D. Genetic Processes · Genetic Processes

Monohybrid Crosses: Punnett Squares, Probability and Test Crosses

D2.3 D2.4 D3.3 D2.1 A1.12About 75 minutes

Learning goals

I can set up and complete a Punnett square for a monohybrid cross.
I can give genotypic and phenotypic ratios as ratios, fractions and percents.
I can use the product and sum rules of probability to solve genetics problems.
I can design and interpret a test cross to find an unknown genotype.

Words to know

Punnett square
A grid that shows all the possible combinations of gametes from two parents, each box equally likely.

genotypic ratio
The ratio of offspring genotypes, for example 1 YY : 2 Yy : 1 yy.

phenotypic ratio
The ratio of offspring phenotypes, for example 3 yellow : 1 green.

probability
How likely an event is: favourable outcomes ÷ total outcomes, from 0 (impossible) to 1 (certain).

product rule
The probability that two independent events both happen = the product of their probabilities.

sum rule
The probability that one OR another of two mutually exclusive events happens = the sum of their probabilities.

test cross
Crossing an individual with the dominant phenotype with a homozygous recessive individual to find out whether it is homozygous or heterozygous.

From peas to predictions

A guinea pig breeder has a black female and wants to know whether her pups could ever be white. A dairy farmer wants to avoid a recessive condition in a herd. A genetic counsellor helps a couple understand a 1-in-4 chance. All of them use the same tool you will learn today: the Punnett square, named after the British geneticist Reginald Punnett, along with a little probability.

Six steps for every genetics problem

Write a key. Choose a letter. Capital = dominant, lower-case = recessive. (B = black fur, b = white fur.)
Write the parents' genotypes. Use the information given; work backward from phenotypes if needed.
List each parent's gametes. By the law of segregation, each gamete gets one allele. Bb makes B and b; BB makes only B.
Draw the Punnett square. One parent's gametes across the top, the other's down the side. Fill each box with one allele from its column and one from its row.
Count the ratios. Genotypic ratio, then phenotypic ratio.
Answer the question in a sentence, with the units it asks for (ratio, fraction, percent or expected number).

A Punnett square for Yy × Yy. Gametes from one parent go across the top, gametes from the other down the side; each box is one equally likely offspring. Shading matches seed colour: three yellow boxes, one green.

Worked example

Example 1: A guinea pig cross

In guinea pigs, black fur (B) is dominant to white fur (b). Two heterozygous black guinea pigs are crossed. They have 8 pups.

(a) What are the expected genotypic and phenotypic ratios? (b) What percent of the pups are expected to be white? (c) About how many of the 8 pups would you expect to be white?

Key: B = black, b = white. Parents: Bb × Bb.
Gametes: each parent makes B and b.
Punnett square boxes: BB, Bb, Bb, bb.
(a) Genotypic ratio 1 BB : 2 Bb : 1 bb. Phenotypic ratio 3 black : 1 white.
(b) 1 box out of 4 is white: 1/4 = 25%.
(c) Expected number = 8 × 1/4 = 2 white pups. (The real litter might have 0, 1, 3 or more; each pup is a separate chance event.)

1 BB : 2 Bb : 1 bb; 3 black : 1 white; 25% white; about 2 of 8.

Probability: the maths behind the square

A Punnett square is really a probability calculation. A Bb parent gives b with probability 1/2. The two parents' gametes are independent, so:

Product rule ("and"): P(b from mother and b from father) = 1/2 × 1/2 = 1/4.
Sum rule ("or"): P(Bb) = P(B from mother and b from father) or P(b from mother and B from father) = 1/4 + 1/4 = 1/2.

Each offspring is a new, independent event. If the first three pups are black, the chance that the fourth is white is still 1/4. Chance has no memory.

The product rule also handles several offspring: P(two pups in a row are both white) = 1/4 × 1/4 = 1/16.

A probability tree for two seeds from a Yy × Yy cross. Multiply along a branch (product rule); add branches that give the same result (sum rule). The four results add up to 16/16 = 1.

Worked example

Example 2: Probability with several offspring

Two Yy pea plants are crossed. Find the probability that (a) two seeds picked at random are both green, (b) exactly one of the two is green.

P(green) = 1/4 and P(yellow) = 3/4 for each seed, independently.
(a) Both green: 1/4 × 1/4 = 1/16.
(b) Exactly one green can happen two ways: green then yellow (1/4 × 3/4 = 3/16) or yellow then green (3/4 × 1/4 = 3/16).
Add the two ways: 3/16 + 3/16 = 6/16 = 3/8.

(a) 1/16 (b) 3/8

The test cross

A black guinea pig could be BB or Bb: you cannot tell by looking. To find out, cross it with a **white (bb)** guinea pig. The bb parent can give only b, so the offspring's phenotypes reveal what the unknown parent gave:

If the unknown is BB, every pup gets B: all black.
If the unknown is Bb, half the pups get b from it: about 1 black : 1 white.

One white pup is enough to prove the parent is Bb. But all-black results only suggest BB: with a small litter, a Bb parent might have all black pups just by chance (for 4 pups the chance is (1/2)⁴ = 1/16). Larger samples give more confidence.

A test cross: cross the unknown (black) animal with a homozygous recessive (white, bb) tester. Shaded boxes = black offspring. Even one white offspring proves the unknown parent is Bb.

Worked example

Example 3: Working backward

In fruit flies, normal wings (V) are dominant to vestigial (small, crumpled) wings (v). Two normal-winged flies produce 62 normal-winged and 21 vestigial-winged offspring. What are the parents' genotypes?

Vestigial offspring are vv, so **each parent gave a v**.
Both parents have normal wings, so each also has a V: both are Vv.
Check: Vv × Vv predicts 3 : 1. 62 ÷ 21 ≈ 2.95, close to 3. ✓

Both parents are Vv.

Just for fun

A geneticist walks into a café and orders four coffees: one black and three with cream.

The barista says, "Let me guess: the black one is recessive?"

Hands-on activity

Coin-toss crosses

You need: 2 coins, a strip of masking tape, a pen and paper.

Put tape on both sides of each coin. Write B on one side and b on the other. Each coin is a Bb parent, and a toss is the gamete it gives.
Toss both coins together 40 times. Record each offspring genotype (BB, Bb or bb).
Make a table of your observed numbers and the expected numbers (10 BB, 20 Bb, 10 bb; 30 black : 10 white).
Combine your results with a family member's 40 tosses (80 total). Are the combined results closer to 1 : 2 : 1?
Test cross: make one coin b on both sides (the bb tester). Toss it with a Bb coin 20 times. Expected: 10 Bb : 10 bb.

What to look for: small samples wander from the expected ratio; bigger samples usually land closer. This is why geneticists count large numbers of offspring.

Your turn

In peas, round seeds (R) are dominant to wrinkled (r). Cross Rr × rr. Give the genotypic and phenotypic ratios.

Cross RR × rr. What fraction of the offspring are wrinkled?

What gametes can a Tt plant make? A TT plant?

Two Bb guinea pigs have 12 pups. How many are expected to be black?

Why do we cross with a homozygous recessive individual in a test cross?

Two Bb guinea pigs have already had 3 black pups. What is the probability the next pup is white?

More practice: warm-up

How many boxes are in a monohybrid Punnett square?

What is 1/2 × 1/2?

Write the ratio 1 : 2 : 1 as fractions of the total.

Which parent genotype is used as the tester in a test cross?

What phenotype ratio does Aa × aa give?

More practice: on your own

In a test cross, a black guinea pig has 7 black pups and 6 white pups. What is its genotype? Explain.

A tall pea plant (T) crossed with a dwarf plant gives 120 offspring, all tall. What is the most likely genotype of the tall parent, and why can't you be completely sure?

Two plants are crossed. Offspring: 48 purple, 52 white. White (p) is recessive. Give the parents' genotypes.

Find the probability that 3 pups from Bb × Bb are all black.

Two Yy plants are crossed. What is the probability that an offspring is homozygous (either YY or yy)?

Challenge

Two Bb guinea pigs have 4 pups. What is the probability that exactly 1 is white?

A black guinea pig whose parents were both Bb is test-crossed. Before you see any pups, what is the probability that it is Bb? (Hint: see the Mendel lesson challenge.)

Continuing: the guinea pig in the last question has 3 pups in a test cross, and all are black. Explain qualitatively whether you are now more or less confident that it is BB.

Check your understanding

Quiz

In a Punnett square for Aa × Aa, what fraction of offspring is expected to be Aa?

A test cross uses which kind of individual as the tester?

A cross produces 50% dominant and 50% recessive phenotypes. The parents are most likely:

The probability that two independent events both happen is found by:

Two Bb guinea pigs have had two white pups. The chance their next pup is white is:

From Rr × Rr, 400 seeds are produced. How many are expected to be wrinkled (rr)?

For parents and teachers

Big idea: This lesson builds the core D2.3 skill (Punnett squares for monohybrid crosses) and the D2.4 skill of using probability and crosses to work out parent genotypes. Students should use the six-step method every time.

How to help: Do the coin-toss activity together; it makes "expected vs observed" real. When checking homework, look for a written key, parent genotypes, gametes, a square, ratios and a sentence answer.

Common difficulties:

Writing gametes with two letters for one gene (a gamete has one allele per gene).
Mixing up genotypic (1 : 2 : 1) and phenotypic (3 : 1) ratios.
Thinking past offspring change the odds for the next one.
Concluding BB from a test cross with only a few all-dominant offspring.