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D. Genetic Processes · Genetic Processes

Dihybrid Crosses and the Law of Independent Assortment

D2.3 D2.4 D3.2 A1.8About 75 minutes
Learning goals
  • I can find all the gametes made by a parent for two genes.
  • I can complete a 16-box Punnett square and get the 9 : 3 : 3 : 1 ratio.
  • I can state the law of independent assortment and link it to meiosis.
  • I can use the product rule as a shortcut for dihybrid problems.

Words to know

dihybrid cross
A cross that follows two genes (two traits) at the same time.
law of independent assortment
Alleles of different genes are sorted into gametes independently of each other (true for genes on different chromosome pairs, or far apart on one).
9 : 3 : 3 : 1
The F2 phenotype ratio from crossing two individuals heterozygous for two independent genes with complete dominance.
linked genes
Genes close together on the same chromosome that tend to be inherited together and so do not assort independently.
product rule shortcut
For independent genes, P(combined phenotype) = P(trait 1) × P(trait 2).

Two traits at once

When Mendel was sure about single traits, he asked a harder question: if a plant inherits round seeds, does that affect whether it also inherits yellow seeds? Or does each trait travel on its own?

He crossed true-breeding round, yellow plants (RRYY) with true-breeding wrinkled, green plants (rryy). Every F1 plant was round and yellow (RrYy). Then he let the F1 self-pollinate and counted 556 F2 seeds:

  • 315 round yellow
  • 108 round green
  • 101 wrinkled yellow
  • 32 wrinkled green

Two of these combinations (round green and wrinkled yellow) were new: neither original parent had them. The traits had separated and recombined. The ratio is close to 9 : 3 : 3 : 1.

The law of independent assortment

Mendel concluded that the alleles for different traits are sorted into gametes independently. Getting R rather than r does not change the chance of getting Y rather than y.

Today we know why: the seed-shape and seed-colour genes are on different chromosome pairs. In metaphase I each pair lines up independently (you saw this in the lesson on meiosis and variation), so all four combinations of alleles end up in gametes equally often.

Important exception: genes that sit close together on the same chromosome are linked and tend to be inherited together. Crossing over can separate them sometimes. In this course, assume genes assort independently unless a question says otherwise.

RrYy1/21/2Rr1/21/2Ygamete RY (1/2 × 1/2 = 1/4)ygamete Ry (1/2 × 1/2 = 1/4)1/21/2Ygamete rY (1/2 × 1/2 = 1/4)ygamete ry (1/2 × 1/2 = 1/4)One allele of each gene goes into every gamete, in every combination.
Finding the gametes of RrYy. The R/r choice and the Y/y choice are independent (they are on different chromosome pairs), so the four gametes RY, Ry, rY, ry are equally likely.

Finding the gametes

Each gamete gets one allele of each gene. A quick method is like FOIL in algebra: for RrYy, pair First (R with Y), Outer (R with y), Inner (r with Y) and Last (r with y): RY, Ry, rY, ry.

  • RrYy makes 4 kinds of gametes.
  • RRYy makes 2 kinds: RY and Ry.
  • rryy makes only 1 kind: ry.

Never put two alleles of the same gene in one gamete (Rr is not a gamete).

gametes from RrYy parentRYRyrYryRYRyrYryRRYYRRYyRrYYRrYyRRYyRRyyRrYyRryyRrYYRrYyrrYYrrYyRrYyRryyrrYyrryyPhenotypesround, yellow (R_Y_)9round, green (R_yy)3wrinkled, yellow (rrY_)3wrinkled, green (rryy)19 : 3 : 3 : 1( _ = either allele )
The F2 of a dihybrid cross, RrYy × RrYy. Each parent makes four kinds of gametes, so the square has 16 equally likely boxes. Boxes are shaded by phenotype; count them to get 9 : 3 : 3 : 1.
Worked example

Example 1: The 16-box square

Show that RrYy × RrYy gives 9 : 3 : 3 : 1, and find what fraction of the offspring are RrYy.

  1. Gametes for each parent: RY, Ry, rY, ry. Write them across the top and down the side.
  2. Fill each box with the two gametes combined. Write the R gene first, dominant allele first: for example RY + ry = RrYy.
  3. Count phenotypes: at least one R and one Y (round yellow): 9 boxes. R_yy (round green): 3. rrY_ (wrinkled yellow): 3. rryy: 1.
  4. Count RrYy boxes: they lie on the diagonal from bottom-left to top-right: 4 of 16 = 1/4.

9 : 3 : 3 : 1; P(RrYy) = 4/16 = 1/4.

The product-rule shortcut

Because the genes assort independently, you can solve each gene separately and multiply.

For RrYy × RrYy:

  • Seed shape (Rr × Rr): 3/4 round, 1/4 wrinkled.
  • Seed colour (Yy × Yy): 3/4 yellow, 1/4 green.

So P(round and green) = 3/4 × 1/4 = 3/16. P(wrinkled and green) = 1/4 × 1/4 = 1/16. That is the whole 9 : 3 : 3 : 1 without drawing 16 boxes. The shortcut is especially handy for crosses like RrYy × rrYy or for three genes (64 boxes!).

Worked example

Example 2: A cross that is not 9 : 3 : 3 : 1

Cross RrYy × rryy (a dihybrid test cross). Give the phenotypic ratio.

  1. Shape: Rr × rr gives 1/2 round, 1/2 wrinkled.
  2. Colour: Yy × yy gives 1/2 yellow, 1/2 green.
  3. Multiply: round yellow 1/4, round green 1/4, wrinkled yellow 1/4, wrinkled green 1/4.
  4. Check with gametes: the RrYy parent makes RY, Ry, rY, ry, the tester makes only ry, so the four offspring types appear equally.

1 : 1 : 1 : 1. (If a dihybrid test cross gives a ratio far from 1 : 1 : 1 : 1, with too many parental types, that is a clue the genes are linked.)

050100150200250300350315313roundyellow108104roundgreen101104wrinkledyellow3235wrinkledgreenobservedexpectedseeds
Mendel's dihybrid F2 (556 seeds): observed counts beside the counts expected from 9 : 3 : 3 : 1 (rounded to whole seeds). They match closely.
Just for fun

Why do geneticists love the number 16?

Because it's the only number that makes a dihybrid cross square up perfectly. (Four gametes times four gametes, every time.)

Hands-on activity

Two-coin dihybrid model

You need: 4 coins (two of one kind, two of another, for example 2 loonies and 2 quarters), masking tape, a pen and paper.

  1. Tape and label: the two loonies are the shape gene (R on one side, r on the other). The two quarters are the colour gene (Y / y). One loonie and one quarter = parent 1; the other pair = parent 2.
  2. Each parent "makes a gamete" by tossing its loonie and its quarter together. Combine both parents' gametes into an offspring genotype.
  3. Do 32 offspring. Record the phenotype of each (round/wrinkled, yellow/green).
  4. Expected: 18 : 6 : 6 : 2 (that is 9 : 3 : 3 : 1 × 2). Compare.
  5. Linkage what-if: tape a loonie and a quarter together so they always land the same way up (R with Y, r with y). Toss again. Which phenotypes disappear? (Round green and wrinkled yellow never appear: linked genes without crossing over.)

What to look for: a ratio roughly like 9 : 3 : 3 : 1 with independent coins, and only two parental types when the coins are taped together.

Your turn

1

List the gametes made by AaBb. By AABb. By aabb.

2

In RrYy × RrYy, what fraction of offspring are wrinkled and yellow?

3

State the law of independent assortment.

4

In which phase of meiosis does independent assortment occur?

5

Cross RrYy × rryy. What fraction are round and green?

6

Why don't linked genes follow the law of independent assortment?

More practice: warm-up

1

How many boxes are in a dihybrid Punnett square for AaBb × AaBb?

2

How many kinds of gametes does AaBb make?

3

What is the F2 phenotype ratio for a dihybrid cross of two heterozygotes?

4

Is Aa a possible gamete?

5

What is 3/4 × 3/4?

More practice: on your own

1

In guinea pigs, black (B) is dominant to white (b) and rough coat (R) is dominant to smooth (r). Cross BbRr × BbRr. What fraction are white and rough?

2

Same genes: cross BbRr × bbrr. Give the phenotypes and ratio.

3

Cross BBRr × bbRr. What fraction are black and smooth?

4

Of 320 offspring from AaBb × AaBb, how many are expected to show both recessive traits?

5

A cross gives a phenotypic ratio of 3 : 3 : 1 : 1. One parent is AaBb. Suggest the other parent's genotype.

Challenge

1

For three independent genes, AaBbCc × AaBbCc: how many kinds of gametes does each parent make, how many boxes would the Punnett square have, and what fraction of offspring show all three dominant traits?

2

Same cross: what fraction of offspring are aabbcc? What fraction are homozygous for all three genes (any homozygous genotype)?

3

A dihybrid test cross (AaBb × aabb) gives 420 AaBb, 410 aabb, 85 Aabb and 85 aaBb. Explain what this tells you about genes A and B.

Check your understanding

Quiz

1.

What gametes can AaBb produce?

2.

The F2 phenotypic ratio from AaBb × AaBb (complete dominance, independent genes) is:

3.

In RrYy × RrYy, the probability of a wrinkled green (rryy) offspring is:

4.

The law of independent assortment is explained by:

5.

AaBb × aabb gives a phenotypic ratio of:

6.

Which pair of genes is LEAST likely to assort independently?

For parents and teachers

Big idea: This lesson completes the dihybrid part of D2.3 and D2.4, and the law of independent assortment in D3.2.

How to help: Let your teen choose either method (16-box square or product rule) but ask them to check one answer with the other. The 16-box square gets messy: suggest writing the gene letters in the same order in every box.

Common difficulties:

  • Writing gametes as Aa or BB (a gamete has one allele per gene).
  • Applying 9 : 3 : 3 : 1 to every dihybrid cross; it only applies to AaBb × AaBb.
  • Forgetting that linked genes do not assort independently.