## D

appear with a frequency of 0.42 (the probability of inheriting a gamete with chromosome _^ from the heterozygous parent) X 1.00 (the probability of inheriting a gamete with chromosome _t_from the recessive parent) = 0.42. The proportions of the other types of F2 progeny can be calculated in a similar manner (see Figure 7.10). This method can be used for predicting the outcome of any cross with linked genes for which the recombination frequency is known.

### Testing for Independent Assortment

In some crosses, the genes are obviously linked because there are clearly more nonrecombinants than recombinants. In other crosses, the difference between independent assortment and linkage is not so obvious. For example, suppose we did a testcross for two pairs of genes, such as AaBb X aabb, and observed the following numbers of progeny: 54 AaBb, 56 aabb, 42 Aabb, and 48 aaBb. Is this outcome a 1:1:1:1 ratio? Not exactly, but it's pretty close. Perhaps these genes are assorting independently and chance produced the slight deviations between the observed numbers and the expected 1: 1: 1: 1 ratio. Alternatively, the genes might be linked, with considerable crossing over taking place between them, and so the number of nonrecombi-nants is only slightly greater than the number of recombinants. How do we distinguish between the roles of chance and of linkage in producing deviations from the results expected with independent assortment?

We encountered a similar problem in crosses in which genes were unlinked — the problem of distinguishing between deviations due to chance and those due to other

7.10 The recombination frequency allows a prediction of the proportions of offspring expected for a cross entailing linked genes.

factors. We addressed this problem (in Chapter 3) with the goodness-of-fit chi-square test, which serves to evaluate the likelihood that chance alone is responsible for deviations between observed and expected numbers. The chi-square test can also be used to test the goodness of fit between observed numbers of progeny and the numbers expected with independent assortment.

Testing for independent assortment between two linked genes requires the calculation of a series of three chi-square tests. To illustrate this analysis, we will examine the data from a cross between German cockroaches, in which yellow body (y) is recessive to brown body (y+) and curved wings (cv) are recessive to straight wings (cv+). A testcross (y+y cv+cv X yy cvcv) produced the following progeny:

77 yy cvcv

28 y+y cvcv

32 yy cv+cv brown body, straight wings yellow body, curved wings brown body, curved wings yellow body, straight wings

### 200 total progeny

Testing ratios at each locus To determine if the genes for body color and wing shape are assorting independently, we must examine each locus separately and determine whether the observed numbers differ from the expected (we will consider why this step is necessary at the end of this section). At the first locus (for body color), the cross between heterozygote and homozygote (y+y X yy) is expected to produce / y+y brown and / yy yellow progeny; so we expect 100 of each. We observe 63 + 28 = 91 brown progeny and 77 + 32 = 109 yellow progeny. Applying the chi-square test (see Chapter 3) to these observed and expected numbers, we obtain:

0 0