Why does a dihybrid cross give a 9:3:3:1 ratio?

Each parent makes four equally likely gametes, so the square has 16 equally likely boxes. Nine of them contain at least one dominant allele of both genes, three show the first dominant trait with the second recessive trait, three show the reverse, and exactly one is recessive for both. You can get the same numbers by multiplying: 3/4 × 3/4 = 9/16, 3/4 × 1/4 = 3/16, and 1/4 × 1/4 = 1/16.

Do I have to draw all 16 boxes?

Not if the question asks for one specific outcome. The multiplication rule handles that in a line of arithmetic. Draw the full square when you need the complete ratio, when you are asked to show genotypes, or when you want to double-check a probability you calculated. Many teachers want to see the square at least once so it is clear you understand where the fractions come from.

How do I know how many gamete types a parent makes?

Count the gene pairs that are heterozygous, call that number n, and the parent makes 2^n gamete types. RrYy has two heterozygous pairs, so 2^2 = 4 gametes. RrYY has one, so 2 gametes. rryy has none, so 2^0 = 1 gamete type.

What if my actual offspring counts do not match 9:3:3:1?

Small samples wander from predicted ratios by chance, just as 20 coin flips rarely split exactly 10 and 10, so ratios sharpen as sample size grows. A large, consistent deviation points to biology instead: the genes may be linked on the same chromosome, one gene may not show simple complete dominance, or a genotype may reduce survival.

BIO-5.3

Dihybrid Crosses & Probability

Learn to list the four gametes of a dihybrid parent, fill a 16-box Punnett square, and use the multiplication rule to find any two-trait probability fast.

What you'll do in this lesson

A voice-first session with the Crimsora tutor on Dihybrid Crosses & Probability, then targeted practice and FRQs — with the tutor adapting to where you get stuck.

Start Dihybrid Crosses & Probability free

What this lesson covers

You already know how to set up a 2-by-2 Punnett square for one trait. Now real organisms inherit thousands of traits at once, so the natural next question is: what happens when we track two genes in the same cross? Because the two gene pairs sort into gametes independently, a heterozygote for both genes can make four different kinds of sperm or egg — not two. Four gamete types on each side means a 4-by-4 grid with 16 boxes, and out of that grid falls one of the most famous numbers in genetics: the

9:3:3:1

phenotype ratio. In this lesson you will learn to generate gametes reliably, fill and read the 16-box square, and then skip the square entirely by multiplying two single-trait probabilities together. Both routes must give the same answer, which makes each a check on the other.

Listing the Four Gamete Types of a Dihybrid Parent

A dihybrid is an individual heterozygous for two genes, such as

RrYy

in peas (

R

= round seed, dominant to

r

= wrinkled;

Y

= yellow seed, dominant to

y

= green). Every gamete gets exactly one allele from each gene pair — one letter from the

R/r

pair and one from the

Y/y

pair. Because the pairs assort independently during meiosis I, each combination is equally likely.

The reliable way to generate them is to pair the first gene's alleles with the second gene's alleles in all combinations, the same pattern as FOIL in algebra:

RY

Ry

rY

ry

. Each of those four gamete types occurs with probability

\frac{1}{4}

.

The number of gamete types depends on how many gene pairs are heterozygous, not on how many genes you are tracking:

Parent genotype	Heterozygous pairs	Gamete types
$RrYy$	2	$RY$ , $Ry$ , $rY$ , $ry$
$RrYY$	1	$RY$ , $rY$
$Rryy$	1	$Ry$ , $ry$
$rryy$	0	$ry$ only

The most common mistake here is writing gametes like

RR

Yy

. A gamete cannot carry two alleles of the same gene — that would be a diploid cell, not a haploid one. Every gamete you write for a two-gene problem must contain exactly one letter from each pair, and by convention you write the genes in the same order every time so the square is easy to read.

Filling and Reading the 16-Box Punnett Square

Put one parent's four gametes across the top and the other parent's four down the side, then combine the row label and column label in each box. Always list the dominant allele first within a pair so identical genotypes look identical (

RrYy

, not

rRyY

). Here is the classic dihybrid cross

RrYy \times RrYy

	$RY$	$Ry$	$rY$	$ry$
$RY$	$RRYY$	$RRYy$	$RrYY$	$RrYy$
$Ry$	$RRYy$	$RRyy$	$RrYy$	$Rryy$
$rY$	$RrYY$	$RrYy$	$rrYY$	$rrYy$
$ry$	$RrYy$	$Rryy$	$rrYy$	$rryy$

Now sort the 16 boxes by phenotype, not genotype. A seed is round if it has at least one

R

, and yellow if it has at least one

Y

. Counting gives 9 round yellow, 3 round green, 3 wrinkled yellow, and 1 wrinkled green — the

9:3:3:1

ratio. Notice that there are nine different genotypes but only four phenotypes; students who report a

9:3:3:1

genotype ratio have mixed up the two levels.

Two checks catch most errors. First, the boxes must total 16; if you have 12 or 20, you duplicated or skipped a gamete. Second, the double-recessive

rryy

should appear exactly once, in the corner where the two

ry

gametes meet. The

9:3:3:1

ratio only appears when both parents are dihybrid and both genes show simple complete dominance.

The Multiplication Rule: A Faster Route to Any Single Probability

Filling 16 boxes is slow, and you rarely need all of them. Because the two genes assort independently, the two traits are independent events, so the probability of a combined outcome is the product of the separate probabilities:

P(A \text{ and } B) = P(A) \times P(B)

Break the dihybrid cross into two monohybrid crosses. From

Rr \times Rr

P(\text{round}) = \frac{3}{4}

and

P(\text{wrinkled}) = \frac{1}{4}

. From

Yy \times Yy

P(\text{yellow}) = \frac{3}{4}

and

P(\text{green}) = \frac{1}{4}

. Multiply the pair you want:

Phenotype	Product	Probability
round, yellow	$\frac{3}{4} \times \frac{3}{4}$	$\frac{9}{16}$
round, green	$\frac{3}{4} \times \frac{1}{4}$	$\frac{3}{16}$
wrinkled, yellow	$\frac{1}{4} \times \frac{3}{4}$	$\frac{3}{16}$
wrinkled, green	$\frac{1}{4} \times \frac{1}{4}$	$\frac{1}{16}$

The four probabilities add to 1, and they reproduce the

9:3:3:1

ratio exactly — strong evidence the square and the math describe the same biology.

The same trick works for genotypes:

P(RrYy) = P(Rr) \times P(Yy) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}

, which is why

RrYy

shows up in four of the sixteen boxes. Two cautions: multiply only when the events are independent, and add (not multiply) when you want "either/or" outcomes, such as

P(\text{round yellow or wrinkled green}) = \frac{9}{16} + \frac{1}{16} = \frac{10}{16}

Crosses That Are Not Dihybrid by Dihybrid

Most homework problems are not the tidy

RrYy \times RrYy

case, and

9:3:3:1

will be a wrong answer for them. Handle any two-gene cross by splitting it into two single-gene crosses, solving each, and then multiplying.

Consider a dihybrid test cross,

RrYy \times rryy

. The first parent makes four gametes; the second makes only

ry

. Each offspring therefore just displays whatever the dihybrid parent contributed, giving

RrYy

Rryy

rrYy

rryy

in a

1:1:1:1

phenotype ratio. You can see this from the multiplication rule too:

Rr \times rr

gives

\frac{1}{2}

round and

\frac{1}{2}

wrinkled,

Yy \times yy

gives

\frac{1}{2}

yellow and

\frac{1}{2}

green, and every product is

\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}

.

For

RrYY \times Rryy

, the

Y

gene contributes no variation at all: every offspring is

Yy

, so all are yellow, and the phenotype ratio is simply

3

yellow round to

1

yellow wrinkled. When one gene has only one possible outcome, the grid shrinks — a 2-by-2 square is enough.

One more limitation worth knowing. Independent assortment holds because the two genes sit on different chromosome pairs. If two genes are close together on the same chromosome (linked), gametes carrying the parental allele combinations are more common than the recombinant ones, and observed offspring counts deviate from the predicted ratios. That deviation is exactly how geneticists first mapped genes to chromosomes, so a data set that stubbornly refuses to fit

9:3:3:1

is informative, not broken.

Key terms

Dihybrid cross.: A cross that tracks two different genes at the same time; strictly, a cross between two individuals heterozygous for both genes, such as $RrYy \times RrYy$ .
Gamete.: A haploid sex cell carrying exactly one allele from each gene pair. For two genes, a dihybrid produces four equally likely gamete types.
Law of independent assortment.: Alleles of genes on different chromosome pairs separate into gametes independently, so inheriting one gene's allele does not affect which allele of the other gene is received.
Multiplication (product) rule.: For independent events, the probability that both occur is the product of their separate probabilities: $P(A \text{ and } B) = P(A) \times P(B)$ .
Addition rule.: For mutually exclusive outcomes, the probability that either occurs is the sum of their probabilities, used for "either/or" questions.
9:3:3:1 ratio.: The expected phenotype ratio among offspring of a dihybrid by dihybrid cross when both genes show complete dominance and assort independently.
Test cross.: A cross with a fully homozygous recessive individual ( $rryy$ ), used to reveal the genotype of the other parent; a dihybrid test cross yields a $1:1:1:1$ ratio.
Linked genes.: Genes located close together on the same chromosome, which do not assort independently and therefore produce offspring ratios that depart from dihybrid predictions.

Worked example

In guinea pigs, black coat (

B

) is dominant to white (

b

), and short hair (

S

) is dominant to long hair (

s

). A black, short-haired guinea pig that is heterozygous for both genes is crossed with a white, short-haired guinea pig that is heterozygous for hair length. Predict the phenotype ratio of the offspring, and find the probability that a single offspring is white with long hair.

Step 1 — write the genotypes. The first parent is heterozygous for both genes:

BbSs

. The second is white, so it must be

bb

, and it is short-haired but heterozygous, so

Ss

. The cross is

BbSs \times bbSs

.

Step 2 — list gametes. Parent 1 is heterozygous for both pairs, so it makes four types:

BS

Bs

bS

bs

. Parent 2 is heterozygous for only one pair, so it makes two types:

bS

and

bs

.

Step 3 — split into two single-gene crosses. Coat color:

Bb \times bb

gives

\frac{1}{2}

Bb

(black) and

\frac{1}{2}

bb

(white). Hair length:

Ss \times Ss

gives

\frac{3}{4}

short and

\frac{1}{4}

long.

Step 4 — multiply for each phenotype combination. Black short:

\frac{1}{2} \times \frac{3}{4} = \frac{3}{8}

. Black long:

\frac{1}{2} \times \frac{1}{4} = \frac{1}{8}

. White short:

\frac{1}{2} \times \frac{3}{4} = \frac{3}{8}

. White long:

\frac{1}{2} \times \frac{1}{4} = \frac{1}{8}

.

Step 5 — state the ratio. Converting to eighths, the ratio is

3:1:3:1

(black short : black long : white short : white long). The probabilities sum to

\frac{8}{8} = 1

, a good check.

Step 6 — answer the specific question.

P(\text{white, long}) = \frac{1}{8}

, or 12.5 percent. Note that this cross is not dihybrid by dihybrid, so

9:3:3:1

would have been wrong here.

Practice questions

In tomatoes, tall (

T

) is dominant to dwarf (

t

) and smooth skin (

P

) is dominant to fuzzy skin (

p

). What is the probability that a cross of

Ttpp \times TtPp

produces a dwarf, fuzzy-skinned plant?

$\frac{1}{16}$
$\frac{1}{8}$
$\frac{3}{16}$
$\frac{1}{4}$

Answer: $\frac{1}{8}$

Split the cross by gene. For height,

Tt \times Tt

gives

\frac{1}{4}

tt

(dwarf). For skin,

pp \times Pp

gives

\frac{1}{2}

pp

(fuzzy). Multiply the independent probabilities:

\frac{1}{4} \times \frac{1}{2} = \frac{1}{8}

. The answer

\frac{1}{16}

is what you get by wrongly assuming both parents are dihybrid, and

\frac{3}{16}

takes two slips at once: treating the skin cross as

Pp \times Pp

(which gives

\frac{1}{4}

fuzzy) and then using

\frac{3}{4}

, the tall class, instead of

\frac{1}{4}

for dwarf.

A dihybrid pea plant (

RrYy

) is crossed with a wrinkled, green plant. List the gamete types each parent can make, predict the offspring phenotype ratio, and explain how this cross could be used to identify an unknown round yellow plant's genotype.

Answer: The dihybrid makes $RY$ , $Ry$ , $rY$ , and $ry$ ; the wrinkled green plant ( $rryy$ ) makes only $ry$ . Offspring are $RrYy$ , $Rryy$ , $rrYy$ , and $rryy$ in a $1:1:1:1$ ratio (round yellow : round green : wrinkled yellow : wrinkled green). Because the recessive parent adds only recessive alleles, each offspring's phenotype directly reveals which gamete the other parent supplied, so this test cross exposes hidden recessive alleles in an unknown round yellow plant.

Only one gamete type comes from

rryy

, so the offspring's phenotype is decided entirely by the dihybrid parent. If the unknown round yellow plant were

RRYY

, every offspring would be round and yellow. Any wrinkled or green offspring proves the unknown parent carried an

r

or a

y

. The proportions of the four offspring classes then tell you which genes were heterozygous — a

1:1:1:1

result means heterozygous for both.

For the cross

AaBb \times AaBb

, what is the probability that an offspring shows at least one dominant trait (that is, is not recessive for both traits)?

Answer: $\frac{15}{16}$

First find the probability of the outcome you want to exclude. Being recessive for both traits means

aabb

P(aa) = \frac{1}{4}

and

P(bb) = \frac{1}{4}

, so

P(aabb) = \frac{1}{4} \times \frac{1}{4} = \frac{1}{16}

— the single corner box of the 16-box square. Since all probabilities total 1, the complement is

1 - \frac{1}{16} = \frac{15}{16}

. Using the complement is much faster than adding

\frac{9}{16} + \frac{3}{16} + \frac{3}{16}

, though both give the same result.

FAQ

Why does a dihybrid cross give a 9:3:3:1 ratio?: Each parent makes four equally likely gametes, so the square has 16 equally likely boxes. Nine of them contain at least one dominant allele of both genes, three show the first dominant trait with the second recessive trait, three show the reverse, and exactly one is recessive for both. You can get the same numbers by multiplying: $\frac{3}{4} \times \frac{3}{4} = \frac{9}{16}$ , $\frac{3}{4} \times \frac{1}{4} = \frac{3}{16}$ , and $\frac{1}{4} \times \frac{1}{4} = \frac{1}{16}$ .
Do I have to draw all 16 boxes?: Not if the question asks for one specific outcome. The multiplication rule handles that in a line of arithmetic. Draw the full square when you need the complete ratio, when you are asked to show genotypes, or when you want to double-check a probability you calculated. Many teachers want to see the square at least once so it is clear you understand where the fractions come from.
How do I know how many gamete types a parent makes?: Count the gene pairs that are heterozygous, call that number $n$ , and the parent makes $2^n$ gamete types. $RrYy$ has two heterozygous pairs, so $2^2 = 4$ gametes. $RrYY$ has one, so 2 gametes. $rryy$ has none, so $2^0 = 1$ gamete type.
What if my actual offspring counts do not match 9:3:3:1?: Small samples wander from predicted ratios by chance, just as 20 coin flips rarely split exactly 10 and 10, so ratios sharpen as sample size grows. A large, consistent deviation points to biology instead: the genes may be linked on the same chromosome, one gene may not show simple complete dominance, or a genotype may reduce survival.

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