One dataset fitting 9:3:3:1 and one failing 1:1:1:1 by a wide margin are tested by chi-square in this SC435 Unit 4 dihybrid analysis, which ends at a map distance. Searches like "sc 435 unit 4 assignment example", "sc435 unit 4 sample" and "sc435 unit 4 example" land here.
What a finished SC435 Unit 4 dihybrid cross analysis looks like
Six pages, two data tables, two chi-square tables and a short results discussion for each dataset. The corn table lists purple smooth, purple wrinkled, yellow smooth and yellow wrinkled kernels at 187, 54, 60 and 19 against expected counts of 180, 60, 60 and 20, and its chi-square table shows each class's contribution before the total of 0.92 with 3 degrees of freedom, well below the 7.81 critical value at 0.05. The fly table lists gray normal-winged, black vestigial-winged, black normal-winged and gray vestigial-winged flies at 412, 398, 91 and 99, against 250 each, and its chi-square reaches about 385. Parental classes are identified as the two largest, recombinants as the two smallest, and the recombination frequency comes out at 19 percent, or 19 map units.
How a SC435 Unit 4 example is structured
The analysis opens with the hypothesis written for each dataset, because a chi-square test is only meaningful against a stated expectation, and both are framed as independent assortment. Parental genotypes and the cross are then drawn, a 4-by-4 grid for the corn F2 and a branch diagram for the test cross, so the expected ratios have visible sources. Calculations follow in tables showing observed, expected, the squared difference over expected for each class and the sum. Degrees of freedom are explained as classes minus one. The corn result is interpreted as consistent with independent assortment, not as proof of it. The fly result rejects the hypothesis, and the discussion explains why linkage is the likeliest reason: parental classes dominate and recombinants are scarce. The recombination frequency is calculated last, with a sentence on why 50 percent is its ceiling.
Hypotheses stated first
Both datasets are tested against independent assortment, written out before any number appears, so each chi-square has a claim to accept or reject.
A grid and a branch diagram
The corn F2 uses a full sixteen-square grid and the test cross uses a branch diagram, each producing the expected ratio its table then tests.
Contributions before the total
Every class shows its own squared difference over expected. On the corn ear no class contributes more than 0.6, which shows the fit is even across classes rather than lucky in one.
Rejecting without overclaiming
At about 385 the fly result rejects independent assortment. The discussion names linkage as the likeliest explanation while noting that the test itself only says the ratio is wrong.
Nineteen map units
Recombinants, 91 plus 99 of 1,000, give a frequency of 19 percent. The analysis explains that frequency approximates map distance and cannot exceed 50 percent between two loci.
Where marks go in SC435 Unit 4
Dihybrid analyses in this course are commonly marked on correct expected values, correct chi-square calculation and interpretation, identification of parental and recombinant classes, and the linkage calculation. Expected values cause the first losses when a 9:3:3:1 is applied to a test cross or a 1:1:1:1 to an F2. Calculation credit needs every class shown, and a lone total with no working earns little even when correct. Interpretation is weighted heavily: a small chi-square means the data are consistent with the hypothesis, not that it is proven, and a large one means it is rejected, not which alternative is true. Degrees of freedom reported as four instead of three draw a deduction. On the fly data, treating the two smallest classes as parental reverses the whole linkage calculation and costs most of that section's credit.
Get a SC435 Unit 4 example written to your instructions
Share the counts and crosses from your SC435 Unit 4 assignment, the problem or lab instructions, the rubric and whichever software is required. A first custom analysis is written free inside 24-48h, with hypotheses stated, each class's chi-square contribution shown and, where the numbers demand it, a recombination frequency worked out.
SC435 Unit 4 questions, answered
How do I know which expected ratio to use?
It depends on the cross. Two double heterozygotes crossed together give 9:3:3:1 if the genes assort independently, while a double heterozygote crossed to a double recessive gives 1:1:1:1. Draw the cross first, work out the gametes each parent makes, then derive the expected proportions rather than recalling them from memory.
What does failing to reject the hypothesis actually mean?
That the observed counts sit close enough to the expected ones that chance could explain the difference. It does not prove independent assortment, only that these data give no reason to doubt it. Graders look for that exact wording, and answers saying the hypothesis is proven usually lose interpretation credit.
Why can't recombination frequency exceed 50 percent?
Because at 50 percent, recombinant and parental gametes are equally common, which is exactly what independent assortment produces. Genes far apart on the same chromosome undergo so many crossovers that they behave as if unlinked. Frequencies near 50 percent therefore cannot distinguish distant linked genes from genes on different chromosomes.