Risk Estimation Flashcards

1
Q

If A and B are two independent events, what is the formula for calculating the probability of:

  1. Either event A OR event B?
  2. Both event A AND event B?
A
  1. P[A] + P[B]

2. P[A] x P[B]

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2
Q

Sickle cell anaemia is a recessive condition. If two parents are both carriers of the sickle cell mutation what is the probability that the unborn child will be unaffected?

A
  • Have to do a punnet square of Ss x Ss
  • Gives 4 possible combos: SS, Ss, Ss or ss
  • Unaffected offspring are either homozygous wild type (ss = 1/4) or heterozygous (Ss = 2/4 or 1/2)
  • Prob of unaffected offspring = 1/4 + 1/2 = 3/4
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3
Q

What is an obligate carrier?

A

An individual who may be clinically unaffected but who must carry a gene mutation base on analysis of the family history

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4
Q

Haemophilia A is an X-linked recessive condition. If mum is a known heterozygous carrier, what is the prob that her unborn child will be affected?

A
  • An affected child must be BOTH male AND must inherit the mutant X allele from mum
  • Prob of male = 1/2 and prob of transmission of mutant allele = 1/2
  • Prob of male child AND transmission of mutant allele = 1/2 x 1/2 = 1/4
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5
Q

What is risk estimation?

A

In clinical practice this provides individuals with information regarding the probability that they (or members of their family) will develop disease in the future

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6
Q

Can additional information be used to modify risk estimates?

A

Yes - this involves Bayes Theorem

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7
Q

What is the basis of Bayes Theorem?

A

Combines information already available about he probability of an event (the prior probability) with new information to provide a revised probability (the posterior probability)

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8
Q

Why does estimating risk of being a carrier in a family with only one affected individual with an X-linked disorder require knowledge of the mutation rate?

A

Because in a number of X-linked recessive disorders, including haemophilia A and duchenne muscular dystrophy, de novo mutations are relatively common

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9
Q

Huntingtons disease is autosomal dominant. If our patient in question has an affected heterozygous mother and an unaffected father what is the risk of him having inherited Huntingtons disease and what is the risk that his children will inherit it?

A
  • His risk stems from punnet square of parents (Aa x aa) which gives the following: Aa, Aa, aa, aa. His risk is therefore 2/4 = 1/2
  • risk for his children involves his risk of being a carrier AND risk of passing it on = 1/2 x 1/2 = 1/4
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10
Q

How do you calculate risk where mode of inheritance is unknown?

A

Rely on empirical tables of recurrence risks to estimate risk of a second affected child e.g. Risk of a second child affected by spina bifida to a couple with one affected child = 4%

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