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Test your knowledge of conditional probability. |
| 1. | Trait A is dominant to trait a and
it shows 80% penetrance in the homozygous state and 50% penetrance in the
heterozygous state. Two heterozygous individuals marry. What is the
probability that their first child would show the a phenotype?
The initial cross is Aa x Aa. There are two phenotypes A and a. The frequency of each is:
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| Of those individuals with the a phenotype
what proportion have the AA genotype, the Aa genotype and
the aa genotype? |
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| Exactly .55 of the individual have the a
phenotype. These individuals can be of three different phenotypes AA,
Aa and aa. To set the frequency of the a
phenotype to 1, divide .55/.55 = 1. The corrected proportional frequencies of the three genotypes are AA = .09 (.05/.55); Aa = .455 (.25/.55 ) and ..455 (.25/.55)
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| 2. | The pole vault competition involves both running
and accurate placement of the pole dead eye (accurate placement) is
dominant to inaccurate. Dead eye, shows 60% penetrance in
the homozygous and heterozygous states. McJumper is an inaccurate
male who is married to a homozygous inaccurate female. McJumper’s
father was heterozygous for dead eye and his mother was homozygous inaccurate. |
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| What is the probability that McJumper is
heterozygous for dead eye?
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McJumper is the product of a marriage of
Aa X
aa (stated in problem). Expected phenotypic frequencies for
McJumper are given below.
McJumper is inaccurate with a probability
of 1 (.70/.70) so he has a 2/7 change of having the genotype Aa
(.20/.70) and a 5/7 change of having the genotype aa (.50/.70). |
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| What is the probability his first child will be a deadeye offspring? | ||||||||||||||||||||||||||
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Prob. of first child being deadeye if McJumper marries a woman who is aa = [Pr(McJumper is Aa) x Prob of having a child who is Aa and who penetrates)] + [Pr(McJumper is aa) x Prob of having a child who is Aa and who penetrates)] = 2/7(3/10) + 5/7(0) = 6/70 = 3/35
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| 3. | In Christmas Jellyfish, flashing lights are dominant to static lights. Flashing lights have 90% penetrance in the homozygote (AA) and 70% penetrance in the heterozygote (Aa). For each of the following different crosses give the distribution of flashing lights to static lights. | |||||||||||||||||||||||||
| Cross | flashing lights | static lights. |
| AA x AA | .9 | .1 |
| AA x Aa | .80 | .20 |
| AA x aa | .70 | .30 |
| Aa x Aa | .575 | .425 |
| Aa x aa | .35 | .65 |
| aa x aa | 0 | 1 |
| 4. | In certain isolated populations of humans in the New Jersey Pine Barrens (especially around Leeds Point, NJ) there is a autosomal dominant trait (A) in which affected individuals are born with the head of a horse, large wings, cloven hooves, horns and a tail. Individuals with this trait have been called various things including the Hoodle-Doodle Bird, "Wozzle Bug", the "Leeds Devil"and most famously, The New Jersey Devil. The recessive phenotype is a normal human. |
| Geneticist Athin Schemata* has studied this extensively and she has shown that Devilism is an autosomal dominant trait with 10% penetrance in the homozygote and 5% penetrance in the heterozygote (which is why NJ Devils are so rare). Two New Jersey Devil’s are both known to be heterozygous (Aa). Their son Eddie is phenotypically normal; that is, Eddie may have the genotype AA, Aa or aa. What is the probability of Eddie having each genotype? | |
| Eddie grows up and marries a normal woman (Lucia Soprano) from North Jersey. This woman does a lot of genealogy and knows that there has never been a NJ Devil born in her family (i.e., her genotype is aa). Eddie and Lucia ask you to calculate the probability that their first child will be a New Jersey Devil. They also want to know the probability of having three normal children. |
4. Eddie's parents are Aa X Aa (given in problem). The genotypes expected among the progeny of this cross are
1/4 Aa + 1/2Aa +1/4aa.
The expected Phenotypes are:
| Genotype | Freq. | Penetrance Devil | Penetrance Normal | Progeny Phenotypes | |
| Devil | Normal | ||||
| AA | 1/4 | 1/10 | 9/10 | 1/40 | 9/40 |
| Aa | 1/2 | 1/20 | 19/20 | 1/40 | 19/40 |
| aa | 1/4 | 0 | 1 | 0 | 10/40 |
| TOTAL | 2/40 | 38/40 | |||
The probability of being normal is before any child is born is
9/40AA + 19/40Aa + 10/40 aa = 38/40
Make the probability of being normal equal to 1.0 since Eddie is normal. Divide all terms by 38/40.
9/38AA + 19/38Aa + 10/38 aa = 38/38 = 1
| Marriage | Freq. | Prob. Aa | Penetrance A | Penetrance a | Prob. Progeny Phenotypes | |
| A | a | |||||
| AA x aa | 9/38 | 1 | 1/10 | 9/10 | 36/1520 | 324/380 |
| Aa x aa | 19/38 | 1/2 | 1/20 | 19/20 | 19/1520 | 741/1520 |
| aa x aa | 10/38 | 0 | 0 | 1 | 0 | 400/1520 |
| TOTAL | 55/1520 | 1465/1520 | ||||
The probability of Eddie and Lucia having a New Jersey Devil is 55/1520 or about 3.6%
The probability of three normal children is (1465/1520)3 = 89.5%
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Penetrance and Conditional Probability |
|
| 1 | Penetrance |
| 2 | Test Your Understanding of Conditional Probability and Penetrance |
| 3 | Answers to "Test Your Understanding ..." |
| 4 | Home |