5 & 6 Problem Solving

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  1. Probability
    the proportion of times a specific outcome occurs in a series of events

    eg. in a pregnancy: 1/2 girl, 1/2 boy
  2. Independent Events
    the outcome of one event does NOT depend on outcome of other event

    eg. a previous pregnancy does NOT affect a future one

    each conception: 1/2 boys, 1/2 girls
  3. Multiplication Rule
    if 2 events are INDEPENDENT of each other, the probability of outcome of both occurrences happening is the product of each outcome

    eg. the chance of having 3 children & all of them be girls = 1/2 * 1/2 * 1/2 = 1/8
  4. Addition Rule
    • the probability of one outcome OR another is the addition of each outcome
    • boy or girl = 1/2 + 1/2 = 1

    • • probability of getting EITHER 2 heads in a row OR 2 tails in a row =
    • (1/2 * 1/2) + (1/2 * 1/2) = 1/4 + 1/4 = 1
  5. p
    frequency of the dominant ALLELE in a population (present in people with both AA & Aa)
  6. q
    frequency of the recessive ALLELE in a population (present in people with both Aa & aa)
  7. p+q =
    1
  8. q2
    how many people have the aa genotype

    • the incidence of homozygous recessive people in a population

    • eg. CF (AR) has an incidence of 1/2500
    • - that means 1 in every 2500 people HAS CF, so q2 = 1/2500
  9. 2pq
    how many people have the Aa genotype

    • the incidence of heterozygous people in a population

    • what you’re looking for if someone asks you what a CARRIER RATE is
  10. p2
    how many people have the AA genotype

    • the incidence of homozygous dominant people in a population
  11. How to calculate the chance that ALL children of two AR disease carriers are affected:
    multiply each of their chances

    eg. for 2 siblings: 1/4 * 1/4 = 1/16 chance that all are affected
  12. How to calculate the chance that ALL children of two AR disease carriers are UNaffected
    multiply each of their chances

    eg. for 2 siblings: 3/4 * 3/4 = 9/16 chance that none of them are affected
  13. How to calculate the chance that AT LEAST 1 child will be affected:
    1 — NONE

    eg. from above example, 1 - 9/16 = 7/16 chance at least 1 child will be AFFECTED
  14. When Penetrance can be Measured
    • eg. if Penetrance = 60%, then there’s a 60% chance each affected individual expresses the trait
  15. What risk of complications/pregnancy loss does Amniocentesis present to a mother?
    ~ 0.5%

    • if a child’s risk for being affected by a condition is LOWER than the risk of Amniocentesis, it’s NOT offered

Card Set Information

Author:
mse263
ID:
322695
Filename:
5 & 6 Problem Solving
Updated:
2016-08-27 02:04:05
Tags:
MedFoundationsI Genetics Exam1
Folders:
MedFoundationsI,Genetics
Description:
Genetics Exam 1
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