Page 164 - Genetics_From_Genes_to_Genomes_6th_FULL_Part1

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

than 0.05 thus suggests that the data show major deviations essential concepts
from predicted values significant enough to reject the null
hypothesis with greater than 95% confidence. More con- • A null hypothesis is a model that leads to a discrete
servative scientists often set the boundary of significance numerical prediction.
at p = 0.01, and they would therefore reject the null hypoth- • The chi-square test for goodness of fit helps determine
esis only if their confidence was greater than 99%. whether two genes are linked by comparing differences
In contrast, p values greater than 0.01 or 0.05 do not nec- between the numbers of progeny of different classes
essarily mean that two genes are unlinked; it may mean only observed in an experiment, and the numbers of progeny
that the sample size is not large enough to provide an answer. of these classes expected from the null hypothesis that
With more data, the p value normally rises if the null hypoth- genes are unlinked and thus assort independently.
esis of no linkage is correct and falls if there is, in fact, linkage. • The probability value (p) measures the likelihood that
Note that in Fig. 5.19 all of the numbers in the second deviations from the predicted values have occurred by
set of data are simply double the numbers in the first set, chance alone; the null hypothesis is rejected when p < 0.05.
with the proportions remaining the same. Thus, by dou-
bling the sample size from 48 to 96 individuals, it was pos-
sible to go from no significant difference to a significant 5.5 Tetrad Analysis in Fungi
difference between the observed and the expected values.
In other words, the larger the sample size, the less the like-
lihood that a certain percentage deviation from expected learning objectives
results happened simply by chance. Bearing this in mind,
you can see that it is not appropriate to use the chi-square 1. Explain the meaning of the term tetrad as applied to the
test when analyzing small samples of less than 10. This is- asci produced by certain fungi.
sue creates a problem for human geneticists because human 2. Differentiate between parental ditype (PD), nonparental
families produce only a small number of children. To ditype (NPD), and tetratype (T).
achieve a reasonable sample size for linkage studies in hu- 3. Describe how the relative numbers of PDs and NPDs
mans, scientists often pool data from a large number of can be used to establish linkage.
family pedigrees and use a different statistical analysis 4. Explain how ordered and unordered tetrad analysis can
called the Lod score (see Chapter 11). map the positions of genes and (for ordered tetrads)
We emphasize again that the chi-square test does not centromeres.
prove linkage or its absence. What it does is provide a
quantitative measure of the likelihood that the data from an
experiment can be explained by a particular hypothesis. With Drosophila, mice, peas, people, and other diploid or-
The chi-square analysis is thus a general statistical test for ganisms, each individual represents only one of the four
significance; it can be used with many different experimen- potential gametes generated by each parent in a single mei-
tal designs and with hypotheses other than the absence of otic event. Thus, until now, our presentation of linkage,
linkage. As long as it is possible to propose a null hypoth- recombination, and mapping has depended on inferences
esis that leads to a predicted set of values for a defined set derived from examining the phenotypes of diploid progeny
of data classes, you can readily determine whether or not resulting from random unions of meiotic products en masse
the observed data are consistent with the hypothesis. (that is, among a large group). For such diploid organisms,
When experiments lead to rejection of a null hypothesis, we do not know which, if any, of the parents’ other progeny
you may need to confirm an alternative. For instance, if you arose from gametes created in the same meiosis. Because
are testing whether two opposing traits result from the segre- of this limitation, the analysis of random products of meio-
gation of two alleles of a single gene, you would expect a sis in diploid organisms must be based on statistical sam-
testcross between an F 1 heterozygote and a recessive homo- plings of large populations.
zygote to produce a 1:1 ratio of the two traits in the offspring. In contrast, various species of fungi provide a unique
If instead, you observe a ratio of 6:4 and the chi-square test opportunity for genetic analysis because they house all four
produces a p value of 0.009, you can reject the null hypoth- haploid products of each meiosis in a sac called an ascus
esis (a 1:1 ratio). But you are still left with the question of (plural, asci). These haploid cells, or ascospores (also
what the absence of a 1:1 ratio means. Two alternatives exist: known as haplospores or simply spores), can germinate
(1) Individuals with the two possible genotypes are not and survive as viable haploid individuals that perpetuate
equally viable, or (2) more than one gene affects the trait. themselves by mitosis. The phenotype of such haploid
The chi-square test cannot tell you which possibility is cor- fungi is a direct representation of their genotype, without
rect, and you would have to study the matter further. The complications of dominance.
problems at the end of this chapter illustrate several applica- Figure 5.20 illustrates the life cycles of two fungal
tions of the chi-square test pertinent to genetics. species that preserve their meiotic products in a sac. One,