Block 1 Pt 2 Pedigree Genetics Dr Larsen

When a normal male mates with a female who is homozygous for an X-linked recessive disorder, all sons will be affected by the disorder. This is because the male only has one X chromosome, which he receives from his mother, and if that X chromosome carries the recessive disorder gene, he will inherit the disorder. On the other hand, all daughters will be carriers of the disorder because they inherit one X chromosome from their father, which is normal, and one X chromosome from their mother, which carries the recessive disorder gene. As a carrier, the daughter will not be affected by the disorder but can pass it on to her own offspring.

Obligatory carriers refers to parents who are carriers of a recessive disease-causing gene and have a child with the disease. This means that both parents have a copy of the mutated gene, but they themselves do not exhibit symptoms of the disease. They are obligated to pass on the mutated gene to their child, resulting in the disease. Therefore, the correct answer is "obligatory carriers."

When an affected person and an unaffected person with an autosomal dominant disorder have an offspring, there is a 50% chance that the offspring will be affected. This is because autosomal dominant disorders only require one copy of the mutated gene for the disorder to be expressed. Therefore, the affected person will always pass on the mutated gene to their offspring, while the unaffected person will always pass on a normal gene. As a result, the offspring has a 50% chance of inheriting the mutated gene and being affected by the disorder.

The correct answer is Autosomal recessive; 1 in 4. This means that the condition is caused by a recessive gene that is not located on the sex chromosomes. Both parents must carry the gene for there to be a 1 in 4 chance of their next child being affected.

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 by the disease. This is because in autosomal recessive inheritance, both parents need to contribute a recessive allele for the disease to be expressed in the child. The affected parent will always contribute a recessive allele, while the carrier parent has a 50% chance of contributing a recessive allele. Therefore, there is a 50% chance of having an affected child.

The risk of I-4's children having hemophilia is 1 in 4 for a son and close to zero for a daughter. This is because hemophilia is an X-linked recessive disorder, meaning it is carried on the X chromosome. Males have one X chromosome and one Y chromosome, while females have two X chromosomes. If I-4 is a carrier of the hemophilia gene, there is a 50% chance that she will pass it on to her sons, resulting in a 1 in 4 risk. However, daughters would need to inherit the hemophilia gene from both parents, making the risk close to zero.

If an affected male mates 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 affected gene is needed for the disease to be expressed. In this case, the affected male must have one copy of the affected gene and passes it on to both the male and female offspring.

Based on the information provided, II3 is a carrier of the hemophilia A gene because she has two brothers with the disorder. Hemophilia A is an X-linked recessive trait, which means it is carried on the X chromosome. Since II3 is a carrier, she has a 50% chance of passing the hemophilia gene to her children. Therefore, the risk of her having a son with hemophilia is 1 in 4, while the risk for her having a daughter with hemophilia is close to zero, as daughters would need to inherit the gene from both parents to have the disorder.

When a normal male mates with a female who is a carrier for an x-linked recessive disorder, there is a 50% chance that each son they have will be affected by the disorder. This is because sons inherit their X chromosome from their mother, and if she is a carrier, there is a 50% chance that she will pass on the affected X chromosome to her son. On the other hand, all daughters will be carriers of the disorder. This is because daughters inherit one X chromosome from each parent, and if the father is normal, they will inherit his normal X chromosome. However, they will also inherit the affected X chromosome from their mother, making them carriers of the disorder.

Based on the given information, the disease is most likely inherited as autosomal dominant. This is because the affected male passes on the disease to both male and female offspring, indicating that the disease is not sex-linked. Additionally, the presence of affected females suggests that the disease is not X-linked recessive, as affected males would not pass the disease to their daughters. The pattern of inheritance is consistent with autosomal dominant inheritance, where the presence of a single copy of the disease-causing allele is enough to express the disease phenotype.

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 affected parent has one dominant allele for the disorder and one normal allele. If the other parent is normal and does not carry the dominant allele, there is a 50% chance that the offspring will inherit the dominant allele and be affected by the disorder.

The probability of having an affected child if one parent is affected with an autosomal recessive disease and the other parent is unaffected is determined by the probability of the unaffected parent being a carrier, multiplied by 1/2. This is because if the unaffected parent is a carrier, there is a 50% chance that they will pass on the affected allele to their child. Therefore, the answer is the probability of the unaffected parent being a carrier multiplied by 1/2.

Based on the given information, the unaffected sister is a carrier of the albino gene, but does not have albinism herself. The parents of the sister are also unaffected, which means they do not carry the albino gene. However, there is a carrier frequency of 2% in the general population. Therefore, the likelihood of the sister's child inheriting the albino gene from both parents (unaffected sister and healthy unrelated man) is 1 in 300.

Based on the given information, the woman (I-1) is deaf due to an autosomal recessive disease. This means that both of her parents must have been carriers of the disease gene. Since she married a hearing man, it is likely that he does not carry the disease gene. However, two out of their four children are deaf at an early age, indicating that both the woman and her husband are carriers of the disease gene. Therefore, both II 1 and II 3, the offspring in generation II, will be carriers of the disease gene.

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If 2% of the general population are carriers of the albinism gene, it means that there is a 2% chance that the healthy unrelated partner of an albino is a carrier. In order for the child to be albino, both parents must pass on the albinism gene. Since the albino parent has the gene, the chance of passing it on is 100%. However, the chance of the healthy unrelated partner passing on the gene is only 2%. Therefore, the overall chance of the child being albino is the product of these two probabilities, which is 1 in 100.

Based on the information provided, 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 age 45. This suggests that there is a possibility that the patient may not develop the disease. Bayes' Formula is used to calculate conditional probabilities, and in this case, it can be used to determine the likelihood of the patient getting the disease. Therefore, the correct answer is 1 in 4 (Bayes' Formula).

Based on the given information, the clinical manifestations of the autosomal dominant form of Marfan's syndrome have been identified in the family as indicated by the filled symbols. This suggests that at least one parent of generation III must be affected by Marfan's syndrome in order for their children to have a chance of inheriting the condition. Since III-1 is affected, it can be concluded that III-1 will be affected by Marfan's syndrome. However, since III-3 is not indicated as being affected, it can be concluded that III-3 will not be affected by Marfan's syndrome. Therefore, the statement "III-1 will be affected and III-3 will not be affected" is true based on the given data.

Based on the information provided, the boy (indicated by the arrow) has skinned his knee and forearm on the cement and also shows unusual and extensive bruising. This suggests that he has a genetic disorder that causes abnormal bleeding and bruising. The Southern blot analysis was performed to determine the genotype of the family members. The correct answer states that III-1 will be a carrier and III-2 will be normal. This suggests that III-1 carries the gene for the disorder but does not show any symptoms, while III-2 does not carry the gene and therefore does not have the disorder.

If the father is affected with an X-linked recessive disease and the mother is a carrier, there is a 50% chance that the mother will pass on the affected X chromosome to her offspring. This means that 50% of the offspring will be affected males. Additionally, there is a 50% chance that the mother will pass on the normal X chromosome to her offspring. In this case, the offspring will not be affected by the disease. Therefore, the proportion of all offspring that are affected females is 25%.

Based on the information given, II-3 is a carrier of the hemophilia gene and has three sons with hemophilia. Hemophilia is an X-linked recessive disorder, meaning it is passed down on the X chromosome. Since II-3 is a carrier, she has a 50% chance of passing the hemophilia gene to each of her children. Therefore, the likelihood that her next child, regardless of gender, will have hemophilia is 50%. However, since the question specifically asks for the likelihood for a son, the answer is 25% since only half of the sons will inherit the hemophilia gene. The likelihood for a daughter to have hemophilia is zero since she would need to inherit the hemophilia gene from both parents, which is not possible in this scenario.

After two healthy sons, the third son being affected by hemophilia suggests that the mother is a carrier of the hemophilia gene. Hemophilia is a sex-linked recessive disorder, meaning that it is carried on the X chromosome. Since males have one X chromosome and females have two, the chance of a son inheriting the hemophilia gene is 1 in 2, as he has a 50% chance of receiving the affected X chromosome from the carrier mother. On the other hand, the chance of a daughter being affected is close to zero, as she would need to receive the affected X chromosome from both parents.

Based on the information provided, II-3 has two brothers with hemophilia and three healthy sons. This indicates that II-3 is the daughter of a carrier mother and a non-carrier father. Using Bayes' formula, the probability of II-3 being a carrier can be calculated. The formula takes into account the prevalence of hemophilia in the population and the likelihood of being a carrier given the family history. Therefore, the correct answer is 1 in 9, indicating that there is a 1 in 9 chance that II-3 is a carrier of the hemophilia gene.

Individual II-2 is a carrier of the cystic fibrosis gene because they have one affected parent (I-1) and one unaffected parent (I-2). Since cystic fibrosis is an autosomal recessive disorder, II-2 can pass on the affected gene to their child if they have a child with a partner who is also a carrier or affected. The risk of II-2 having an affected child is 1/4 (50% chance of passing on the affected gene) multiplied by 1/4 (50% chance of their partner being a carrier) multiplied by 1/2 (50% chance of their partner passing on the affected gene). This equals 1/32. However, since the prevalence of cystic fibrosis is 1 in 2500 individuals, the risk is further reduced to 1/32 multiplied by 1/2500, which simplifies to 1/100.

Based on the information given, we know that two-thirds of people your age who carry the gene have an abnormal vision test, while one-third have a normal vision test. Since your vision test was normal, it is more likely that you do not carry the gene. Therefore, the likelihood of you carrying the gene is 1 in 4.

The question mentions that I-4 has two healthy sons. This information is crucial because it indicates that I-4 is not affected by the genetic condition in question. Bayes' theorem is used to calculate the probability of an event given prior knowledge or information. In this case, the probability of I-4 being a carrier can be calculated using Bayes' theorem by taking into account the prevalence of carriers in the general population and the fact that I-4 has two healthy sons. The correct answer, 20%, suggests that there is a 20% chance that I-4 is a carrier based on the given information.

Based on the information provided, the onset of the disease is by 40 years of age. All of the individuals shown are at least 45 years old, except for III-2 who is only 30 years old. This suggests that III-2 may be a carrier of the disease but has not yet developed symptoms. Since III-2 is the parent of IV-1 (the fetus in question), there is a 50% chance that IV-1 will inherit the disease-causing gene from III-2. Therefore, the probability that the fetus will be affected is 50%.

If your mother is a carrier of alpha-antitrypsin deficiency, an autosomal recessive disorder, there is a 50% probability that you have inherited the disease gene. As a carrier, you have one normal copy of the gene and one copy of the disease-causing gene. When you have a child, there is a 50% chance that you will pass on the disease-causing gene to them, making them a carrier as well. Therefore, the probability that your child is also a carrier of this disease gene is 1/2 or 50%.