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    in this pedigree, a genotype of i-1 is aa. if iii-3 marries an unaffected man, what fraction of their children will be affected?

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    If a genetic disorder runs in my family, what are the chances that my children will have the condition?: MedlinePlus Genetics

    It is hard to predict if your children will inherit a genetic disorder. Learn about the factors that impact the chances of developing a genetic condition.

    If a genetic disorder runs in my family, what are the chances that my children will have the condition?

    When a genetic disorder is diagnosed in a family, family members often want to know the likelihood that they or their children will develop the condition. This can be difficult to predict in some cases because many factors influence a person's chances of developing a genetic condition. One important factor is how the condition is inherited. For example:

    Autosomal dominant inheritance: A person affected by an autosomal dominant disorder has a 50 percent chance of passing the altered gene to each child. The chance that a child will not inherit the altered gene is also 50 percent. However, in some cases an autosomal dominant disorder results from a new (de novo) variant that occurs during the formation of egg or sperm cells or early in embryonic development. In these cases, the child's parents are unaffected, but the child may pass on the condition to his or her own children.

    Autosomal recessive inheritance: Two unaffected people who each carry one copy of the altered gene for an autosomal recessive disorder (carriers) have a 25 percent chance with each pregnancy of having a child affected by the disorder. The chance with each pregnancy of having an unaffected child who is a carrier of the disorder is 50 percent, and the chance that a child will not have the disorder and will not be a carrier is 25 percent. If only one parent is a carrier of the altered gene and the other parent does not carry the variant, none of their children will develop the condition, and the chance with each pregnancy of having an unaffected child who is a carrier is 50 percent.

    X-linked dominant inheritance: The chance of passing on an X-linked dominant condition differs between men and women because men have one X chromosome and one Y chromosome, while women have two X chromosomes. A man passes on his Y chromosome to all of his sons and his X chromosome to all of his daughters. Therefore, the sons of a man with an X-linked dominant disorder will not be affected, but all of his daughters will inherit the condition. A woman passes on one or the other of her X chromosomes to each child. Therefore, a woman with an X-linked dominant disorder has a 50 percent chance of having an affected daughter or son with each pregnancy.

    X-linked recessive inheritance: Because of the difference in sex chromosomes, the probability of passing on an X-linked recessive disorder also differs between men and women. The sons of a man with an X-linked recessive disorder will not be affected, and his daughters will carry one copy of the altered gene. With each pregnancy, a woman who carries an altered gene for X-linked recessive has a 50 percent chance of having sons who are affected and a 50 percent chance of having daughters who carry one copy of the altered gene. Females with one gene variant associated with an X-linked recessive disorder typically have no or very mild signs or symptoms of the condition.

    X-linked: Because the inheritance pattern of many X-linked disorders is not clearly dominant or recessive, some experts suggest that conditions be considered X-linked rather than X-linked dominant or X-linked recessive. As above, the probability of passing on an X-linked disorder differs between men and women. The sons of a man with an X-linked disorder will not be affected, but all of his daughters will inherit the altered gene and may develop signs and symptoms of the condition. A woman passes on one or the other of her X chromosomes to each child. Therefore, with each pregnancy, a woman with an X-linked disorder has a 50 percent chance of having a child with the altered gene. An affected daughter may have milder signs and symptoms than an affected son.

    Y-linked inheritance: Because only males have a Y chromosome, only males can be affected by and pass on Y-linked disorders. All sons of a man with a Y-linked disorder will inherit the condition from their father.

    Codominant inheritance: In codominant inheritance, each parent contributes a different version of a particular gene, and both versions influence the resulting genetic trait. The chance of developing a genetic condition with codominant inheritance, and the characteristic features of that condition, depend on which versions of the gene are passed from parents to their child.

    Mitochondrial inheritance: Mitochondria, which are the energy-producing centers inside cells, each contain a small amount of DNA. Disorders with mitochondrial inheritance result from variants in mitochondrial DNA. Although these disorders can affect both males and females, only females can pass variants in mitochondrial DNA to their children. A woman with a disorder caused by changes in mitochondrial DNA will pass the variants to all of her daughters and sons, but the children of a man with such a disorder will not inherit the variant.

    It is important to note that the chance of passing on a genetic condition applies equally to each pregnancy. For example, if a couple has a child with an autosomal recessive disorder, the chance of having another child with the disorder is still 25 percent (or 1 in 4). Having one child with a disorder does not “protect” future children from inheriting the condition. Conversely, having a child without the condition does not mean that future children will definitely be affected.

    Although the chances of inheriting a genetic condition appear straightforward, factors such as a person's family history and the results of genetic testing can sometimes modify those chances. In addition, some people with a disease-causing variant never develop any health problems or may experience only mild symptoms of the disorder. If a disease that runs in a family does not have a clear-cut inheritance pattern, predicting the likelihood that a person will develop the condition can be particularly difficult.

    Source : medlineplus.gov

    Oxford University Press

    Young: Medical Genetics

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    Risk calculation in pedigrees

    Calculating probability and risks in pedigree analysis:

    Elementary principles

    The pedigree shows the occurrence of an autosomal recessive trait, where the black squares have genotype aa.  We wish to calculate the probability that IV-1 (shown as ?) will be either affected (aa), or a carrier heterozygote (Aa).

    (1) For IV-1 to be an affected recessive homozygote, s/he must inherit an a allele from the father the mother.  Given that II-1 must be aa, both great-grandparents (I-1 and I-2) must be Aa.  II-2 shows the dominant phenotype, and therefore has at least one A allele: the probability that the other is a is 1/2.  II-3 is from outside the affected pedigree and can be assumed to be AA.  Like his father, III-1 shows the dominant phenotype, and therefore has at least one A. Then, the probability that III-1 is Aa is the probability that II-2 is heterozygous passed the a allele to III-1 : (1/2) x (1/2) = 1/4. The same reasoning leads to the conclusion that III-2 is heterozygous with a probability of 1/4. Thus, for IV-1 to be aa, both parents must be Aa, they must both pass the a allele to their offspring: 1/4 x 1/4 x 1/4 = 1/64

    (2) Alternatively, for IV-1 to be a heterozygous, carrier, either s/he most inherit an a allele from the father, or from the mother. The chance of either parent being a heterozygote is 1/4, as calculated above. Then, the probability that both parents are heterozygotes, and the probability that two heterozygotes will have a heterozygous child, is 1/4 x 1/4 x 1/2 = 1/32.

    (3) Finally, the probability that IV-1 is a dominant homozygote is 1 - 1/64 - 1/32 = (64 - 1 - 2)/64 = 61/64. This can also be calculated more tediously by summing the alternative probabilities at each of the steps above.

    The calculations in this example involved a distinction between a priori and a posteriori probability, which are often presented incorrectly in elementary genetics textbooks. To take a simple case: the a priori probability of getting heads on a single toss of a penny is 1/2, since there are two equal possibilities, H or T. Then, given two pennies tossed at random, HH, HT, TH, and TT are all equally likely. The a priori probability of getting at least one head is 3/4. The a priori probability that any combination with at least one head will have two tails (HT or TH vs HH) is 2/3.

    However, consider an experiment in which I have tossed two pennies. I show you that one is H, and ask, What is the probability that the other is also H? The a posteriori probability is 1/2 : given the knowledge that one coin is H, the other is H or T with equal probability. In anticipation of the experiment, the probability of HT given H- is 2/3. In analyzing the results of any particular experiment, the added information changes probabilities a posteriori.

    In the above example, we know that I-1 and I-2 are heterozygotes and II-2 shows the dominant phenotype. We therefore know a posteriori that he has inherited a dominant alleles from one parent, and the probability that he will inherit a recessive alleles from the other parent and be heterozygous is 1/2. It is incorrect to reason that, since 2/3 of all unaffected children (that is, all non-aa) are heterozygotes a priori, his individual risk is also 2/3. Stated another way, by knowing the nature of one allele, we have lost one statistical degree of freedom.

    [In the simplest case: the probability that the next child of I-1 and I-2 will be a boy is 1/2: once the child is born, the probability is either 0 or 1].

    Two further extensions of this idea. For this scenario, assume that a genetic test is available to distinguish AA from Aa, but II-5 is deceased and III-2 will not take the test.

    (4) Suppose III-2 has a heterozygous sibling. How does this change the calculation IV-I's risk? This would mean that II-5 must be a heterozygote with a probability of 1, not 1/2 as before. Then, the probability that III-2 is a heterozygote is 1/2, the probability that the father (III-1) is a heterozygote remains 1/4, and the probability that IV-1 is aa is 1/2 x 1/4 x 1/4 = 1/32.

    (5) Suppose III-2 has one or more siblings who test as unaffected homozygotes (AA). How does this change the calculation of IV-1's risk? Note that, whereas the birth of a heterozygous sibling proves that the mother (II-5) is a heterozygote, the birth of unaffected homozygous offspring cannot prove that she is a homozygote. However, multiple births of unaffected siblings do decrease the probability that she is a heterozygote, as follows. The probability that a heterozygote will not pass the a allele to an offspring is 1/2. Then, the probability that she will not pass it to either of two offspring is (1/2)(1/2) = 0.52 = 1/4. The probability that she will pass it to none of three offspring is 0.53 = 1/8, to none of four is 1/16, and so on. Less than 0.1% of all families with ten children would have known with an a alleles, if II-5 were a heterozygote. In other words, this is strong a posteriori evidence that II-5 is a , which if true means that IV-1 cannot be affected. Of course, the birth of an eleventh child who is Aa immediately proves that II-5 is heterozygous, and returns IV-1's risk calculation to 1/16, as in (4) above..

    All figure & text material ©2020 by Steven M. Carr

    Source : www.mun.ca

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