Conditional Probabilities Flashcards

1
Q

What is the probability that II.5 is red head and female

A

1/8

Because the probability of the child being red-head and female are indepent – can multiple their indivdiual probabilities

1/4 X 1/2 = 1/8

P(red head) = 1/4 – because parents are Aa X Aa
P(Femailes) = 1/2

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2
Q

What is the probability that II.5 is a carrier

A

2/3 – tempting to say 50% because Aa X Aa shows that Aa is 50% of fofspring BUT we know that is is NOT affecets so we KNOW that she is NOT aa = know that she is either AA or Aa = only look at those options for genotypes

2Aa: 1 AA = 2/3 Aa and 1/3 AA –> P(Aa) = 2/3

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3
Q

What is the probability that III.1 is a red head

A

There is only one way for III.1 to be a red head –> can only happen if both of the parents are Aa BUT we don’t know what the genotypes of the parents = NEED TO USE CONDITIONAL PROBABILITY

The parents = can be AA OR Aa
P(mom is Aa) = 2/3 (because we know she is NOT aa = 2/3)

P(dad is Aa) = 2/3

P(III.1) is red requires that mom and dad is Aa

THEN use conditional poprbability

P(III.1 is aa given mom is Aa AND dad is Aa) = Proabbility that III.1 is aa (assuming that mom is Aa and dad is Aa) X probvaility that mom AND dad are Aa = 1/4 X 4/9 = 4/36 = 1/9

***Overall – do the probaility assuming that your asusumption is right X the probability that your assumiption actually occurs

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4
Q

What is the probability that III.1 is a carrier

A

Step #1 – need to look at all of the ways that this can happen

HERE: there are 3 ways that this can happen
1. Mom and Dad can be Aa
OR
2. Mom can be Aa and dad can be AA
3. Mom can be AA and dad can be Aa

EACH of these have a conditional probability – need to find the probabilities of each and add the three together

ANSWER: 4/9

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5
Q

How to do conditional probability

A
  1. Look to see if there are multiple ways your event can occur (list them out)
  2. Do the probaility assuming that your asusumption is right X the probability that your assumiption actually occurs
  3. If there were multiple ways that each probability can occur then add of the individual probilities for each way together
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6
Q

Stratagies for solving Probabilities from pedigrees

A
  1. Determine the mode of inheritance
  2. Assign the known genotypes
  3. Determine all of the possible paths that could yeild the desired outcome and write down the proabbility eqautions
  4. calculate the probabilities of each conditional genotype
  5. Solve the total proababilty of the desired outcome
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7
Q

What traits can have incomplete dominance

A

You can have incomplete dominamnce for any pedigree – more common in mtDNA

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8
Q

Strategies for solving pedigrees

A

USE process of elimintaion – walk through each probability
- Sometimes there is a mode of inhertance that fits but another mode is more likley

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9
Q

What is the mode of inheritance for this pedigree?

A

A. Autosomal recesisve

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10
Q

What is the mode of inheritance for this pedigree?

A

B. X-link Dominenet

Key = Affected males pass down to ALL of the daughters and no sons

IF X link dominent = pass to all daughters and no son
IF Autosmal dominent = passes equally to males and females

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11
Q

What is the mode of inheritance for this pedigree?

A

D. Xlink dominent

NOTES:
- Affected males pass only to daughters and no sones = more likley X lin dominenet = better fit

***Technically Autosomal dominent also works BUT since only daughteer het it from males = more likley X link dominent

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12
Q

What if theer are multiple modes of inheritance

A

Choose the option that takes the simplest explaintion with the fewest number of assumptions and be willing to reeavulate as more data is presented

***Use parisimony

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13
Q

Pedigrees follwoing Mendelian ratios

A

Pedigrees rarley follow Mendelian ratios perfectly

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14
Q

Parisomy

A

Choosing the option that takes the simplest explaintion with the fewest number of assumptions

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15
Q

What is the probability of the child of the probands being affected?

A

Need to know:
1. Allele relationships – need to know the mode of inheritance
- Know that it is recessive
2. Need to know the parent genotypes
A_ X Aa
- Parents can be AA X Aa or Aa X Aa (Because we know that the unaffceted indioviauk is AA or Aa but don’t know which they are)
***Know that the dad has to be Aa because his mom needs to be aa and we assume that the dad is AA because arre trait
P(IV.1 Affected)
Have 1 options that works to have an aa child:
Aa X Aa
- Parents making an offspring aa depends on the gebotype of teh mom (if she is Aa or AA – we don’t know which she is) = KNOW THAT IT IS CONDITIONAL PROBABILITY
- Probability that the offspring will be affected is conditionla on teh genotype of the mother

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16
Q

What do we need to know to answer the questions

A
17
Q

Probability work

A

Need to look at the possible crosses:

AA X Aa OR Aa X Aa

THEN need to pick the crosses that work – we know that only Aa X Aa can give an aa child

NOW calculate the conditional probability:
P(child is aa GIVEN than mom is Aa) = Pribability that child is aa assuming that mom is Aa) X the proabbility that mom is Aa

P(Mom Aa) = 2/3 – because Aa X Aa parents and know she is NOT aa = she can only be AA or Aa = 2/3

Pribability that child is aa assuming that mom is Aa = 1/4

P(aa) = 2/3 X 1/4 = 2/12 = 1/6

18
Q

When do you have a conditional probability

A

When there is a conditional genotype – if an outcome is dependent on the genotype of the parent being a specific one but there are multiple options for that parents genotype

19
Q

Conditional probability general equation

A

P(A given B) = P(A assuming B occurs) X P(B)

IF there are multiple ways for the outcome to occur – find the conditional probabilities for each of them and add individual probabilities together

20
Q

Low of total probabilities

A

Probability of A given B X the porbability of B – sum all of the ways that you can produce an outcome you want

21
Q

Ruling out genotypes + Probabilities

A

IF you KNOW that one of the genotypes is NOT possible = can rule it out and change the probability

Example – if you know an indiviual is not affected by recesisve trait = know that they are AA or Aa = can rule out aa = can just take into account AA an d Aa

Aa X Aa –> Possible Genotypes are AA or Aa –>
AA = 1/3
Aa = 2/3

22
Q

Example of incomplete Penetrance

A

Because it does not fit any of the standrad modes
- The mode it fits most is Autosomal Recessive BUT that requires too many assumptions – because it requires too many assumptions we say incplomete penetrance

SINCE Kids have unaffected parents = incomplete penetrance for autosomal Dominant

23
Q

What can lead to incomplete penetrance

A

Can have different gene expression level – can push you in different direction

24
Q

Example Incomplete penetrance disease

A

Mafan Syndomre

HAVE a range of phenotypes – can be very functional or can be very affected
- Low end = might not have typical phenotypes = incopmpete penetrance
- Because it is a bog pathway = not all or nothig you can have otehr genes that make up for it – can have otehr things at play (can have many genes)

Becasuse there is this range in phenotypes = incomplete penetrance (Not all or nothing)

25
Q

Penetrance

A

The proportion of individuals with a particlay genotype that alsp express the trait

- The proprotion with the genotype that actually express the trait
26
Q

Inomplete penetrance

A

When some indovdiuals with a partcular genotype express the trait but others do not

27
Q

What is the penetrance of this Autosomal Dominant Trait

A

5/6 = 83%

28
Q

In a Chi Squared GOF test the greater the difference between the observed and expected values the _____ the P-Value

A

Lower

As deviations between the O and E values increase the Chi Square valued increases – as teh Ch- Square value increases the more condifentley you can reject the null = P-value decreases