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Solving Crosses
Steps for Solving a Genetics
Problem:
Trait – dominant = A (AA or Aa)
Trait – recessive = a (aa)
 ___________ x ___________
 Punnett Square

____ ____
____
____

Answer questions based on results from
Punnett Square
Monohybrid Crosses
Cross that involves one pair of
contrasting traits
 Solve using Punnett Square
 Sample problems:

– Rr x rr
– RR x rr
– Rr x Rr
– Rr x RR
Example 1: Monohybrid

Short hair (L) is dominant to long hair (l)
in mice. What is the genotype and
phenotype ratio of a heterozygous
short-haired mouse crossed with a longhaired mouse?
Example 1: Monohybrid





Short hair = dominant = L (LL or Ll)
long hair = recssive = l
Ll x ll
(heterozygote parent = Ll)
L
l
Punnett Square:
l
Ll ll
l
Ll ll
Genotype ratio: ½ Ll: ½ ll
Phenotype ratio: ½ short hair: ½ long hair
Dihybrid Crosses

Involves two pairs of contrasting traits
– Pea shape and pea color
– Coat length and coat color in rodents
– Plant height and flower color
Example 2: Dihybrid

In guinea pigs, the allele for short hair
(S) is dominant to long hair (s), and the
allele for black hair (B) is dominant over
the allele for brown hair (b). What is the
probable offspring phenotype ratio for a
cross involving two parents that are
heterozygotes for both traits?
Example 2: Dihybrid
Short hair = dominant = SS or Ss
Long Hair = recessive = ss
Black coat = dominant = BB or Bb
Brown coat = recessive = bb
 SsBb x SsBb (gametes done by the
FOIL method)

– SB, Sb, sB, sb and SB, Sb, sB, sb
Example 2: Punnett Square
SB
Sb
sB
sb
SB
SSBB SSBb SsBB SsBb
Sb
SSBb SSbb SsBb
Ssbb
sB
SsBB SsBb
ssBB
ssBb
sb
SsBb
ssBb
ssbb
Ssbb
Example 2: Answer the Question

What is the probable offspring
phenotype ratio for a cross involving two
parents that are heterozygotes for both
traits?
– 9/16 Black, short coats
– 3/16 Black, long coats
– 3/16 Brown, short coats
– 1/16 Brown, long coats
Incomplete Dominance

blending of traits in heterozygote; trait is
controlled by both alleles
– Pink flowers
• RR = red
• Rr = pink
• rr = white
Ex: Japanese 4 o’clock (Mirabilis)
RR X WW = RW (pink)
http://www.hobart.k12.in.us/jkousen/Biology/incc
odom.htm
Codominance

can see both alleles at the same time.
ex: Roan coats in horses
• Some white hairs, some red hairs
ex: erminette
• Black & white speckeld chickens; heterozygotes
ex: human protein for cholesterol level
• Heterozygotes produce 2 forms of the protein
Multiple Alleles
-still only get 2 alleles, but more possibilities
of the forms of a gene for a trait

Blood Types in Humans
– Single gene, but four phenotypes
•
•
•
•
Type A  can be AA or Ao
Type B  can be BB or Bo
Type AB  only AB (codominant pattern here)
Type O  only oo (both recessive)
– All 3 blood types are dominant to O
Continuous Variation/Polygenic Traits

Multiple genes and often, environmental
factors, are involved in an interaction of these
genes to produce a trait

Examples:
– @ least 3 genes involved in the reddish-brown pigment in the eyes
of fruitflies
– Skin color, wide range in humans produced by more than 4 genes
– Milk Yield in cows
– Height, weight, shoe size, hand span
**different combos of alleles = very different
phenotypes
Sex-linked Genes
Present on the X chromosome
 More common in males
 When would a female have this
phenotype?
 Examples:

– Baldness
– Hemophilia (X-linked recessive)
Some Human Genetic Disorders
Of Interest
Cystic Fibrosis
 Sickle-cell Anemia
 Tay-Sachs Disease
 Phenylketonuria (PKU)
 Hemophilia
 Huntington’s Disease
 Muscular Dystrophy
