probability in genetics

mikau · 2007-10-24 10:01:50

this is the root cause for my last thread.

In genetics, as some of you may know, we often inherit recessive traits from our parents, but they are not manifested if we get a dominant trait from our other parent. However, the recessive traits that are not manifested are still contained in us and can be passed on to our children, thus given them characteristics which we don't have.

Basically the way it works is this. For blue eyes and brown eyes, brown eyes is a dominant trait (B) and blue eyes are a recessive trait (b).

Now there are 3 possible combinations for blue and brown traits within a person:

BB, Bb, bb (note, order is not considered)

if a person has BB or Bb, they posess the dominant brown eyed trait B. The only way a person can be blue eyed is if they are both blue eyed traits (bb).

Now here's where the probabity kicks in. If a person with one Brown eyed trait and one blue eyed trait (Bb, which means he/she has brown eyes) mates with a person who is also Bb, then what happens is one random gene from the first parent is paired with one random gene fom the second parent, and this pair is given to the offspring. To work out the probability and possibility for certain combinations, they usually draw what they call a punett square
_B_b
B
b
then consider the intersections of each row and collumn as a possible combination. From this we see the possible combinations are
BB, Bb, Bb, and bb. There is therefore a 75% chance that the offspring will be brown eyed, and 25% that they will be blue eyed. Moreover, there is a 25% chance they will be BB specifically, a 50% chance they will be Bb, and a 25% chance they will be bb.

Its all rather simple really. Hope that makes sense. They've been discussing this for like 2 weeks in my biology class but it seems there is little more to it than this.

What i'm wondering is what do you do if you have a problem like this:
suppose a brown eyed man and a blue eyed woman have a child. What is the probability that the child will be BB, bb, or Bb?

This problem is a bit difficult because while we know the mother is bb, (else she wouldn't have blue eyes) the father, being brown eyed, could be either BB, or Bb. So what is the probability?

If the father is BB, then we end up with a 100% chance of Bb, (BB crossed with bb, take one from each pair and you always get bb)

If the father is Bb, then we end up with a 50% chance of bb, and a 50% chance of Bb, (Bb crossed with bb, take one from each pair and you either get Bb or bb)

so the question is, 100% chance of Bb if the father is BB, and 50/50 chance of Bb or bb if the father is Bb. Note, there is a 50/50 chance of the father being BB or BB.

so now what do we do? We have (Bb) or (bb) evenly distributed on one half of the scale, and only Bb on the other half. But both halfs can't happen at once. So what do we do now?

Moreover, suppose it says both parents are brown eyed. Well that could mean parent 1 is Bb while parent two is Bb, or it could mean parent 1 is BB and parent 2 is BB OR it could mean parent 1 is BB or parent 2 is Bb! Each of these scenarios brings with them their own set of probabilities for their respective offspring. So what is the probability all four considered? That is, what is the probability that a given combination (such as Bb) will occur?

there will probably not be questions of this calibur on the test but, i want to be ready for it just in case.

Last edited by mikau (2007-10-24 10:18:27)

bossk171 · 2007-10-24 10:08:34

Personally I would just make all the Punett squares needed the write down all the possible answers. It's tedious and probably the longest possible method, but if you can spare the time (ie the test is not timed), it's a good way to know for sure you're right.

mikau · 2007-10-24 10:23:22

All the possible answers you say? You mean like this?

If its THIS probability is this, if its THAT probability is that, etc. ???

My gut feeling is that i should expand every possibility and write all the occurances of a possible trait divided by the total number of possibilities. BUT since only parent combination can occur, i'm not sure if this counts.

Probability is really my greatest weakness.

John E. Franklin · 2007-10-24 10:49:47

Don't forget that you need to know what is the genetic-trait percentages for the racial group you are considering. If you don't know race or gender or other traits, then you need some statistical average for the population. Because where we are now is best described by our present data.
For brown eyed people, what is the BB and Bb breakdown?
For hazel eyed people, what is the breakdown?

Last edited by John E. Franklin (2007-10-24 10:50:36)

bossk171 · 2007-10-24 11:01:29

Have you ever taken a formal probability class? I haven't, what little I know is ether intuitive, or I've read about it somewhere.

You have two possible punett squares:

Square A:                            

      |  b   |  b
  ----+------+-----
    B |  Bb  |  Bb
  ----+------+-----    
    b |  bb  |  bb

Square B:                            

      |  b   |  b
  ----+------+-----
    B |  Bb  |  Bb
  ----+------+-----    
    B |  Bb  |  Bb

So the odds are

(6/8) = 75% Bb (Brown)
and
(2/8) = 25% bb (Blue)

Assuming, that is, that there is a 50%-50% chance that the Dad is brown either dominate or recessive.

mathsyperson · 2007-10-24 11:46:50

That's a pretty big assumption though.
The best you can do is use the eye-colours of the father's ancestors to estimate the probability distribution, and even then you'll never get an entirely accurate answer.

bossk171 · 2007-10-24 11:55:14

I agree, that's a huge assumption. He did however say, "Note, there is a 50/50 chance of the father being BB or Bb."

He actually said "BB or BB" but I'm assuming he meant "BB or Bb"

mikau · 2007-10-24 15:53:45

well suppose we are willing to assume that all variations that satisfy the condition for the parent are evenly distrubted. (that is, if the parent is Brown eyed, we assume a 50/50 chance of Bb or BB).

My teacher stated repeatedly that dominance and recessiveness has nothing to do with how many exist in a community. That seems a bit contrary to reason for me, (certainly, it can't have NOTHING to do) but if I use his argument in my calculations it shouldn't be marked wrong.

Now the way i'm approach it is, say two parents have Brown eyes, there are therefore, 3 possibilities for parents
BB-BB
Bb-Bb
BB-bb

crossing each produces
BB,BB,BB,BB

BB,Bb,Bb, bb

Bb,Bb,Bb, Bb
respectively (in these proportions)

therefore, the probability of BB should be 1/3(4/4 + 1/4 + 0) = 5/12

of Bb, should be 1/3(0 + 2/4 + 4/4) = 6/12

and bb should be 1/3(0 + 1/4 + 0) = 1/12

this results in 12/12 alltogether.

Seems right, doesn't it? Assuming the parent selection is totally random?

Last edited by mikau (2007-10-24 15:56:25)

bossk171 · 2007-10-24 16:17:55

It seems right to me, although like I said before, I have no formal training in statistics. It certainly "feels" right though.

mathsyperson · 2007-10-24 23:42:49

mikau wrote:

Now the way i'm approach it is, say two parents have Brown eyes, there are therefore, 3 possibilities for parents
BB-BB
Bb-Bb
BB-bb

That last one should be BB-Bb. Plus, that one is twice as likely as the other two, so it needs to be weighted.

So then you get:

BB, BB, BB, BB
BB, Bb, Bb, bb
2(BB, BB, Bb, Bb)

Which gives P(BB) = 9/16, P(Bb) = 3/8 and P(bb) = 1/16.

mikau · 2007-10-25 02:22:00

oops, yeah it was supposed to be Bb.

Hmm.. so BB-Bb is twice as likely you say?

BB with (BB or Bb)
or
Bb with (BB or Bb)

BB-BB
BB-Bb
Bb-BB
Bb-Bb

you're right!

so lets see now if we have blood types A (A0 or AA) and B (B0 or BB), we have
A0 with (BB or B0)
or
AA with (BB or B0)

so
A0-BB
A0-B0
AA-B0
AA-BB

so each set of probabilities counts for 1/4 the total probability. Correct?
If so thats pretty much everything I needed to know.

mikau · 2007-10-26 02:04:51

WELL i took the test yesterday and i believe i may have aced it, assuming i didn't make any computational errors.

Thanks for your help, guys!

It just goes to show you having an edge in math can help you out in more places than one. Incidently, next weeks lab activity was changed because the original activity required "too much math" according to the teacher. So instead we are going to to be dissecting pig fetuses.

See what mathematical illiteracy leads to?

sunil verma · 2007-11-09 18:48:33

bossk171 wrote:

Have you ever taken a formal probability class? I haven't, what little I know is ether intuitive, or I've read about it somewhere.
You have two possible punett squares:
Square A:                            

      |  b   |  b
  ----+------+-----
    B |  Bb  |  Bb
  ----+------+-----    
    b |  bb  |  bb

Square B:                            

      |  b   |  b
  ----+------+-----
    B |  Bb  |  Bb
  ----+------+-----    
    B |  Bb  |  Bb
So the odds are
(6/8) = 75% Bb (Brown)
and
(2/8) = 25% bb (Blue)
Assuming, that is, that there is a 50%-50% chance that the Dad is brown either dominate or recessive.

Math Is Fun Forum

#1 2007-10-24 10:01:50

probability in genetics

#2 2007-10-24 10:08:34

Re: probability in genetics

#3 2007-10-24 10:23:22

Re: probability in genetics

#4 2007-10-24 10:49:47

Re: probability in genetics

#5 2007-10-24 11:01:29

Re: probability in genetics

#6 2007-10-24 11:46:50

Re: probability in genetics

#7 2007-10-24 11:55:14

Re: probability in genetics

#8 2007-10-24 15:53:45

Re: probability in genetics

#9 2007-10-24 16:17:55

Re: probability in genetics

#10 2007-10-24 23:42:49

Re: probability in genetics

#11 2007-10-25 02:22:00

Re: probability in genetics

#12 2007-10-26 02:04:51

Re: probability in genetics

#13 2007-11-09 18:48:33

Re: probability in genetics

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