Dihybrid Problem Help
Ms. Gaynor/H. Genetics
What is a dihybrid problem?
 A cross (or mating) between 2 parents, who have a geneotype for 2 different types of genes
 Example: HhBb x HHBb
o The first gene is the “H” gene and it codes for hair color.
 Brown Hair is dominant = HH or Hh , Blonde Hair is recessive = hh
o The second gene is the “B” gene and it codes for eye color.
 Brown eyes are dominant = BB or Bb, Blue eyes are recessive = bb
How do I solve a dihybrid problem?
 You need to solve it in steps:
1. Write out a key
2. Write out the DIHYBRID genotypes of each parent (Ex: HhBb x HHBb)
3. Separate each gene in the genotype into a MONOHYBRID problem
a. Ex: Hh x HH and Bb x Bb
i. NOTE: You are using the genes (allele letters) listed in the parental genotypes (see
step #2)
4. Write out the genotype and phenotype ratios using FRACTIONS ( ¼ , ½ , ¾ , or 4/4) for each
monohybrid cross in #3.
a. Ex: Hh x HH 
H
h
H
HH
Hh
H
HH
Hh
A
a
A
AA
Aa
a
Aa
aa
b. Ex: Aa x Aa 
Genotype ratio = ½ HH and ½ Hh
Phenotype ratio = 4/4 ( or 1) Brown hair
Genotype ratio = ¼ AA ½ Aa ¼ aa
Phenotype ratio = ¾ Brown eyes ¼ Blue eyes
5. Now…determine the DIHYBRID (2 trait) combination that are possible in the offspring/
a. Ex: The offspring could be
i. Genotype ratio = ½ HH
¼ AA
½ Hh
½ Aa
¼ aa
Possible Genotypes: HHAA, HHAa, HHaa, HhAA, HhAa, and Hhaa
ii. Phenotype ratio = 4/4 Brown hair
¾ Brown eyes
¼ Blue eyes
Possible Phenotypes: Brown Hair & Brown eyes OR Brown Hair & Blue eyes
6. Finally…multiply fractions to determine the possibilities (ratios) of getting each genotype or
phenotype combination in step #5. Use the fractions next to each SINGLE genotype/phenotype
from step #4.
a. Example-GENOTYPE RATIOS (putting it all together)
¼ x ¼ = 1/16 chance of getting HHAA genotype
½ x ¼ = 1/8 chance of getting HhAA genotype
1/
¼ x ½ = 8 chance of getting HHAa genotype
½ x ½ = ¼ chance of getting HhAa genotype
¼ x ¼ = 1/16 chance of getting HHaa genotype
½ x ¼ = 1/16 chance of getting Hhaa genotype
b. Example-PHENOTYPE RATIOS (putting it all together)
4/4 x ¾ = 12/16 = ¾ (75%) chance of getting brown hair and brown eyes phenotype (2 traits together)
4/4 x ¼ = 4/16 = ¼ (25%) chance of getting brown hair and blue eyes phenotype (2 traits together)
How do I find the possible gametes in a dihybrid problem?

You need use F.O.I.L
o

First, Outer, Inner, Last
Ex: female = HhBb x
o
male = HHBb
Female with this genotype can make the following eggs: HB, Hb, hB, hb
(mom can make 4 different types of eggs..genetically speaking)
o
o

First allele of each gene: H and B

Outer allele of each gene: H and b

Inner allele of each gene: h and B

Last allele of each gene: h and b
Males with this genotype can make the following sperm: HB, Hb, hb

(dad can only make 3 different types of sperm..genetically speaking because HB is repeated )

First allele of each gene: H and B

Outer allele of each gene: H and b

Inner allele of each gene: H and B

Last allele of each gene: h and b
How do I solve a dihybrid problem using SHORT CUTS ?

You can only use the SHORT CUTS on HETEROZYGOUS crosses that are monohybrid or dihybrid!
o
If 2 heterozygotes are crossed in a monohybrid (single trait) cross (Ex: Aa x Aa), then you may memorize the
following and use it EACH time you encounter 2 heterozygous individuals that are being crossed (NOTE: THIS
ONLY WORKS WITH GENES THAT FOLLOW COMPLETE DOMINANCE!)

Aa x Aa (assume “A” is hitchhiker thumb and “a” is straight thumb)

Genotype ratio will always be a 1:2:1 ratio as follows:
o
¼ homozygous dominant


2/4
Heterozygous ¼ homozygous recessive
Aa ¼ aa, which means 1 AA : 2 Aa : 1 aa
Phenotype ratio will always be a 3:1 ratio as follows:
o
¾ dominant version of trait ¼ recessive version of trait

o
Ex: ¼ AA
2/4
Ex: ¾ Hitchhiker thumb ¼ Straight thumb
If 2 heterozygotes are crossed in a dihybrid (2 traits at the SAME time) cross (Ex: AaBb x AaBb), then you
may memorize the following and use it EACH time you encounter 2 heterozygous individuals that are being
crossed (NOTE: THIS ONLY WORKS WITH GENES THAT FOLLOW COMPLETE DOMINANCE!)

AaBb x AaBb (assume “A” is hitchhiker thumb/“a” is straight thumb; “B” is brown eyes /“b” is blue eyes)
Phenotype ratio will always be a 9:3:3:1 ratio as follows:

9/16 dominant version of gene #1 (A gene) and dominant version of gene #2 (B gene)

3/16 dominant version of gene #1 (A gene) and recessive version of gene #2 (B gene)

3/16 recessive version of gene #1 (A gene) and dominant version of gene #2 (B gene)

1/16 recessive version of gene #1 (A gene) and recessive version of gene #2 (B gene)

Ex:
9/16 hitchhiker thumb and brown eyes
3/16 straight thumb and brown eyes
3/16 hitchhiker thumb and blue eyes
1/16 straight thumb and blue eyes
How do I solve a multi-hybrid problem?

You need to follow the steps for solving a dihybrid (see #1-5 on page 2) then multiple fractions.
o
Ex: AaBbCcDdEeFf

x
AABBCCddEeFf
What is the possibility of getting the following genotype? AABBCCDDEEFF

You need to determine the possibility in FRACTIONS of getting: AA separately, then BB, then
CC, etc., from each of the parental genotypes.

You need to do a Punnett Square in your head (or on paper ) for EACH gene (the A gene,
the B gene, the C gene, the D gene, the E gene and the F gene)

The question is actually asking “what is the possibility of getting AA and BB and CC and DD
and EE and FF all at one time?”

Next, you need to multiply all the fractions together because “AND” means multiply.

Example:
o
AA
BB
CC
DD
EE
FF = ??
½ x ½ x ½ x ½ x ¼ x ¼ = 1/ 256
o
There is a 1/256 (0.40%) chance that you will get AABBCCDDEEFF genotype from a
cross between 2 parents with the following genotypes: AaBbCcDdEeFf x
AABBCCddEeFf