have 10.)
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The second one requires 3 cards of suit a, that is 3 from 10
Number of combinations = 10C3 = 10! / (7! x 3!) = 10 x 9 x 8 / 3 x 2 x 1 . . . (call this result A)
You then will have 7 cards from the remaining 3 suits, that gives 30C7 combinations, which
= 30! / (23! x 7!) = 30 x 29 x . . . . 24 / 7 x 6 x 5 x . . . . 1 . . . . .(call this result B)
And the total number of combinations will be any one from A with any one from B, that is A x B.
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"Less that 2 number 7's" means 0 sevens or just 1 seven, and you have to deal with each possibility separately.
If you get no sevens, that means choosing 9 cards from the other 36, so the combinations number
36C9 = 36! / (27! x 9!) = 36 x 35 x . . . . . 28 / 9 x 8 x . . . . 1 . . . . . .(result C)
If you have just one 7, that means 1 from 4, so the combinations are
4C1 = 4! / (3! x 1!) = 4 . . . . . . .(result D)
and you then need 8 cards from the 36 cards, giving
36C8 = 36! / (28! x 8!) = 36 x 35 x . . . . 29 / 8 x 7 x 6 x . . . 1 . . . . . .(result E)
So total number of ways of getting one 7 = (D x E)
And total number of ways of getting either zero or one 7 = C + (D x E)
(You seem to be on the right track with your solution, but seem to be trying to combine the solutions to the different parts into one calculation, which does not work. Keep them all separate, and just combine them at the end, like above.)
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Your thinking for the solution to the last one is correct. It does not matter what the number on the card is, the result will be the same.
Since you know the result for less than two 7's, the result for two or more will be total number of combinations minus the result for less than two.
And the same result will apply to cards of any specific number, whether 7, or 8, or any other number.
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| Q. | OBSESSION WITH THE NEW ...please share your thoughts? | Related Search:
Philosophy | | | To demand that information tell the truth is to revert to a pre-industrial mode. Today there is no reality, or everything is real and everything is unreal. Today the object no longer refers to the real nor to information. Both are already the result of a selection, a montage, a taking of views. The role of messages is no longer information but a test - of success at interpreting the code according to the code for the perpetuation of the code. Thus the control problem is not one of surveillance, propaganda or paranoia. It is one of subjective influence, consent and extension to all possible spheres of life. The incorporation of the code into the corpse itself (Cf. Baudrillard: the "leucemisation" of all social substance, 103).
Modernity is not the transmutation of values - the myth of progress and change - but a commutation, combinatory and ambiguous.
In this process art and reality come to simulate each other. The dichotomy between real and imaginary collapse and commerce, the political, and the scientific become immersed more in aesthetics than in reality in the old sense. Symbols everywhere - ideologies, personalities, publicity - the new form of power. In politics as in art and culture, the obsession with the "new" is always limited to the rate of change tolerable without altering the essential order. And our lives, as works of art created by this public/ity code, participate in the same re-production.
"It remains to be seen if this operationality is not itself a myth, if DNA is not itself a myth." (Jean Baudrillard, L'Echange symbolique et la Mort, p94)
We are living at the beginning of an epoch which history will come to know as another Dark Age. But unlike the first one characterized by the concealment of information, we suffer from an almost opposite problem
- information overload.
The demand for more information is not radical - it is to demand exactly what the system already inundates us with.
| | A. | Louis, thru overload of manufactured information the rulers of our world shape and form the new all frightened human being. We are scared for our lives as crime rate escalates to new levels every day, we are anxious about tomorrow as the planet goes thru climate changes, we are worried about inflation as petrol price is sky-rocketing, we live in a constant manufactured fear because the elite few decision makers and power brokers move to impose upon us the new world order.
They don’t want us to know what really goes on right now therefore they stage stories, create news, conceal the actual events and keep us in the dark whichever way fits their agendas best.
New and change is encoded in our genes so is the drive to learn new things and change views as new data is added to our experience and awareness.
It is our solemn duty towards our self and fellow human beings to maintain the harmonics of love so as to clear our memory banks and allow true knowledge to surface.
The truth is always btw the lines, into the unsaid… | | | |
| Q. | How do you call the subject "sciences industrielles"in American education? | Related Search:
Other - Education | | | I am a french student and i want to know how do you call in the United States the subject which is called in education at High school or at University in France : "Sciences Industrielles"(in french) or "technologie"(in french).
It is a subject that addresses automatics, mechanics, kinematics, statics, construction (design ...).... etc.
I leave you the link to an exercise in cinematic that I posted a thread in a U.S. forum, ill-suited I think :
[Link] ...
There's no real forum in this area specializes in French gift I went to see the side of the American forums and I 've yet to find, may be you can help me.
I'm not sure to know how you call the subject which speak about automatons and binary code, combinatory and sequential logic.
I think you call this "automatics" like in France.
| | A. | j'crois pa k'ils ont cette matiére là, "technologie"
en secondaire en amerique, je croi k'ils choisi lé matiére k'ils veuleut étudier et la technoligie C pa une matiére necessaire
franchement, chui pa sur
mé....voila!!
hope i hepled
good luck | | | |
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