Two genes, four gametes, sixteen boxes. Learn to read a ratio backwards: and the cross that exposes a hidden genotype.
A double heterozygote RrTt makes four gamete types, RT, Rt, rT, rt, because unlinked genes assort independently in metaphase I.
Cross two double heterozygotes and the F2 phenotypes fall out as 9 : 3 : 3 : 1. Nine boxes show both dominant traits; one shows both recessive.
Cross anything with a homozygous recessive and you have a test cross. The offspring read out the unknown parent's gametes directly.
Seed shape: round (R) is dominant to wrinkled (r). Plant height: tall (T) is dominant to dwarf (t). Tap a phenotype class to light up its boxes.
Each parent makes four gamete types, so the grid is 4 × 4 = 16 equally likely combinations. Tap a class to see which boxes produce it.
A round, tall plant could be four different genotypes. Cross it with rrtt and the offspring give it away: because the recessive parent contributes nothing but rt.
Choose one of the four possible genotypes for the round, tall parent: or hit “Mystery plant” and work backwards from the offspring.
Five steps, then two crosses solved in full. Try each one on paper first: then open the solution.
The method is identical whether the question asks for 9:3:3:1 or for an unknown genotype. Only the second parent changes.
In a tomato variety, tall stems (T) are dominant to dwarf (t), and hairy stems (H) are dominant to smooth (h). The two genes are unlinked. Two plants that are heterozygous for both genes are crossed. Determine the expected phenotypic ratio of the offspring.
Tall T > dwarf t; hairy H > smooth h.
Heterozygous for both means TtHh × TtHh. Alleles are paired gene by gene: TtHh, never THth.
Each parent gives four types: TH, Th, tH, th. Take one allele of each gene, in every combination.
4 × 4 = 16 equally likely boxes.
Group the boxes by appearance: 9 have at least one T and one H; 3 are T_hh; 3 are ttH_; 1 is tthh.
In a breed of dog, black coat (B) is dominant to brown (b) and short hair (S) is dominant to long (s). A black, short-haired dog is test crossed with a brown, long-haired dog. Across several litters the puppies are: 12 black short-haired, 14 black long-haired, 11 brown short-haired, 13 brown long-haired. Deduce the genotype of the black, short-haired parent.
Brown and long-haired is the double recessive, bbss, so it can only donate bs. Every dominant trait in a puppy must have come from the other parent: so the puppies are a direct readout of that parent's gametes.
Black 12 + 14 = 26, brown 11 + 13 = 24. Roughly 1 : 1, so coat colour segregated and the unknown parent is heterozygous: Bb.
Short 12 + 11 = 23, long 14 + 13 = 27. Again roughly 1 : 1, so hair length segregated too: Ss.
BbSs × bbss gives four gametes (BS, Bs, bS, bs) against bs only → 1 BbSs (black short) : 1 Bbss (black long) : 1 bbSs (brown short) : 1 bbss (brown long). Four phenotypes in equal numbers, 1 : 1 : 1 : 1, which is what the data show.
The four classes appearing at all is the giveaway. Had the parent been BBSs, no brown puppies could have appeared; had it been BBSS, every puppy would have been black and short-haired.
Note: ratios are probabilities · real litters and real F2 populations scatter around them
Tap a concept to light up how it connects.
Independent assortment in meiosis I is what makes the 9:3:3:1 ratio possible; when a cross fails to give it, the genes were probably never independent in the first place.
Tap any concept to trace its connections · tap the background to reset
Drag each term into the gap it belongs in. Two terms are traps.
Single best answer, Paper 1 style. Pick one: you'll see why.