Genes on the same chromosome travel together. The offspring stop obeying Mendel: and that broken ratio turns out to be the most useful thing on the page.
Genes with loci on the same chromosome do not assort independently. They tend to be inherited as a unit, so a dihybrid cross behaves more like a monohybrid one.
Crossing over in prophase I can separate linked alleles. Offspring carrying combinations not present in the parents are recombinants: and they are always the minority classes.
The further apart two loci sit, the more likely a chiasma forms between them. Crossover frequency is a measure of distance: the basis of Morgan's first gene maps.
Green came from one parent, pink from the other. A chromatid that ends up two-toned is a recombinant. Tap through the parts.
Two homologues, each already replicated into two sister chromatids. A single chiasma between the two loci affects only two of the four chromatids: which is why recombinants can never exceed 50% of the offspring.
A heterozygote AB/ab test crossed with ab/ab. Drag the loci closer and watch 1:1:1:1 collapse: and watch χ² decide, in real time, whether the difference is big enough to believe.
At 50% the two loci behave as though they were on different chromosomes entirely: all four classes equal, χ² near zero, no evidence of linkage. Now start dragging.
χ² asks a single question: is the gap between what you observed and what you expected too large to blame on chance?
The null hypothesis is always the boring one: there is no significant difference between the observed and expected frequencies, in other words, the genes are unlinked.
Type your own observed counts. Everything else is calculated live: including the verdict.
The values loaded are Mendel's real counts from 556 F₂ pea seeds. Change any number and watch χ² respond: then try making the fit bad enough to cross 7.815.
Each token is one offspring from a test cross. A green box means the dominant allele of gene A, a filled dot means the dominant allele of gene B. Tally the four types, type your counts in, then check them: and only then find out whether these genes were linked.
For every dihybrid cross in this course df = number of phenotype classes − 1 = 3
Do them on paper first, then open the solution. You can check every line in the calculator above.
In tomato plants, purple stem (P) is dominant to green stem (p) and hairy stem (H) is dominant to smooth (h). Two plants heterozygous for both genes were crossed and 160 offspring were scored: 92 purple hairy, 26 purple smooth, 34 green hairy, 8 green smooth. Use a chi-squared test to determine whether these genes are linked.
H₀: there is no significant difference between the observed and expected frequencies, the genes are unlinked. H₁: there is a significant difference, the genes are linked.
Expected = predicted ratio × total. With 160 offspring: 9/16 × 160 = 90, 3/16 × 160 = 30, 3/16 × 160 = 30, 1/16 × 160 = 10.
χ² = 0.044 + 0.533 + 0.533 + 0.400 = 1.51. There are four phenotype classes, so df = 4 − 1 = 3.
At p = 0.05 with df = 3 the critical value is 7.815. Since 1.51 is well below it, p > 0.05 and the difference is not statistically significant.
A fruit fly heterozygous for body colour and wing shape was test crossed with a black, vestigial-winged fly. The 300 offspring were: 142 grey normal, 12 grey vestigial, 15 black normal, 131 black vestigial. Determine whether the genes are linked, and if so, calculate the recombination frequency.
A test cross on unlinked genes predicts 1 : 1 : 1 : 1, so each expected value is 300 ÷ 4 = 75. This is the prediction H₀ is built on.
χ² = 59.85 + 52.92 + 48.00 + 41.81 = 202.6, with df = 3. That is enormously greater than the critical value of 7.815, so p is far below 0.05.
The two large classes, grey normal and black vestigial, are the parental types, so the heterozygous parent carried the alleles in that arrangement. The two small classes are the recombinants: 12 + 15 = 27.
RF = recombinants ÷ total × 100 = 27 ÷ 300 × 100 = 9%, which places the two loci roughly 9 map units apart on the same chromosome.
A common slip: χ² is calculated from counts, never from percentages or ratios
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Linkage is invisible in a single organism. You can only see it in the numbers: which is why a statistical test sits at the centre of this topic rather than at the end of it.
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Drag each term into the gap it belongs in. Two terms are traps.
Single best answer, Paper 1 style. Have the critical value table handy.