Inheritance: What is the chance of the next calf being amputated

  • Thread starter Swetasuria
  • Start date
Okay - it looks like you got the idea.Now - what is the probability of not getting an amputated calf from a single mating?The probability of not getting an amputated calf from a single mating would be 3/4, since there are 3 out of 4 possible combinations that do not result in an amputated calf.
  • #1
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Homework Statement


The absence of legs in cattle (amputated) has been attributed to a completely 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?


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.
 
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  • #2


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


Simon Bridge said:
How did you get 2/8 ?

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


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 let's 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:[tex]P(K=k) = \binom{n}{k}p^k(1-p)^{n-k}[/tex] so what is the probability of getting n "successes" out of n trials?
 
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  • #5
Is k=2 (since we need two ampuated calves) and n=2 (parents breed two times)?
What is the value for p?1/4?:confused:

When I substitute these values, I get P=1/16 :smile:
 
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  • #6
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
Simon Bridge said:
...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 the offsprings did not inherit the bad gene, the probability of getting a dead calf is zero?


Simon Bridge said:
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?

P=1/2 Correct?
 
  • #8
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 - let's 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.
 
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  • #9
Simon Bridge said:
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?

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
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?

Take care to make sure your answer matches the question.
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?
 
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  • #11
Okay. So the answers are 24, 8, 24 and 16/24=2/3 respectively.
 
  • #12
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.
 

What is inheritance?

Inheritance is the passing down of genetic traits from parents to offspring. This includes physical characteristics, such as eye color and height, as well as potential health conditions.

What is the chance of inheriting a specific trait?

The chance of inheriting a specific trait depends on the type of trait and the genetic makeup of the parents. Some traits, such as eye color, are determined by a single gene and have a predictable inheritance pattern. Other traits, like height, are influenced by multiple genes and can be more difficult to predict.

What is the likelihood of inheriting a health condition?

The likelihood of inheriting a health condition depends on several factors, including the presence of the condition in the parents, the type of inheritance pattern, and environmental factors. Some conditions, like cystic fibrosis, are caused by mutations in a single gene and have a 25% chance of being inherited if both parents carry the mutated gene.

Can inheritance be influenced by environmental factors?

Yes, environmental factors can play a role in how traits are expressed. For example, a person may have a genetic predisposition for a certain health condition, but if they lead a healthy lifestyle and avoid certain environmental triggers, they may never develop the condition.

Is it possible to predict the inheritance of a specific trait in offspring?

Predicting the inheritance of a specific trait can be challenging, as it depends on various factors including the genes of the parents, the type of inheritance, and environmental factors. While some traits can be predicted with a high level of accuracy, others may be more difficult to determine.

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