Researchers cross a corn plant that is pure-breeding for the dominant traits col

Question

Researchers cross a corn plant that is pure-breeding for the dominant traits colored aleurone (C1), full kernel (Sh). and waxy endosperm (Wx) to a pure-breeding plant with the recessive traits colorless aleurone (c1). shrunken kernel (sh), and starchy (wx). The resulting F_1 plants were crossed to pure-breeding colorless, shrunken, starchy plants. Counting the kernels from about 30 ears of corn yields the following data: a. Perform a chi-square test to determine if these data conform to the expected results if we assume independent assortment. Provide your chi-square value, degrees of freedom, P value, and if independent assortment is supported or not. b. What is the order of these genes in corn? c. What is the recombination frequency between each gene pair? C1-Sh = ______ C1-Wx = _______ Sh-Wx = ________

Explanation / Answer

(a) make a chi square table including observed and expected values .

so ,The chi-square value is 10,649. P value is below 0.01, which indicates that the difference between the expected and observed results is highly statistically significant .

degree of freedom = n-1 = 8-1 = 7

The lack of the 1:1:1:1:1:1:1:1 ratio suggests linkage among the three genes. You can verify this through a chi-square test in less obvious cases. The parental types of the fully heterozygous parent must be: C s W and c S w because they occur at the highest frequencies. At this point though, we do not know the order (i.e., phase) in which these genes occur along the chromosome.

(b)

Since 21.7 cM is the largest distance, we conclude that C and W are the

outside genes, with S being in the middle:

____|______|___________________|_____

| 3.5 | 18.4 |

C SW

Note that 3.5 + 18.4 = 21.9 cM is slightly larger than the 21.7 cM distance calculated above. The

combined distance between C and S and S and W is a more accurate measure of distance

between C and W because it compensates for undetectable double-crossovers in each interval.

Finally, it is also possible to determine gene order without calculating map distances. The two

least frequent phenotype classes (Colored, plump, starchy and colorless, shrunken, waxy) had to

have been produced by two crossovers: one between C and S and another between S and W.

Because we know the parental types are C s W and c S w, a crossover in each interval (between

C and S and S and W) results in CSw and csw, which are the phenotypes of the two least

frequent progeny classes. Thus, S must be in the middle

so The gene order is Cl-Sh-Wx.

(c)

C1 and Sh: 100 * [(113+116+4+2)/6708] = 3.5 cM

Sh and Wx: 100 * [(626+601+4+2)/6708] = 18.4 cM

C1 and Wx: 100 * [(113+116+626+601)/6708] = 21.7 cM

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