What is the probability of two heterozygous parents having a recessive offspring?
9-14-05PROBABILITY/STATISTICS IN INHERITANCEWe know that when two people who are both heterozygous for a simple Mendelian autosomal gene alpha have a child, the probability that the child will show the dominant phenotype is 3/4. Let's ask a somewhat more complex question. If this couple has a total of four children, what is the probability that 3 of the 4 will show the dominant phenotype? To answer this, we will first derive the appropriate formula and then use it to calculate the numerical answer. The same formula allows us to understand the expected statistical distribution of the various possible phenotype patterns in four-child ( or any other size) families in a large population.1. Review: How can some simple overall probabilities be calculated by combining the multiplication and addition "rules" we covered earlier? Show
Let's start with a very simple case: asking about gender probabilities in families of three children. What is the probability that all three children in a family will be the same gender? What is the probability that a three-child family is two girls and one boy? This allows us to write the overall gender probability distribution for families of three children as follows: Consider the numerators of the fractions in the above three-child family gender equation: Figure 4.21, "Pascal's triangle", shows these coefficients for the expansion of (p + q) raised to any power up to 10. The numbers in any row can be used just as described above. For example, assume that over the next two decades you have 6 kids. There are 64 possible gender birth orders, with 20 of these resulting in you having three girls and three boys. The terms p and q are the individual probabilities for a specific outcome from a single "event". For "gender" calculations, the probabilities p and q are equal, both = 1/2 (the equal probabilities of male and female births). For "dominant : recessive phenotype" type of calculations, p and q will usually not be equal. For a simple Mendelian inheritance from two heterozygotic parents, p will = 3/4 (if AA and Aa give dominant phenotype) and q will = 1/4 (aa gives recessive phenotype). Generalizing this, we get to the formula on page 161 in your text that is "the general rule for repeated trials of events with constant probabilities". The term (n!/s!t!) is the number of possible ways (orders) of getting a certain net outcome ( "a total of n; with s of one and t of the other"). This number can either be calculated or taken directly from Pascal's triangle. If we were looking at thousands of such families, we know that the overall ratio of dominant to recessive phenotypes in the children would average out to 3:1, as shown by a simple Punnett square. But for one couple having four children, what's the probability, P(3D,1r)? To calculate P(3D,1r), we use formula 1 for the case n=4, s=3, t=1, p=3/4, q=1/4. By also calculating the other four possibilities, we can construct a graph that shows the statistical distribution you would expect to see in a large population. Problem S-4: "Heterozygous parents; have three children". Modify the sample problem above to do the calculations for you and your mate (both Aa) having THREE children. Do the calculations for the probabilities that all three, two, one, or none of the three children will show the dominant phenotype from gene alpha. Construct the graph and compare the result with the "four children" graph done in class. What is the probability of a heterozygous recessive offspring?If two heterozygotes are crossed, the probability that an offspring will show the dominant trait is 75% or 0.75. The probability that an offspring will show the recessive trait is 25% or 0.25.
What is the probability that two heterozygous parents?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 happens if you breed 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.
What is the probability that two parents who are heterozygous for the recessive trait of albinism will have all albino offspring assuming they have four children?1 Answer. Mandira P. When both parents are heterozygous (Cc) carrier, in each pregnancy there is 25% chance of birth of an albino i.e. 1 in 4. So, in each pregnancy, there is 75% chance of birth of a normal (phenotypic) child i.e. 3 in 4.
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