# Homework Help: Inheritance: What is the chance of the next calf being amputated

1. Aug 29, 2012

### Swetasuria

1. The problem statement, all variables and given/known data
The absence of legs in cattle (amputated) has been attributed to a completly recessive lethal gene. A normal bull is mated with a normal cow and they produce an amputad calf (usually dead at birth). The same parents are mated again.
1) What is the chance of the next calf being amputated?
2) What is the chance of these parents having two calves both of which are amputated?

3. The attempt at a solution
1) 1/4
2) The chance of getting two amputated calves is 2/8. So answer is 1/4. My teacher says its 1/16 but I don't get it.

2. Aug 29, 2012

### Simon Bridge

Re: Inheritance

two independent trials with a 1/4 probability of something happening gives 1/16 probability of it happening both times.
To understand it - draw the tree.

How did you get 2/8 ?

3. Aug 29, 2012

### Swetasuria

Re: Inheritance

While having their first child, [favourable outcome (ie., amputated calf)]/[total number of outcomes]=1/4
Same chances with second child.

So, the probability of having 2 amputated calves= [total number of favourable outcomes]/[total number of outcomes]=2/8=1/4

4. Aug 29, 2012

### Simon Bridge

Re: Inheritance

1. I would hesitate to describe an amputated (and, therefore, dead) calf as a "favorable" outcome.

2. You have miscalculated the total number of possible outcomes as well as the number of those outcomes which are of interest.

lets say the first calf dies - this is one path to the final family.
then the final family is either
2 dead calves (second calf dies too) OR 1 dead and 1 alive

there are three ways the 2nd calf can live - it inherits the bad gene from dad alone, from mum alone, or from neither.
This makes a total of four possible families from the case that the first calf dies.
One of those possible families involves two dead calves.

Now lets say the first calf does not die. There are three ways this can happen.
There are still four possible families for each of these paths - some of them include a single dead calf because the second calf can still die. But none of them can result in 2 dead calves.

Since there are three paths leading to four families each - that is 3x4=12 possible families.

With the four from the path where the first calf dies, that is a total of 16 possible ways the family could end up.

Only one of those combinations has two dead calves.

Therefore - probability of 2 out of 2 dead (amputated) calves is 1/16.

This is a binomial distribrution: probability of getting k "successes" out of n trials with probability p per trial is:$$P(K=k) = \binom{n}{k}p^k(1-p)^{n-k}$$ so what is the probability of getting n "successes" out of n trials?

Last edited: Aug 29, 2012
5. Aug 29, 2012

### Swetasuria

Is k=2 (since we need two ampuated calves) and n=2 (parents breed two times)?
What is the value for p?1/4?

When I substitute these values, I get P=1/16

Last edited: Aug 29, 2012
6. Aug 29, 2012

### Simon Bridge

Cool - from that you should be able to figure out the probability of getting exactly one dead calf - of getting at least one dead calf - of a family where neither offspring inherited the bad gene? If you mated a new cow with one of the surviving offspring - what is the probability the cow is mating with a carrier of the bad gene?

Notice that 1/16 is also the probability of getting four heads out of four coin tosses.
It is worth spending some effort getting used to the way these things work.

7. Aug 30, 2012

### Swetasuria

If the offsprings did not inherit the bad gene, the probability of getting a dead calf is zero?

P=1/2 Correct?

8. Aug 30, 2012

### Simon Bridge

I think you need to practice combinations and permutations.

Go back to basics: each possible calf-type can be distinguished genetically by which parent it got it's gene from - lets designate the bad gene by x and any good gene by o and each calf is described by an ordered pair of genes "dad/mom" ... then I can denote a family of mom dad and two calves as two pairs of genes. The number of different possible families are:

xx xx, xx xo, xx ox, xx oo,

xo xx, xo xo, xo ox, xo oo,

ox xx, ox xo, ox ox, ox oo,

oo xx, oo xo, oo ox, oo oo.

Count them up - there are 16 possible pairs of calves.
One of the pairs is (xx xx) - which is both calves dead/amputated.
So - number of instances (refusing to call it a "success") divided by number of possibilities is 1/16. => 1/16 chance both calves are dead.

Exactly one dead calf is either (xx xo) (xx ox) (xx oo) and the reverse.
Count them all up - there's three where xx is first-born, and another three where xx is second born = 6. So the probability of exactly one dead calf is 6/16.

For "at least one dead calf" includes the possibility of two dead calves - probability is 7/16.

Since the number of pairs is 16, then the number of individual calves is 16x2=32. (note: I've counted a lot of combinations several times in this - because there are several paths to get those combinations.)

However, you can only mate live cows - how many live cows? (xx = dead)
How many did not inherit the bad gene (combination oo)?
Therefore - how many did inherit the x gene?
What is the probability of mating with a carrier of the x gene?

But we could have done it thus:
If the cow is alive then it is either xo ox or oo - for three total. Two of them have the x gene, so the probability is 2/3 ... however, the combinations may not be equally likely. Fortunately we have more information: we know about their parentage... just to be sure.

Last edited: Aug 30, 2012
9. Aug 30, 2012

### Swetasuria

Answers are 24/32=3/4, 8/32=1/4, 24/32=3/4 and 16/32=1/2 respectively.

I hope these are right. I notice that these are the answers I get if the family had only one child.

10. Aug 30, 2012

### Simon Bridge

So your answer for "how many live cows" is 24/32 ... so there is less than one live cow out of all 32 possible cows?

Also be careful not to include dead cows in your possible mating partners.
Note: your fractions are telling you that 3/4 of all cows born to these parents will live, 1/4 will not have the x gene, etc ... which, as you say, is not surprising since these are the odds you started out with.

But what you want to work out is the probability for mating a new cow with a partner who has the x gene.
Focus on what that means - does the new cow have 32 possible mating partners?

Last edited: Aug 30, 2012
11. Aug 30, 2012

### Swetasuria

Okay. So the answers are 24, 8, 24 and 16/24=2/3 respectively.

12. Aug 30, 2012

### Simon Bridge

There you go :)

You could also have pointed out that each of the possible options could be arrived at by 8 ways, so they each do actually have equal probability.

A good discipline when you get a number is to turn the question into the start of an answer and see if the number makes sense in it. eg. if the question was "how much wood would a woodchuck chuck?" and your number is 5kP, writing or just thinking to yourself: "A woodchuck would chuck 5 kiloPascals of wood." would alert you that something may be wrong and you need to check if you really did measure wood in terms of pressure.