Pedigree Quiz: Test Your Knowledge About The Pedigree Diagram

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Pedigree Quiz: Test Your Knowledge About The Pedigree Diagram - Quiz

Think you are a Biology buff? Do you want to test your genetic knowledge? This quiz is based on the pedigree diagram that represents biological relationships between an organism and its ancestors. Ever wondered how much you know about the pedigree diagram. Let's play this quiz and prove yourself!


Questions and Answers
  • 1. 

    II-3 in the pedigree below has two brothers with hemophilia A, a bleeding disorder that is inherited as an X‑linked recessive trait. What is the risk of hemophilia for her children?  

    • A.

      1 in 4 for a son, close to zero for a daughter

    • B.

      1 in 2 both for sons and daughters

    • C.

      1 in 2 for a son and 1 in 4 for a daughter

    • D.

      1 in 2 for a son, close to zero for a daughter.

    • E.

      1 in 4 both for sons and daughters

    Correct Answer
    D. 1 in 2 for a son, close to zero for a daughter.
    Explanation
    The correct answer is: 1 in 2 for a son, close to zero for a daughter.
    If a mother is a carrier for hemophilia and the father does not have hemophilia, each son has a 1 in 2 (50%) chance of inheriting his mother’s hemophilia allele and having hemophilia. Each daughter also has a 1 in 2 (50%) chance of inheriting her mother’s hemophilia allele and being a carrier. However, females with one hemophilia allele usually have a normal allele on their other X chromosome that can produce normal clotting factor, so they have some protection against having hemophilia. Therefore, the risk for a daughter to have hemophilia is close to zero.

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  • 2. 

    II-3 in the below family has two brothers and three sons with classical hemophilia (factor VIII deficiency). Now she is pregnant again. How likely is it that this child will also have hemophilia?

    • A.

      100% for a son and 50% for a daughter

    • B.

      100% for a son, zero for a daughter

    • C.

      50% for a son, zero for a daughter

    • D.

      50% for both sons and daughters

    • E.

      25% for a son and zero for a daughter

    Correct Answer
    E. 25% for a son and zero for a daughter
    Explanation
    Based on the information given, II-3 has two brothers and three sons with classical hemophilia. Hemophilia is an X-linked recessive disorder, which means that it is carried on the X chromosome. Males have one X chromosome and females have two X chromosomes. Since II-3 is a female carrier of hemophilia, she has a 50% chance of passing on the hemophilia gene to each of her children. Therefore, the likelihood of her child being a son with hemophilia is 50%, while the likelihood of her child being a daughter with hemophilia is 0%. Hence, the correct answer is "25% for a son and zero for a daughter".

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  • 3. 

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

    • A.

      1 in 4

    • B.

      1 in 2

    • C.

      1 in 16

    • D.

      1 in 3

    • E.

      1 in 9 (Bayes’ Formula)

    Correct Answer
    E. 1 in 9 (Bayes’ Formula)
    Explanation
    Based on the information given, II-3 has two brothers with hemophilia and three healthy sons. Hemophilia is a genetic disorder that is typically passed down through the X chromosome. Since II-3 does not have hemophilia herself but has two brothers with the condition, it is likely that she is a carrier of the hemophilia gene. The probability of being a carrier can be calculated using Bayes' formula, which takes into account the prevalence of the condition in the population. Therefore, the likelihood that II-3 is a carrier of the hemophilia gene is 1 in 9.

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  • 4. 

    In the below family, a child has been born with Acheiropodia (congenital absence of hands and feet). Assuming that this is a genetic problem, what is the MOST LIKELY inheritance pattern and how LIKELY is it that a next child of III3 and III4 will be affected?

    • A.

      X linked recessive; 1 in 2 for a son and 1 in 4 for a daughter

    • B.

      Autosomal recessive; 1 in 2

    • C.

      Autosomal dominant; 1 in 2

    • D.

      Autosomal recessive; 1 in 4

    • E.

      Mitochondrial; 1 in 2

    Correct Answer
    D. Autosomal recessive; 1 in 4
    Explanation
    The most likely inheritance pattern for Acheiropodia in this family is autosomal recessive. This is because the condition is present in a child, indicating that both parents are carriers of the recessive allele. The likelihood of a next child of III3 and III4 being affected is 1 in 4, as there is a 25% chance that both parents will pass on the recessive allele to the child.

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  • 5. 

    Given that the onset of the disease is by 40 years of age and given that all of the individuals shown are at least 45-years-old except III‑2 who is only 30-years-old, what is the probability that the fetus (IV­-1) will be affected?  

    • A.

      75%

    • B.

      50%

    • C.

      25%

    • D.

      100%

    • E.

      Virtually zero

    Correct Answer
    B. 50%
    Explanation
    Based on the information provided, the onset of the disease is by 40 years of age. Therefore, individuals who are at least 45-years-old are expected to show symptoms of the disease if they are affected. However, individual III-2 is only 30-years-old and already affected, indicating that the disease can manifest earlier than expected. This suggests that there is a 50% probability that the fetus (IV-1) will be affected, as there is a chance that the disease can manifest earlier than 40 years of age.

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  • 6. 

    In the family below, two brothers have classical hemophilia. Now I-4 asks you about the risk that her children will have hemophilia. You can tell her that the risk is:

    • A.

      Close to zero for both sons and daughters

    • B.

      1 in 2 for a son, close to zero for a daughter

    • C.

      1 in 4 for both sons and daughters

    • D.

      1 in 4 for a son, close to zero for a daughter

    • E.

      1 in 8 for a son, close to zero for a daughter

    Correct Answer
    D. 1 in 4 for a son, close to zero for a daughter
    Explanation
    The risk that I-4's children will have hemophilia is 1 in 4 for a son, close to zero for a daughter. This is because hemophilia is an X-linked recessive disorder, meaning it is carried on the X chromosome. Since I-4 is a carrier of the hemophilia gene, she has a 50% chance of passing on the affected X chromosome to her children. If she has a son, he will have a 50% chance of inheriting the affected X chromosome from her and therefore having hemophilia. However, if she has a daughter, the daughter would need to inherit the affected X chromosome from both I-4 and her father, which is less likely.

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  • 7. 

      It turns out that after two healthy sons, the third son of  I-4 in the family above has hemophilia. After this sad event, what is the chance that a next child will be affected?

    • A.

      1 in 4 for both sons and daughters

    • B.

      1 in 4 for a son, close to zero for a daughter

    • C.

      1 in 2 for a son, close to zero for a daughter

    • D.

      1 in 8 for a son, close to zero for a daughter

    • E.

      1 in 2 for both sons and daughters

    Correct Answer
    C. 1 in 2 for a son, close to zero for a daughter
    Explanation
    After two healthy sons, the third son being affected by hemophilia suggests that the mother is a carrier of the hemophilia gene, while the father is not affected. Hemophilia is an X-linked recessive disorder, meaning it is carried on the X chromosome. Since females have two X chromosomes, the chance of a daughter inheriting the hemophilia gene from the carrier mother is close to zero. However, for a son, there is a 50% chance of inheriting the affected X chromosome from the mother. Therefore, the chance of the next child being affected by hemophilia is 1 in 2 for a son and close to zero for a daughter.

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  • 8. 

    Now assume that I-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%
  • 9. 

    Albinism is a harmless autosomal recessive trait. If we assume that 2% of the general population are carriers of the albinism gene, how likely is it that the child of an albino with a healthy unrelated partner is an albino?

    • A.

      1 in 4

    • B.

      1 in 200

    • C.

      1 in 50

    • D.

      1 in 100

    • E.

      1 in 25

    Correct Answer
    D. 1 in 100
    Explanation
    If 2% of the general population are carriers of the albinism gene, it means that the probability of an individual being a carrier is 0.02. Since albinism is an autosomal recessive trait, both parents must be carriers in order for their child to have a 25% chance of being albino. Therefore, the probability of the child of an albino with a healthy unrelated partner being albino is 0.02 * 0.02 = 0.0004, which is equivalent to 1 in 100.

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  • 10. 

    Now assume that the unaffected sister of an albino asks you how likely it is that her child with a healthy unrelated man will be an albino. Her parents are both unaffected. Assume again a carrier frequency of 2% in the general population.

    • A.

      1 in 800

    • B.

      1 in 75

    • C.

      1 in 300

    • D.

      1 in 50

    • E.

      1 in 150

    Correct Answer
    D. 1 in 50
    Explanation
    To determine the likelihood of the child being an albino, we need to consider the inheritance of the albinism trait. Albinism is typically inherited as an autosomal recessive trait, which means both parents must carry at least one copy of the albino gene (albinism is caused by mutations in various genes) for their child to have a chance of being albino.
    Given the information provided:
    The unaffected sister (presumably a carrier) has one normal allele and one albino allele (Aa).
    The healthy unrelated man is presumed to have two normal alleles (aa).
    To calculate the probability of their child being albino (having two copies of the albino allele), we can use a Punnett square:

       A   a
    A  Aa  Aa
    a  Aa  Aa
     

     

    In this case, the unaffected sister (Aa) and the healthy unrelated man (aa) have a 50% chance of having a child with one normal allele (Aa) and a 50% chance of having a child with two normal alleles (aa).
    Since the carrier frequency in the general population is 2%, we can estimate that the sister has a 2% chance of being a carrier (Aa). Therefore, the probability that their child will be albino (aa) is the complement of the probability of having two normal alleles, which is 1 - 0.02 = 0.98.
    So, the likelihood of their child being an albino is 1 in 50 (1/0.02).

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  • 11. 

    The pedigree below represents a family with cystic fibrosis (autosomal recessive). What is the risk that individual II‑2 will have an affected child if the prevalence of cystic fibrosis is 1 in 2500 individuals?

    • A.

      1/100

    • B.

      1/150

    • C.

      1/25

    • D.

      1/50

    Correct Answer
    A. 1/100
    Explanation
    Individual II-2 is a carrier of the cystic fibrosis gene because they have one affected parent (I-2) and one unaffected parent (I-1). This means that there is a 50% chance that II-2 will pass on the cystic fibrosis gene to their child. The prevalence of cystic fibrosis is 1 in 2500 individuals, so the probability that II-2 will have an affected child is 1/2 multiplied by 1/2500, which equals 1/5000. Therefore, the risk that individual II-2 will have an affected child is 1/100.

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  • 12. 

    A woman (I‑1) is deaf from an autosomal recessive disease. She marries a hearing man and has four children, two of the four children are deaf at an early age. Genomic DNA was isolated from peripheral blood lymphocytes from all family members and subjected to Southern blot analysis using a radiolabeled DNA probe known to be strictly linked to the disease gene. Which of the following statements is TRUE about the offspring in generation II?

    • A.

      II 1 will be a carrier and II 3 will be normal

    • B.

      II 1 will be normal and II 3 will be a carrier

    • C.

      Both II 1 and II 3 will be carriers

    • D.

      Both II 1 and II 3 will be normal

    • E.

      II 1 will be affected and II 3 will be normal

    Correct Answer
    C. Both II 1 and II 3 will be carriers
    Explanation
    Based on the information provided, the woman (I-1) is deaf due to an autosomal recessive disease. This means that she has two copies of the disease-causing gene, one from each parent. When she marries a hearing man, who does not carry the disease-causing gene, their offspring in generation II have a 50% chance of inheriting one copy of the disease-causing gene and becoming carriers like II 1 and II 3. Since both II 1 and II 3 have one parent who is affected by the disease, they both have a 50% chance of being carriers. Therefore, the statement "both II 1 and II 3 will be carriers" is true.

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  • 13. 

    A 4‑year‑old boy was brought to the emergency room by his mother after falling three steps. The boy had skinned his knee and forearm on the cement and showed unusual and extensive bruising. To determine the genotype of the family members, genomic DNA was isolated from peripheral blood lymphocytes and subjected to Southern blot analysis using a radiolabeled DNA probe known to be closely linked to the disease gene. The boy is indicated by the arrow. Which of the following statements is TRUE about his siblings in generation III?

    • A.

      III -1 will be a carrier and III - 2 will be normal

    • B.

      Both III -1 and III- 2 will be affected

    • C.

      Both III- 1 and III -2 will be normal

    • D.

      III -1 will be affected and III -2 will be a carrier

    • E.

      Both III -1 and III -2 will be carriers

    Correct Answer
    A. III -1 will be a carrier and III - 2 will be normal
    Explanation
    Based on the given information, the boy has skinned his knee and forearm and shows unusual and extensive bruising. This suggests that he has a genetic disorder that causes abnormal bruising. The Southern blot analysis was performed to determine the genotype of the family members, indicating that the disorder is likely inherited. The correct answer suggests that III-1 will be a carrier, meaning they have one copy of the disease gene but do not show symptoms, and III-2 will be normal, meaning they do not have the disease gene. Therefore, III-1 will pass on the disease gene to their offspring, while III-2 will not.

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  • 14. 

    Genomic DNA was isolated from peripheral blood lymphocytes from the indicated family members subjected to Southern blot analysis using a radiolabeled DNA probe known to be strictly linked to the disease gene. The clinical manifestations of the autosomal dominant form of Martin's syndrome have been identified in this family, as indicated by the filled symbols. Given that the population frequency of Marfan's syndrome is 1 in 10,000, which of the following statements is TRUE about the siblings in generation III?

    • A.

      III -1 will not be affected and III -3 will be affected

    • B.

      Both III - 1 and III- 3 will be affected

    • C.

      III -1 and III -3 will NOT be affected

    • D.

      III -1 will be affected and III -3 will not be affected

    • E.

      It can not be determined whether III 1 and III 3 will be affected based on the data presented

    Correct Answer
    D. III -1 will be affected and III -3 will not be affected
    Explanation
    Based on the given information, the genomic DNA from the family members was analyzed using a DNA probe known to be linked to the disease gene. The clinical manifestations of the autosomal dominant form of Martin's syndrome have been identified in the family, indicated by the filled symbols. Since Martin's syndrome is an autosomal dominant disorder, it means that if one parent has the disease gene, there is a 50% chance of passing it on to each child. Therefore, III-1 will be affected because they inherited the disease gene from an affected parent, while III-3 will not be affected because they did not inherit the disease gene from the affected parent.

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  • 15. 

    The figure below shows the pedigree of a family with Huntington disease. Which of the following statements best describes the linkage analysis for the person indicated by the arrow (III‑2)?

    • A.

      This is an x- linked recessive disease, the mutant allele was received from the mother (II- 2), and she will be a carrier

    • B.

      This is an x linked recessive disease, the mutant allele was received from the father (II- 1), and she will see a carrier

    • C.

      This is an autosomal dominant disease, the mutant allele was received from the mother (II- 2), and she will be affected

    • D.

      This is an autosomal dominant disease, the mutant allele was received from the father (II- 1), and she will be affected

    • E.

      This is an autosomal dominant disease, the mutant allele was not inherited from either the mother or father, and she will not be affected

    Correct Answer
    E. This is an autosomal dominant disease, the mutant allele was not inherited from either the mother or father, and she will not be affected
  • 16. 

    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 analysis was normal. How LIKELY is it that you carry the gene? Use Bayes' theorem.

    • A.

      1 in 2

    • B.

      2 in 3

    • C.

      1 in 16

    • D.

      1 in 4

    • E.

      1 i n 6

    Correct Answer
    D. 1 in 4
    Explanation
    Based on the information given, we can use Bayes' theorem to calculate the likelihood of carrying the gene. Let's assume that the population consists of 100 people your age who carry the gene.

    Out of these 100 people, two-thirds (2/3) or approximately 67 people would have an abnormal vision test, while one-third (1/3) or approximately 33 people would still have a normal vision test.

    Since your analysis was normal, you can only belong to the group of 33 people who have a normal vision test.

    Therefore, the likelihood of you carrying the gene is 33 out of 100, which simplifies to 1 in 3.

    Hence, the given answer of 1 in 4 is incorrect.

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  • 17. 

    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 (Bayes’ Formula)

    • E.

      1 in 6

    Correct Answer
    D. 1 in 4 (Bayes’ Formula)
    Explanation
    Based on the information provided, the patient's father died of Huntington's disease, and two-thirds of all patients who carry the Huntington's gene show signs of the disease by age 45. However, the patient is still healthy at 45 years old. This suggests that the patient may not have inherited the gene for Huntington's disease, as they have not yet shown any signs of the disease. Therefore, it is likely that the patient's chances of getting the disease are 1 in 4, as determined by Bayes' formula.

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  • 18. 

    A normal male mates with a female who is homozygous for an x‑linked recessive disorder. Of the offspring that they may have, it is probable that:

    • A.

      No sons will be affected, and all daughters will be carriers

    • B.

      All sons will be affected, and all daughters will be affected

    • C.

      All sons will be affected, and all daughters will be carriers

    • D.

      No sons will be affected, and all daughters will be affected all sons will be affected, and 1/2 the daughters will be carriers

    Correct Answer
    C. All sons will be affected, and all daughters will be carriers
    Explanation
    When a normal male mates with a female who is homozygous for an X-linked recessive disorder, all sons will be affected. This is because the male will pass on his Y chromosome to his sons, which does not carry the recessive disorder gene. However, the female will pass on one of her X chromosomes to her sons, which does carry the recessive disorder gene. As a result, all sons will inherit the disorder. On the other hand, all daughters will be carriers because they will inherit one X chromosome with the recessive disorder gene from their mother, but their other X chromosome from their father will not carry the disorder gene. Therefore, all daughters will carry the gene but not show symptoms of the disorder.

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  • 19. 

    What is the probability of having an affected child if one parent is affected with an autosomal recessive disease and the other parent is unaffected?

    • A.

      1/4

    • B.

      1/4 X 1/4 X 1/4

    • C.

      Probability of the unaffected parent being a carrier X 1/2

    • D.

      Pretty close to 100% because new mutations are frequent

    • E.

      50% if the other parent is normal

    Correct Answer
    C. Probability of the unaffected parent being a carrier X 1/2
    Explanation
    If one parent is affected with an autosomal recessive disease and the other parent is unaffected, the unaffected parent has a 2/3 chance of being a carrier of the disease. The unaffected parent can pass on either the disease allele or the normal allele to their child. Therefore, the probability of having an affected child is the probability of the unaffected parent being a carrier (2/3) multiplied by the probability of passing on the disease allele (1/2), which is equal to 1/3 or approximately 33.3%.

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  • 20. 

    Parents of a child with an autosomal recessive disease are:

    • A.

      Likely to have mutations in different genes, locus heterogeneity

    • B.

      Likely to have more affected sons than daughters

    • C.

      Obligatory carriers

    • D.

      Most likely homozygous normal and the disease was caused by a new mutation

    • E.

      Not at risk of having another child with the disease

    Correct Answer
    C. Obligatory carriers
    Explanation
    Obligatory carriers refers to parents who are carriers of a recessive disease gene and have a child with the disease. In this case, both parents are carriers of the autosomal recessive disease gene, which means they are obligatory carriers. This suggests that they each have one copy of the mutated gene but do not show any symptoms of the disease themselves.

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  • 21. 

    If one parent has an autosomal dominant disorder and is a heterozygote, what are the chances that they will have an affected offspring?

    • A.

      1/4 X 1/4 X 1/4

    • B.

      Pretty close to 100% because new mutations are frequent

    • C.

      1/4

    • D.

      50%, if the other parent is normal

    • E.

      Probability of the unaffected parent being a carrier X 1/4

    Correct Answer
    D. 50%, if the other parent is normal
    Explanation
    If one parent has an autosomal dominant disorder and is a heterozygote, the chances that they will have an affected offspring is 50%, if the other parent is normal. This is because the offspring has a 50% chance of inheriting the dominant allele from the affected parent, which would result in them being affected by the disorder.

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  • 22. 

    What is the probability that an offspring of an affected person and an unaffected person with an autosomal dominant disorder will be affected?

    • A.

      1/3

    • B.

      1/4

    • C.

      2/3

    • D.

      1/2

    • E.

      Virtually 0

    Correct Answer
    D. 1/2
    Explanation
    The probability that an offspring of an affected person and an unaffected person with an autosomal dominant disorder will be affected is 1/2. This is because autosomal dominant disorders only require one copy of the affected gene to be inherited in order for the disorder to be expressed. Since the affected person will pass on one copy of the affected gene and the unaffected person will pass on one copy of the unaffected gene, there is a 50% chance that the offspring will inherit the affected gene and be affected by the disorder.

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  • 23. 

    What proportion of all offspring are affected females if the father is affected with an X‑linked recessive disease and the mother is a carrier of the same disease?

    • A.

      25%

    • B.

      100%.

    • C.

      75%

    • D.

      50%

    • E.

      Virtually 0

    Correct Answer
    A. 25%
    Explanation
    If the father is affected with an X-linked recessive disease and the mother is a carrier, there is a 25% chance that the offspring will be affected females. This is because the father can only pass on the disease to his daughters, as they receive his X chromosome. The mother, being a carrier, has a 50% chance of passing on the disease-causing gene to her children. Therefore, the probability of having an affected female offspring is 50% of the 50% chance from the mother, which equals 25%.

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  • 24. 

    If an affected male mates with a normal female and has four offspring: one normal male, one affected male, and two affected female, the disease is MOST LIKELY inherited as:

    • A.

      X linked dominant

    • B.

      X linked recessive

    • C.

      Autosomal dominant

    • D.

      Autosomal recessive

    • E.

      Y linked

    Correct Answer
    C. Autosomal dominant
    Explanation
    In this scenario, the disease is most likely inherited as autosomal dominant. This is because the affected male has four offspring, and both males and females are affected. In autosomal dominant inheritance, a single copy of the mutated gene from one parent is enough to cause the disease. This means that the affected male must have the disease-causing gene on one of his non-sex chromosomes (autosomes) and has a 50% chance of passing it on to each of his offspring. The fact that both males and females are affected suggests that the disease is not sex-linked, ruling out X linked dominant or X linked recessive inheritance.

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  • 25. 

    If your mother was known to be a carrier of alpha‑antitrypsin deficiency, an autosomal recessive disorder, what is the probability that your child is also a carrier of this disease gene?

    • A.

      1/4

    • B.

      100%

    • C.

      2/3

    • D.

      1/2

    • E.

      Virtually 0

    Correct Answer
    D. 1/2
    Explanation
    If your mother is a carrier of alpha-antitrypsin deficiency, an autosomal recessive disorder, it means she has one normal copy of the gene and one copy with the disease-causing mutation. Since it is an autosomal recessive disorder, in order for your child to be a carrier, they would need to inherit the disease-causing mutation from both you and your mother. The probability of this happening is 1/2, as there is a 50% chance that you would pass on the disease-causing mutation to your child.

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  • 26. 

    A normal male mates with a female who is a carrier for an x‑linked recessive disorder. Of the offspring that they may have, it is probable that:

    • A.

      All sons will be affected, and all daughters will be carriers

    • B.

      No sons will be affected, and all daughters will be affected

    • C.

      1/2 the sons will be affected, and all daughters will be carriers

    • D.

      All sons will be affected, and all daughters will be affected

    • E.

      1/2 the sons will be affected, and 1/2 the daughters will be carriers

    Correct Answer
    E. 1/2 the sons will be affected, and 1/2 the daughters will be carriers
    Explanation
    In this scenario, the male is normal and does not carry the x-linked recessive disorder. However, the female is a carrier, which means she carries one copy of the mutated gene for the disorder on one of her X chromosomes. The female has two X chromosomes, and the male has one X and one Y chromosome.

    When they have offspring, there is a 50% chance that a son will inherit the mutated gene from the carrier mother and be affected by the disorder. Therefore, 1/2 the sons will be affected.

    For daughters, there is a 50% chance that they will inherit the mutated gene from the carrier mother and become carriers themselves. Therefore, 1/2 the daughters will be carriers.

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  • 27. 

    What is the probability of having an affected child if one parent is affected with an autosomal recessive disease and the other parent is a carrier?

    • A.

      1/2

    • B.

      2/3

    • C.

      Pretty close to 100% because new mutations are frequent

    • D.

      1/4

    • E.

      Virtually 0

    Correct Answer
    A. 1/2
    Explanation
    If one parent is affected with an autosomal recessive disease and the other parent is a carrier, there is a 50% chance that their child will be affected. This is because in an autosomal recessive inheritance pattern, both copies of the gene must be abnormal for the disease to be expressed. The affected parent will have two abnormal copies, while the carrier parent will have one normal and one abnormal copy. Therefore, when they have a child, there is a 50% chance that the child will inherit the abnormal copy from the affected parent and a 50% chance that the child will inherit the normal copy from the carrier parent, resulting in a carrier status.

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  • 28. 

    If an affected male mate with a phenotypically normal female and has both an affected male and an affected female offspring, the disease is MOST LIKELY inherited as:

    • A.

      Autosomal dominant

    • B.

      X linked recessive

    • C.

      X linked dominant

    • D.

      Y linked

    • E.

      Autosomal recessive

    Correct Answer
    A. Autosomal dominant
    Explanation
    If an affected male mate with a phenotypically normal female and has both an affected male and an affected female offspring, the disease is MOST LIKELY inherited as autosomal dominant. This is because autosomal dominant inheritance means that only one copy of the mutated gene is needed to inherit the disease. In this scenario, the affected male must have one copy of the mutated gene, which he passes on to both his male and female offspring. Since the disease is present in both sexes, it is not X-linked dominant or Y-linked. Additionally, it is not X-linked recessive because affected males can have affected female offspring. Autosomal recessive inheritance requires both parents to be carriers of the mutated gene, which is not the case here.

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