Page 88 - Genetics_From_Genes_to_Genomes_6th_FULL_Part3

382 Chapter 11 Analyzing Genomic Variation

the T allele of the SNP. II-2, the unaffected mate of II-1, family sizes are small, making it difficult to obtain suffi-
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does not have the disease (so she is NF /NF ), and the cient data for precise mapping.
DNA analysis indicates that she is homozygous for the G
allele of SNP1. The children in generation III of the pedi- The phase problem
gree are thus in effect the progeny of a testcross. Examin- You should note that the calculation just presented for the
ing each child both for the presence or absence of the map distance between the neurofibromatosis gene and
disease and the SNP genotype will reveal whether the child SNP1 does not include one of the crosses shown in the
obtained a parental-type (nonrecombinant) or a recombinant- pedigree: the mating of I-1 and I-2 (see Fig. 11.20). The
type sperm from the doubly heterozygous father II-1. reason is that we don’t know the configuration of the
As Fig. 11.20a shows, seven out of eight progeny in
generation III resulted from parental-type sperm: either alleles (or phase) in the affected grandmother I-1. It is
possible that her mutant NF allele was on the same chro-
the combination of NF and G seen in the grandmother I-1, mosome 17 as the G allele of the SNP, but it is also pos-
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or the combination of NF and T in the grandfather I-2. sible that the SNP allele on this NF-bearing chromosome
Only one of the children in generation III was the product was instead T. Because we cannot say with certainty
of a recombinant sperm (III-8, an affected son with NF whether II-1 (her affected son) results from a parental-
and the T allele of the SNP). The data thus strongly sug- type or recombinant-type egg, we did not consider this
gest that the neurofibromatosis gene and this particular gamete in calculating the map distance.
SNP are genetically linked, separated by a map distance The phase problem can be resolved in either of two
(1/8) × 100 = 12.5 cM. circumstances. First, if you know the genotypes of two loci
A family size of eight children is large by today’s stan-
dards, but this number of progeny is still not sufficiently in both of a person’s parents, sometimes you can determine
which alleles came from one parent and which from
large to achieve great statistical significance. Nonetheless, the other. This is precisely how we knew that the phase in
the most straightforward interpretation of the data is that the double heterozygous father II-1 was NF G/NF T
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the neurofibromatosis gene is located on chromosome 17, (Fig. 11.20). So to determine the phase in I-1 by this
within a region that extends roughly 12.5 Mb to either side method, we would need to have genotyping information
of the SNP1 locus. Fig. 11.20b shows graphically how you about her affected parent. Second, if the two loci are suffi-
can easily interpret the cross between II-1 and II-2 based on ciently close together, you can infer the probable phase be-
these provisional conclusions.
cause the linked alleles should segregate with each other
more often than not. In Fig. 11.20, where the parental
Multipoint analysis classes considerably outnumber the recombinants in gen-
Microarray data include not just two-point information eration III, you could have inferred the likely arrangement
of alleles in the doubly heterozygous II-1 even if you had
about linkage relationships between a disease gene and no information about his parents.
individual DNA loci, but also multipoint information
about the behavior of millions of DNA loci with respect
to each other. You saw in Chapter 5 the power of three- Informative and noninformative crosses
point crosses in fruit flies, but microarrays are like three- Even if you know the phase in a doubly heterozygous
point crosses “on steroids.” Researchers can now pinpoint parent, a mating may not provide any useful information
particular crossovers occurring during the production of a about whether the two loci are linked. One example is seen
single gamete to positions between two polymorphisms in Fig. 11.21a, which again examines the mating between
on the same chromosome. As you will see in Section 11.5, male II-1 with neurofibromatosis and his unaffected part-
this information in turn allows researchers to pinpoint dis- ner II-2. However, we are now looking at a different SNP
ease genes to short regions defined by mapped recombi- locus (SNP2), where both parents are heterozygotes for the
nation events. alleles A and C. If a child were also a heterozygote for A
and C, you would not be able to tell which parent contrib-
uted the A and which the C, so you could not determine if
Positional Cloning Has the child was the result of a parental or recombinant gamete.
Several Limitations But if the child were a homozygote for either SNP2 allele,
say of genotype AA, then you know both the egg and sperm
For traits expressed in plants or small animals, researchers must have carried the A allele.
can easily set up crosses that generate hundreds of progeny You should remember from Chapter 5 that the basic
so as to allow accurate genetic mapping. But scientists do requirement for genetic mapping is that at least one parent
not direct human breeding, so not every mating between must be a double heterozygote. Figure 11.21b emphasizes
two people provides interpretable information about the this crucial point by showing that if neither parent is a
relative positions of any two given loci. In addition, human double heterozygote, the cross cannot be informative.