Lesson 3 of 9 / Inheritance and phenotype
Build a two-gene cross from gametes
When is 9:3:3:1 justified?
I can solve dihybrid crosses with explicit independent-assortment and dominance assumptions.
I can solve dihybrid crosses including codominance, incomplete dominance, multiple alleles and sex linkage.Syllabus
Syllabus 8876, Core 2(o). Solve dihybrid crosses including codominance, incomplete dominance, multiple alleles and sex linkage.
About 6 min
Make a guess. You are not marked.
Can a gamete from an AaBb plant be Aa?
Show the answer
A gamete has one allele per gene.
Gametes carry one allele from each locus: AB, Ab, aB or ab.
The key idea
List complete gametes before combining them. AaBb can produce AB, Ab, aB and ab equally when the loci assort independently.
Explore the idea
Build and classify the 16 combinations
AaBb x AaBb
Assumes independent assortment, complete dominance at both loci, and equal gamete/offspring viability. Row and column labels are gametes, not parent genotypes.
| Gamete | AB | Ab | aB | ab |
|---|---|---|---|---|
| AB | AABB | AABb | AaBB | AaBb |
| Ab | AABb | AAbb | AaBb | Aabb |
| aB | AaBB | AaBb | aaBB | aaBb |
| ab | AaBb | Aabb | aaBb | aabb |
Explanation
A dihybrid problem follows two loci. For AaBb, each gamete contains one allele from the A locus and one from the B locus. With independent assortment, the four types AB, Ab, aB and ab each have expected probability 1/4. AA and Bb are not complete gametes for this two-locus problem.
Crossing AaBb with AaBb gives 16 equally weighted gamete combinations. At each locus, complete dominance gives a 3/4 dominant-phenotype and 1/4 recessive-phenotype probability. Independence allows the two locus probabilities to be multiplied.
The joint phenotypes are A_B_ = 9/16, A_bb = 3/16, aaB_ = 3/16 and aabb = 1/16. The underscore means either allele can occupy that second position while the dominant allele is present. This gives the familiar 9:3:3:1 expectation only.
A true-breeding (pure-breeding) organism is homozygous for the alleles in question. When true-breeding organisms are crossed with each other, every offspring shows the same phenotype as the parents. Crossing a true-breeding purple, tall plant (PPTT) with a true-breeding white, dwarf plant (pptt) therefore gives an F1 that is all PpTt.
Mark schemes give a mark for each row of a genetic diagram, so set it out in full and label every row: a key to the symbols, parental phenotypes, parental genotypes, gametes, offspring genotypes, and offspring phenotypes with their ratio. Circle each gamete; schemes deduct a mark when gametes are not circled.
To turn a ratio into expected numbers, divide the total by the sum of the ratio, then multiply by each part. Of 320 F2 plants, 320 / 16 = 20, so the expected numbers are 180 : 60 : 60 : 20. Observed numbers rarely match exactly, because fertilisation is random and chance has a larger effect in small samples. A large, repeated difference suggests the model is wrong, for example because the genes are linked or some genotypes survive less well.
Step by step
- 1
Key
P = purple, p = white; T = tall, t = dwarf. State that the genes assort independently.
- 2
Parental phenotypes
Purple, tall x purple, tall (two F1 plants).
- 3
Parental genotypes
PpTt x PpTt
- 4
Gametes, each circled
(PT) (Pt) (pT) (pt) from each parent. Each gamete has one allele of each gene.
- 5
Offspring genotypes
Draw a 4 x 4 Punnett square: one parent's gametes across the top, the other's down the side. Fill in all 16 genotypes.
- 6
Offspring phenotypes and ratio
9 purple, tall : 3 purple, dwarf : 3 white, tall : 1 white, dwarf.
Worked example
One dominant, one recessive
For AaBb x AaBb with independent assortment and complete dominance, what is the chance of A_bb?
One way to explain it
(3/4 for A_) x (1/4 for bb) = 3/16. The A locus may be AA or Aa, while the B locus must be bb.
Why this answer works
- Specify the acceptable genotypes at each locus.
- Multiply only because the model gives independence.
Watch out for this
A student says: "A dihybrid gamete can be Aa or Bb." What is wrong with this?
Show the answer
A gamete must have one allele from each locus, such as AB or ab, not both alleles from just one locus.
Check your understanding
What is P(aabb) in the stated AaBb x AaBb independent cross?
Show the answer
1/16.
P(aa) x P(bb) = 1/4 x 1/4.
Exam-style questions
Check your understanding
AaBb x AaBb, with independent assortment and complete dominance, gives 320 offspring. About how many show both recessive phenotypes?
Show the answer
20
P(aabb) = 1/4 x 1/4 = 1/16, and 320 / 16 = 20.
Quick recall
When is 9:3:3:1 justified?