014 CAL clinical genetics scenarios

Question 1

what is the definition of penetrance?

Answer 1

• the proportion of people who show symptoms of the disease when they have the appropriate genotype
• e.g. penetrance of 70% = 70% of people with the genotype will get disease symptoms

Question 2

what is the Bayesian Theorem?

Answer 2

Bayes’ Theorem states that the conditional probability of an event, based on the occurrence of another event, is equal to the likelihood of the second event given the first event multiplied by the probability of the first event.

Question 3

what is a genetic prior probability?

Answer 3

the probability of an event occurring before new data is collected. In other words, it represents the best rational assessment of the probability of a particular outcome based on current knowledge before an experiment is performed.
e.g. prior probability of Jasmine being heterozygous = 0.5

Question 4

what is genetic conditional probability?

Answer 4

the likelihood of an event or outcome occurring, based on the occurrence of a previous event or outcome
- e.g. if penetrance = 70%, there is a 30% chance jasmine is a symptomless carrier (0.3)

Question 5

what is genetic joint probability?

Answer 5

• the probabiloty of the co-occurance of the prior and conditional probability ( multiply together)
• e.g. for jasmine prior = 0.5, conditional = 0.3 so joint probability = 0.5 x 0.3 = 0.15

Question 6

what is genetic posterior probability?

Answer 6

• the final probability
  = divide the joint probability that jasmine is a carrier by the sum of the 2 joint probabilities ( is and isn’t a carrier)
• e.gg. 0.15 / 0.15 + 0.5 = 0.23
  (23% chance jasmine is a carrier of the disease with no symptoms)

Question 7

what is the probability of jasmine having an affected child?

Answer 7

• jasmine = symptomless carrier = 23% chance
• punnet square risk of having a child with the autosomal dominant disease when 1 parent is carrier = 50% chance
• 70% penetrance if the child does have dominant allele
• total chance = 0.23 x 0.5 x 0.7 = 8% chance

Question 8

what would be the karyotype of a female child with Down syndrome due to an unbalanced Robertsonian Translocation (chromosome 21 - chromosome 14)?

Answer 8

• 46, XX, t(14q:21q)

Question 9

would it be possible to have phenotypically healthy carries of translocations?

Answer 9

• yes, carriers may have balanced translocations, so the cell still has complete genetic information of the correct amount so could be phenotypically normal

Question 10

what is the frequency of carriers of cystic fibrosis if the prevalence of CF is 1/1600?

Answer 10

• hardy weinburg = p^2 + 2pq + q^2
• q^2 = 1/1600
• q = 0.025
• p = 1-0.025 = 0.975

carriers = 2pq = 2 x 0.975 x 0.025 = 0.04875 = 0.05