16.3 Linkage and Sex Linkage
Key Takeaways
Linked genes lie close together on the same chromosome and tend to be inherited together, so close linkage does not produce a 9:3:3:1 phenotype ratio.
Crossing over in prophase I produces recombinant gametes, and recombination frequency equals recombinant offspring divided by total offspring.
One percent recombination equals one map unit, and genes very far apart on one chromosome can approach 50 percent recombination and look unlinked.
Thomas Hunt Morgan connected a white-eye allele in fruit flies to the X chromosome, and sons do not inherit their father's X-linked allele.
An unaffected carrier mother and an unaffected father give each son a 1/2 chance of showing a recessive X-linked trait; an affected father and a non-carrier mother have carrier daughters and unaffected sons.
16.3 Linkage and Sex Linkage
Independent assortment is the expectation for genes on different chromosomes. Linkage is the exception. Genes that sit close together on the same chromosome tend to be inherited together, because a gamete usually receives that stretch of chromosome in one piece. Linkage leaves segregation in place. Each gene still contributes one allele to a gamete. What changes is the combination of alleles from neighboring genes. Close linkage does not give a 9:3:3:1 phenotype ratio, because that ratio requires independent assortment.
Parental Combinations and Crossing Over
Consider two genes on one chromosome in a common arrangement: A and B together on one homolog, and a and b together on the partner. The allele combinations already together on a chromosome are the parental combinations. Without an exchange, the gametes are AB and ab.
Crossing over in prophase I produces recombinant gametes. Homologous chromosomes pair, and nonsister chromatids can exchange matching segments. A crossover between the two genes yields Ab and aB gametes. Those genotypes were not the combinations on the parent's original homologs. In a test cross, offspring that receive a recombinant gamete from the dihybrid parent are recombinant offspring. Offspring that receive AB or ab are parental offspring.
Recombination frequency equals recombinant offspring divided by total offspring. Suppose 12 of 200 test-cross offspring are recombinant. The recombination frequency is 12/200, which is 6 percent. One percent recombination is one map unit, also called a centimorgan, so 6 percent recombination is 6 map units.
Genes very far apart on one chromosome can recombine often enough to look unlinked, near 50 percent. Repeated exchanges between distant loci mix combinations until a test cross approaches four equal classes. A result near 50 percent therefore leaves open the possibility that the genes are far apart on one chromosome. A result far below 50 percent, such as 6 percent, is evidence that the genes are linked and relatively close.
Completely linked genes, with no crossover in the sample, behave as one unit. A test cross of a dihybrid that carries AB on one homolog and ab on the other then yields two phenotype classes, the parental classes, in about equal numbers. That result is not 1:1:1:1, and it is not 9:3:3:1.
Morgan Connected White Eyes to the X Chromosome
Thomas Hunt Morgan's fruit-fly work connected a white-eye allele to the X chromosome. Red eyes were the common phenotype, and white eyes appeared as a variant. A male has one X, so an X-linked allele in a male has no second copy on the Y. Sons do not inherit their father's X-linked allele. A father passes his Y chromosome to his sons and his X chromosome to his daughters.
A white-eyed female crossed with a red-eyed male shows the pattern in one generation. The sons are white-eyed because they receive the mother's X. The daughters are red-eyed because they receive the father's X, which carries the common allele. An autosomal gene would not sort the first-generation phenotypes by sex in this criss-cross way. Morgan's conclusion was chromosomal: the white-eye allele is carried on the X chromosome.
| Situation | What is transmitted or counted | Result to remember |
|---|---|---|
| Genes close together on one chromosome | Mostly parental allele combinations | Not a 9:3:3:1 phenotype ratio |
| Crossing over in prophase I | Recombinant gametes | New combinations of linked alleles |
| Recombinant offspring divided by total offspring | Recombination frequency | 1 percent equals 1 map unit |
| Genes very far apart on one chromosome | Recombination can approach 50 percent | They can look unlinked |
| Father with an X-linked allele | X to daughters and Y to sons | Sons do not inherit his X-linked allele |
Recessive X-Linked Probabilities
Start with a recessive X-linked trait, an unaffected carrier mother, and an unaffected father. The mother has one trait allele and one typical allele. The father has a typical allele on his single X. Sons receive their single X from the mother and a Y from the father. Half of the mother's X chromosomes carry the trait allele, so each son has a 1/2 chance of being affected. Daughters receive the father's normal X. A daughter is an unaffected carrier only if she also receives the mother's trait allele. In this mating her chance of being a carrier is 1/2, and she is unaffected, because the father's X carries the typical allele. An affected daughter needs the trait allele from both parents.
A color-blind father does not pass his X to his sons. He passes his Y to sons and his X to all daughters. If the mother is not a carrier, the sons are unaffected, because their X comes from her. Every daughter is a carrier, because each daughter received the father's X-linked allele and the mother's typical allele.
Y-linked inheritance passes a trait from a father to all of his sons and to none of his daughters. Such traits are rare. Red-green color blindness follows the X-linked pattern above. In an X-linked recessive family, affected males are connected through carrier females, and an affected father passes the trait allele to daughters rather than to sons.
One Pedigree Told in Words
An affected father carries a recessive X-linked allele on his only X. The mother is unaffected and is not a carrier, so both of her X chromosomes carry the typical allele. Every daughter receives the father's X and one typical X from the mother. All daughters are unaffected carriers. They inherited the allele, and the recessive phenotype stays hidden. Every son receives the father's Y and a typical X from the mother. No son is affected. No son inherited the father's X-linked allele.
Ask which children inherit the allele, not only which children show the trait. In this family the carriers are the daughters, and none of the children is affected.
Warning
Close linkage does not give a 9:3:3:1 ratio, because that phenotype ratio requires unlinked genes. Sons do not inherit their father's X-linked allele. A father passes his X to daughters and his Y to sons. Recombination frequency is recombinant offspring divided by total offspring, and one percent recombination is one map unit.
A test cross produces 200 offspring. Eighteen of them are recombinant, and the other 182 match the parental allele combinations. The loci are on the same chromosome. Which statement is correct?
The recombination frequency is 182/200, because the parental offspring are the ones produced by crossing over in prophase I.
The recombination frequency is 18/200, or 9 percent, so the loci are 9 map units apart; genes very far apart on one chromosome can look unlinked, near 50 percent recombination.
The recombination frequency is 18 map units, and any two genes on one chromosome produce a 9:3:3:1 phenotype ratio.
The recombination frequency is 1/2 for every linked pair, because one map unit is defined as 50 percent recombination.
An unaffected mother carries a recessive X-linked allele, and the father is unaffected. What is the chance that a particular son is affected, and what does a color-blind father pass to his sons?
Every son is affected, and a color-blind father passes his X-linked allele to every son.
Each son has a 1/2 chance of being affected because he gets his single X from his mother; a color-blind father passes his Y to his sons and his X to his daughters.
No son can be affected, because a son inherits his single X from his father.
Each son has a 1/4 chance of being affected, and a color-blind father passes his X to sons and his Y to daughters.
An affected father and an unaffected mother who is not a carrier have several children. The trait is recessive and X-linked. Which conclusion is sound?
Every son is affected, no daughter inherits the allele, and close linkage produces a 9:3:3:1 phenotype ratio.
No child can inherit the allele, because a father transmits an X-linked allele only when the mother is also a carrier.
All daughters are unaffected carriers and no son is affected, because sons do not inherit their father's X-linked allele; close linkage also does not give a 9:3:3:1 ratio.
Sons inherit the father's X-linked allele, daughters inherit his Y chromosome, and linked genes cannot recombine.
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