Page 176 - Genetics_From_Genes_to_Genomes_6th_FULL_Part1

168 Chapter 5 Linkage, Recombination, and the Mapping of Genes on Chromosomes

from the red-green color blindness locus that the two genes Figure 5.31 A genetic map of part of the human
recombine often. The color blindness and hemophilia B X chromosome.
genes may appear to be genetically unlinked in a small sam-
ple (as in Fig. 5.1b), but the actual recombination distance
separating the two genes is about 36 m.u. Pedigrees point-
ing to two different forms of hemophilia, one very closely
linked to color blindness, the other almost not linked at all,
provided one of several indications that hemophilia is deter- Hunter syndrome
mined by more than one gene (Fig. 5.31). Hemophilia B
Linkage and recombination are universal among life- Fragile X syndrome
forms and must therefore confer important advantages to
living organisms. Geneticists reason that linkage provides Hemophilia A
the potential for transmitting favorable combinations of
genes intact to successive generations, while recombina- G6PD deficiency: Favism
tion produces great flexibility in generating new combina- Drug-sensitive anemia
Chronic hemolytic anemia
tions of alleles. Some new combinations may help a species
adapt to changing environmental conditions, whereas the Color blindness (several forms)
inheritance of successfully tested combinations can pre- Dyskeratosis congenita
serve what has worked in the past. Deafness with stapes fixation
TKCR syndrome
Thus far, this book has examined how genes and chro-
mosomes are transmitted. As important and useful as this Adrenoleukodystrophy
knowledge is, it tells us very little about the structure and Adrenomyeloneuropathy
mode of action of the genetic material. In the next section Emery muscular dystrophy
(Chapters 6–8), we carry our analysis to the level of DNA, SED tarda
the molecule of heredity. Spastic paraplegia, X-linked

SOLVED PROBLEMS

I. The XG locus on the human X chromosome has two during meiosis affects the transmission of alleles: in
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alleles, XG and XG. The XG allele causes the pres- this problem, the daughter. The X chromosome she
ence of the Xg surface antigen on red blood cells, inherited from her father (who had icthyosis and no
while the recessive XG allele does not allow antigen Xg antigen) must be STS XG. (No recombination
to appear. The XG locus is 10 m.u. from the STS could have separated the genes during meiosis in her
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locus. The STS allele produces normal activity of father since he has only one X chromosome.) Because
the enzyme steroid sulfatase, while the recessive STS the daughter is normal and has the Xg antigen, her
allele results in the lack of steroid sulfatase activity other X chromosome (inherited from her mother)
and the disease ichthyosis (scaly skin). A man with must contain the STS and XG alleles. Her X
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ichthyosis and no Xg antigen has a normal daughter chromosomes can be diagrammed as:
with Xg antigen. This daughter is expecting a child.
a. If the child is a son, what is the probability he will STS XG
lack Xg antigen and have ichthyosis?
b. What is the probability that a son would have both STS + XG +
the antigen and ichthyosis? Because the STS and XG loci are 10 m.u. apart on the
c. If the child is a son with ichthyosis, what is the chromosome, the recombination frequency is 10%.
probability he will have Xg antigen? Ninety percent of the gametes will be parental: STS
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XG or STS XG (45% of each type) and 10% will be
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recombinant: STS XG or STS XG (5% of each type).
Answer The phenotype of a son directly reflects the genotype
a. This problem requires an understanding of how link- of the X chromosome from his mother. Therefore, the
age affects the proportions of gametes. First designate probability that he will lack the Xg antigen and have
the genotype of the individual in which recombination icthyosis (genotype: STS XG / Y) is 45/100.