Conditional Probabilities Flashcards
What is the probability that II.5 is red head and female
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
What is the probability that II.5 is a carrier
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
What is the probability that III.1 is a red head
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
What is the probability that III.1 is a carrier
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
How to do conditional probability
- Look to see if there are multiple ways your event can occur (list them out)
- Do the probaility assuming that your asusumption is right X the probability that your assumiption actually occurs
- If there were multiple ways that each probability can occur then add of the individual probilities for each way together
Stratagies for solving Probabilities from pedigrees
- Determine the mode of inheritance
- Assign the known genotypes
- Determine all of the possible paths that could yeild the desired outcome and write down the proabbility eqautions
- calculate the probabilities of each conditional genotype
- Solve the total proababilty of the desired outcome
What traits can have incomplete dominance
You can have incomplete dominamnce for any pedigree – more common in mtDNA
Strategies for solving pedigrees
USE process of elimintaion – walk through each probability
- Sometimes there is a mode of inhertance that fits but another mode is more likley
What is the mode of inheritance for this pedigree?
A. Autosomal recesisve
What is the mode of inheritance for this pedigree?
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
What is the mode of inheritance for this pedigree?
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
What if theer are multiple modes of inheritance
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
Pedigrees follwoing Mendelian ratios
Pedigrees rarley follow Mendelian ratios perfectly
Parisomy
Choosing the option that takes the simplest explaintion with the fewest number of assumptions
What is the probability of the child of the probands being affected?
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
Probability work
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
Low of total probabilities
Probability of A given B X the porbability of B – sum all of the ways that you can produce an outcome you want
Example of incomplete Penetrance
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
What can lead to incomplete penetrance
Can have different gene expression level – can push you in different direction
What is the penetrance of this Autosomal Dominant Trait
5/6 = 83%
