Lesson 6 of 8 / Inheritance and phenotype
Track the X chromosome and a second gene
Is the probability among sons or among all children?
In this lesson: Explain sex linkage and solve a dihybrid cross combining an X-linked and an autosomal locus.
About 6 min
The key ideaFor a usual X-linked recessive model, sons receive their X from the mother. Combine that inheritance with the second locus and state the denominator.
Explore the idea
Use all offspring as the denominator
XNXnPp x XNYpp
Xn is recessive in the usual XY model. P/p is an invented independent marker. Daughters in this particular cross are unaffected, including carriers. Row and column labels are gametes, not parent genotypes.
| Gamete | XNp | Yp |
|---|---|---|
| XNP | XNXNPp | XNYPp |
| XNp | XNXNpp | XNYpp |
| XnP | XNXnPp | XnYPp |
| Xnp | XNXnpp | XnYpp |
Explanation
Sex linkage means a locus is on a sex chromosome. In the usual XY model, a male has one X and one Y; for many X-linked loci, the Y has no corresponding allele. An X-linked recessive allele can therefore be expressed in a male with that single allele on his X.
Use XN for an X bearing the normal-vision allele and Xn for the recessive colour-vision allele in this teaching model. A heterozygous XNXn mother and XNY father give equal expected categories XNXN, XNXn, XNY and XnY. Among sons, half are XnY; among all offspring, one quarter are XnY.
Add an invented unlinked autosomal marker P/p, with P dominant. For XNXnPp x XNYpp, the mother makes XNP, XNp, XnP and Xnp gametes; the father makes XNp and Yp. The marker cross gives pp in half the offspring independently of the X-linked outcome.
The probability of an affected son with pp among all offspring is 1/4 x 1/2 = 1/8. Among sons only, it is 1/2 x 1/2 = 1/4. State whether the question conditions on sex; confusing these denominators doubles or halves the result incorrectly.
Step by step
- 1
Keep X alleles attached to X
Do not place an X-linked allele on the Y in this model.
- 2
Make full gametes
Each gamete also receives one allele at the autosomal marker locus.
- 3
State the denominator
All offspring and sons only are different sample spaces.
Worked example
A conditional probability
In the stated cross, what fraction of sons are expected to be XnY and pp?
One way to explain it
Among sons, P(XnY) = 1/2 and P(pp) = 1/2, giving 1/4 of sons. The corresponding fraction of all offspring is 1/8 under equal expected sex proportions.
Why this answer works
- Condition on being a son first.
- Then combine the independent marker probability.
Is this true? "A son receives his father's X-linked allele because both are male."
In the usual XY model, a son receives Y from his father and X from his mother.