Block 7 Repro Bayes Questions MCQ's

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Block 7 Repro Bayes Questions MCQ

Questions and Answers
  • 1. 

    II-3 in the family below  has two brothers with hemophllia and three healthy sons. How ikely is it that she is a carrier of the hemophilia gene?

    • A.

      1in 4

    • B.

      1 in 2

    • C.

      1 in 16

    • D.

      1 in 3

    • E.

      1 in 9

    Correct Answer
    E. 1 in 9
    Explanation
    Prior risk 1/2 then chance of having normal child 1/2. Because she had 3 normal it is 1/8.
    so then 1/2 x 1/8 gives 1/16, the other colum give 8/16 so 1/16/(1/16)+(8/16) gives 1/9

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  • 2. 

    In the family below, two brothers have classical hemophilia, assume that 1-4 (above) has two healthy sons. In that case you can use Bayes' theorem to figure out that her chance of being a carrier is

    • A.

      10%

    • B.

      20%

    • C.

      50%

    • D.

      12.5%

    • E.

      33%

    Correct Answer
    B. 20%
    Explanation
    Based on the information given, it is stated that two brothers have classical hemophilia. Classical hemophilia is an X-linked recessive genetic disorder, which means it is carried on the X chromosome. Since the two brothers have hemophilia, it can be inferred that their mother must be a carrier of the gene. Therefore, the chance of the mother being a carrier is 50%. However, since the question specifically asks for the chance of her being a carrier, we need to consider that she has two healthy sons. This information reduces the probability by half, resulting in a 20% chance of her being a carrier.

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  • 3. 

    Your father has a dominantly-inherited form of retinitis pigmentosa, an eye disease that leads to blindness in middle-aged people. Now you are 25 years old, and you took a vision test to check whether you are developing this disease. In two thirds of all people your age who carry the gene, the vision test is abnormal while in one third it is still normal. Your test was normal. How LIKELY is it that you carry the gene?

    • A.

      1 in 2

    • B.

      2 in 3

    • C.

      1 in 16

    • D.

      1 in 4

    • E.

      1 in 6

    Correct Answer
    D. 1 in 4
    Explanation
    prior risk of 1/2, you have a 1/3 chance of being normal if you carry the gene. therefore you multiply your prior risk of 1/2 by 1/3 to give you 1/6 (so you know that is not the final answer) then the right colum is 3/6. so divide the 1/6 by the sum of 1/6 and 3/6 to give you 1/4.....ta da!!!!!

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  • 4. 

    Your patient's father died of Huntington's disease. Now the patient is 45 years old and asks you whether he can still get the disease. Two thirds of all patients who carry the Huntington's gene already show signs of the disease by age 45, but your patient is still healthy. Based on this information, how likely is it that he will get the disease?

    • A.

      1 in 2

    • B.

      1 in 3

    • C.

      1 in 9

    • D.

      1 in 4

    • E.

      1 in 6

    Correct Answer
    D. 1 in 4
    Explanation
    Based on the information provided, it states that two-thirds of all patients who carry the Huntington's gene show signs of the disease by age 45. However, the patient in question is still healthy at 45 years old. This suggests that the patient falls into the one-third of individuals who do not show signs of the disease by this age. Therefore, the likelihood of the patient getting the disease can be estimated as 1 in 4.

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  • 5. 

    The pedigree below segregates for Duchenne muscular dystrophy (DMD), an X-linked recessive condition lethal in males. What is the risk for III-3 of being a carrier for DMD ?

    • A.

      1 in 2

    • B.

      1 in 3

    • C.

      1 in 9

    • D.

      1 in 10

    • E.

      1 in 6

    Correct Answer
    D. 1 in 10
    Explanation
    moms risk is 1/5 therefore her risk is 1/2 of that.

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  • 6. 

    Retinoblastoma (Rb), a malignant tumor of the eye may be inherited as an autosomal dominant trait. Penetrance is about 4/5 (80%). Michael's paternal grandmother, father and brother had Rb, but Michael is unaffected. What is the risk that Michael's child may develop retinoblastoma?

    • A.

      1 in 2

    • B.

      1 in 3

    • C.

      1 in 9

    • D.

      1 in 15

    • E.

      1 in 6

    Correct Answer
    D. 1 in 15
    Explanation
    Michael's risk is 1/6 so multiply by 1/2 (chance of passing the gene) and 4/5 (penetrance)
    1/6 x 1/2 x 4/5 = 4/60= 1/15

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  • 7. 

    Sydney's paternal grandfather died of Huntington disease (HD). Her father is 50 years old and is unaffected. Her brother was diagnosed at 32 years of age with HD. Sydney is now 30 and wants to know her chance to develop HD. (Approximately 10% of HD gene carriers are clinically affected by 30, 50% by age 50, and almost 100% by age 70.)

    • A.

      3/26

    • B.

      3/92

    • C.

      1/6

    • D.

      1/352

    • E.

      9/19

    • F.

      4/22

    Correct Answer
    E. 9/19
    Explanation
    Based on the information given, Sydney's paternal grandfather had Huntington disease (HD), her father is unaffected, and her brother was diagnosed with HD at the age of 32. It is mentioned that approximately 10% of HD gene carriers are clinically affected by the age of 30. Therefore, Sydney has a 10% chance of developing HD by the age of 30. The answer 9/19 suggests that Sydney has a higher chance (9 out of 19) of developing HD by the age of 30, based on the given statistics.

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  • 8. 

    Kim and Peter have two healthy children but Kim's sister has a child recently diagnosed with CF. Peter has no family history of CF. The population carrier frequency is 1/25. What is the chance that Kim and Peter's next child will have CF?

    • A.

      3/26

    • B.

      3/92

    • C.

      1/6

    • D.

      1/352

    • E.

      9/19

    • F.

      4/22

    Correct Answer
    D. 1/352
    Explanation
    __________Both Carriers___________ Not Both Carriers_____________
    Prior.............1/25 x 1/2=1/50.....................1 - 1/50=49/50
    Conditional ....(3/4)2=9/16 ......................................16/16
    Joint .........................9/800...................................784/800
    Posterior....................9/793 (= ~1/88*)..................784/793
    ________________________________________________________________
    risk of having an affected child therefore is 1/88 x 1/4 = 1/352

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  • 9. 

    Jane is 28 weeks pregnant and consults a genetic counselor because her brother died of severe X-linked hydrocephalus (aqueductal stenosis). She has two unaffected brothers. An ultrasound found that her fetus is a male. Jane already has one unaffected son and one unaffected daughter. What is the chance that Jane's fetus will be affected? (Mother of an isolated case of an X-Inked lethal condition has 2/3 chance of being a carrier (1/3 isolated cases are new mutations).

    • A.

      1/22

    • B.

      3/92

    • C.

      1/6

    • D.

      1/352

    • E.

      9/19

    • F.

      4/22

    Correct Answer
    A. 1/22
    Explanation
    The chance that Jane's fetus will be affected is 1/22. This is because the question states that Jane's brother died of severe X-linked hydrocephalus, which implies that he was affected by the condition. Since Jane has two unaffected brothers, it is likely that her mother is a carrier of the X-linked gene. The chance of Jane being a carrier is 2/3, and since she already has one unaffected son and one unaffected daughter, it is likely that she is a carrier. Therefore, the chance of Jane passing on the affected gene to her fetus is 1/2, and the chance of the fetus being affected is 1/2 x 2/3 x 1/2 = 1/6. However, the question also states that 1/3 of isolated cases are new mutations, so the chance of the fetus being affected by the inherited gene is 2/3 x 1/6 = 1/22.

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  • 10. 

    A family is affected with Huntington disease, an adult-onset neurologic disorder. Fifty percent (50%) of people who inherit the HD gene show some symptoms of HD by the age of 50.  Betty is 50 and asymptomatic, but her father died of HD. What is the probability that Betty's son, John, inherited the HD gene?

    • A.

      1/22

    • B.

      3/92

    • C.

      1/6

    • D.

      1/352

    • E.

      9/19

    • F.

      4/22

    Correct Answer
    C. 1/6
  • 11. 

    Amanda has a son with Duchenne muscular dystrophy (DMD). There is no other family history of DMD. She has one healthy son and one healthy daughter. She also had a normal CPK level (70% of DMD carriers have a high CPK). What is amanda's daughter's chance to be a DMD carrier?

    • A.

      1/22

    • B.

      3/92

    • C.

      1/6

    • D.

      3/26

    • E.

      9/19

    • F.

      4/22

    Correct Answer
    D. 3/26
    Explanation
    _________________Carrier____________ Non-Carrier
    Prior..............................2/3 ................................1/3
    Conditional #1 .............3/10.............................10/10
    Conditional #2 .............1/2............................... 2/2
    Joint ............................6/60 ...........................20/60
    Posterior ....................6/26=3/13* .................10/13

    3/13 x 1/2 = 3/26

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  • 12. 

    Carol has a son with Duchenne muscular dystrophy. There is no other family history of DMD. The woman's CPK IS normal (70% of obligate DMD carriers have high CPK's). The woman also has two healthy sons. What is the chance that Carol's next child will have DMD?

    • A.

      1/22

    • B.

      3/92

    • C.

      1/6

    • D.

      1/352

    • E.

      9/19

    • F.

      4/22

    Correct Answer
    B. 3/92
    Explanation
    Based on the information provided, the chance that Carol's next child will have DMD is 3/92. This can be inferred from the fact that Carol has a son with DMD and there is no other family history of the condition. The woman's normal CPK levels also suggest a lower likelihood of being an obligate carrier. Therefore, the chance of having a child with DMD is relatively low, resulting in a probability of 3/92.

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  • 13. 

    Rebecca has a sister with cystic fibrosis. Her husband, Paul, has no family history of CF. The couple already has one healthy child . What is the chance that their next child could have CF?  

    • A.

      1/22

    • B.

      3/92

    • C.

      1/6

    • D.

      1/352

    • E.

      1/199

    • F.

      4/22

    Correct Answer
    E. 1/199
    Explanation
    ________________Both Carriers_______________ Not Both Carriers_______
    Prior ..........................1/25 x 2/3=2/75..........................1 - 2/75=73/75
    Conditional ........................3/4 .....................................................4/4
    Joint ..................................6/300..............................................292/300
    Posterior ...........................6/298 (= 3/149) ..............................292/298
    ________________________________________________________________
    risk of having an affected child therefore is 3/149 x 1/4 = 3/596
    approximately 1/199

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