Once the mode of inheritance of a disease or trait is identified, some inferences about the genotype of individuals in a pedigree can be made, based on their phenotypes and where they appear in the family tree. Given these genotypes, it is possible to calculate the probability of a particular genotype being inherited in subsequent generations. This can be useful in genetic counseling, for example when prospective parents wish to know the likelihood of their offspring
inheriting a disease for which they have a family history. Probabilities in pedigrees are calculated using knowledge of Mendelian inheritance and the same basic methods as are used in other fields. The first formula is the product rule: the joint probability of two independent events is the product of their individual probabilities; this is the probability of one event AND another event occurring. For example, the probability of a rolling a “five” with a single throw of a
single six-sided die is 1/6, and the probability of rolling “five” in each of three successive rolls is 1/6 x 1/6 x 1/6 = 1/216. The second useful formula is the sum rule, which states that the combined probability of two independent events is the sum of their individual probabilities. This is the probability of one event OR another event occurring. For example, the probability of rolling a five or six in a single throw of a dice is 1/6 + 1/6 = 1/3. With these rules in
mind, we can calculate the probability that two carriers (i.e. heterozygotes) of an AR disease will have a child affected with the disease as ½ x ½ = ¼, since for each parent, the probability of any gametes carrying the disease allele is ½. This is consistent with what we already know from calculating probabilities using a Punnett Square (e.g. in a monohybrid cross Aa x Aa, ¼ of the offspring are aa). We can likewise calculate probabilities in the more complex
pedigree shown in Figure \(\PageIndex{11}\). Assuming the disease has an AR pattern of inheritance, what is the probability that individual
14 will be affected? We can assume that individuals #1, #2, #3 and #4 are heterozygotes (Aa), because they each had at least one affected (aa) child, but they are not affected themselves. This means that there is a 2/3 chance that individual #6 is also Aa. This is because according to Mendelian inheritance, when two heterozygotes mate, there is a 1:2:1 distribution of genotypes AA:Aa:aa. However, because #6 is unaffected, he can’t be
aa, so he is either Aa or AA, but the probability of him being Aa is twice as likely as AA. By the same reasoning, there is likewise a 2/3 chance that #9 is a heterozygous carrier of the disease allele. If individual 6 is a heterozygous for the disease allele, then there is a ½ chance that #12 will also be a heterozygote (i.e. if the mating of #6 and #7 is Aa × AA, half of the progeny will be Aa; we are also assuming that #7, who is unrelated, does not carry any disease alleles). Therefore, the combined probability that #12 is also a heterozygote is 2/3 x 1/2 = 1/3. This reasoning also applies to individual #13, i.e. there is a 1/3 probability that he is a heterozygote for the disease. Thus, the overall probability that both individual #12 and #13 are heterozygous, and that a particular offspring of theirs will be homozygous for the disease alleles is 1/3 x 1/3 x 1/4 = 1/36. It may come as a surprise that our genes and probabilities have some things in common. Due to the random nature of cell meiosis, some aspects to the study of genetics is really applied probability. We will see how to calculate the probabilities associated with dihybrid crosses. Definitions and AssumptionsBefore we calculate any probabilities, we will define the terms that we use and state the assumptions that we will work with.
Monohybrid CrossBefore determining the probabilities for a dihybrid cross, we need to know the probabilities for a monohybrid cross. Suppose that two parents who are heterozygous for a trait produce an offspring. The father has a probability of 50% of passing on either of his two alleles. In the same way, the mother has a probability of 50% of passing on either of her two alleles. We can use a table called a Punnett square to calculate the probabilities, or we can simply think through the possibilities. Each parent has a genotype Dd, in which each allele is equally likely to be passed down to an offspring. So there is a probability of 50% that a parent contributes the dominant allele D and a 50% probability that the recessive allele d is contributed. The possibilities are summarized:
So for parents who both have genotype Dd, there is a 25% probability that their offspring is DD, a 25% probability that the offspring is dd, and a 50% probability that the offspring is Dd. These probabilities will be important in what follows. Dihybrid Crosses and GenotypesWe now consider a dihybrid cross. This time there are two sets of alleles for parents to pass on to their offspring. We will denote these by A and a for the dominant and recessive allele for the first set, and B and b for the dominant and recessive allele of the second set. Both parents are heterozygous and so they have the genotype of AaBb. Since they both have dominant genes, they will have phenotypes consisting of the dominant traits. As we have said previously, we are only considering pairs of alleles that are not linked to one another, and are inherited independently. This independence allows us to use the multiplication rule in probability. We can consider each pair of alleles separately from each other. Using the probabilities from the monohybrid cross we see:
The first three genotypes are independent of the last three in the above list. So we multiply 3 x 3 = 9 and see that there are these many possible ways to combine the first three with the last three. This is the same ideas as using a tree diagram to calculate the possible ways to combine these items. For example, since Aa has probability 50% and Bb has a probability of 50%, there is a 50% x 50% = 25% probability that the offspring has a genotype of AaBb. The list below is a complete description of the genotypes that are possible, along with their probabilities.
Dihybrid Crosses and PhenotypesSome of these genotypes will produce the same phenotypes. For example, the genotypes of AaBb, AaBB, AABb, and AABB are all different from each other, yet will all produce the same phenotype. Any individuals with any of these genotypes will exhibit dominant traits for both traits under consideration. We may then add the probabilities of each of these outcomes together: 25% + 12.5% + 12.5% + 6.25% = 56.25%. This is the probability that both traits are the dominant ones. In a similar way we could look at the probability that both traits are recessive. The only way for this to occur is to have the genotype aabb. This has a probability of 6.25% of occurring. We now consider the probability that the offspring exhibits a dominant trait for A and a recessive trait for B. This can occur with genotypes of Aabb and AAbb. We add the probabilities for these genotypes together and have18.75%. Next, we look at the probability that the offspring has a recessive trait for A and a dominant trait for B. The genotypes are aaBB and aaBb. We add the probabilities for these genotypes together and have a probability of 18.75%. Alternately we could have argued that this scenario is symmetric to the early one with a dominant A trait and a recessive B trait. Hence the probability for this outcomes should be identical. Dihybrid Crosses and RatiosAnother way to look at these outcomes is to calculate the ratios that each phenotype occurs. We saw the following probabilities:
Instead of looking at these probabilities, we can consider their respective ratios. Divide each by 6.25% and we have the ratios 9:3:1. When we consider that there are two different traits under consideration, the actual ratios are 9:3:3:1. What this means is that if we know that we have two heterozygous parents, if the offspring occur with phenotypes that have ratios deviating from 9:3:3:1, then the two traits we are considering do not work according to classical Mendelian inheritance. Instead, we would need to consider a different model of heredity. What is the probability of heterozygous offspring if both parents are heterozygous?In another example (shown below), if the parent plants both have heterozygous (YG) genotypes, there will be 25% YY, 50% YG, and 25% GG offspring on average.
What is the percentage of two heterozygous offspring?If we cross two heterozygous individuals (Aa), then the percentage of offsprings with recessive phenotype will be: Aa x Aa = AA Aa Aa aa, so recessive phenotypes would be 1/4 = 25%. Thus, the correct answer is option D.
What is the result of two heterozygous parents?If both parents are heterozygous (Ww), there is a 75% chance that any one of their offspring will have a widow's peak (see figure). A Punnett square can be used to determine all possible genotypic combinations in the parents. A pedigree that depicts a dominantly inherited trait has a few key distinctions.
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