Download DIHYBRID CROSSES

DIHYBRID CROSSES
 a cross involving two genes consisting of
nonidentical alleles
 The same rules for monohybrid questions apply.
The Law of Independent Assortment: if genes are
located on separate chromosomes, they are inherited
independently of each other
Inheritance of one trait is not affected by the other
Ex. Green pea, smooth coat
Green pea, wrinkled coat
See figure 1 pg. 150 and figure 2 pg. 151
The results of homozygous dihybrid crosses
See figure 3 pg. 151
Four possible
combinations
for the gametes
of the genotype
YyRr
See figure 4 pg. 152
F2 Generation
Punnett square
for a Dihybrid
cross
Probability:
 what is the likelihood a specific trait will by inherited
in regards to both phenotype and genotype
 the number of ways a specific event can occur (total
number of possible genetic outcomes)
Rules:
1. past outcomes have no effect on the future outcomes
2. the probability of independent events occurring
together is equal to the product of those effects
occurring separately
Read pgs. 150 – 154
Pg. 154 # 1-4
Pg. 156 # 2,3
Chapter Review
Pg. 158 – 159 # 6 – 17
DIHYBRID CROSSES
1. A pea plant that is homozygous for yellow and wrinkled peas is crossed
with a plant that is homozygous for green and round peas. Green and
round peas are dominant. Determine the possible offspring which may
result.
2. In horses, black is dependant upon a dominant gene (B), and chestnut
upon its recessive allele (b). The trotting gait is due to a dominant gene
(T), the pacing gait to its recessive allele (t). If a homozygous black
pacer is mated to a homozygous chestnut trotter, what will be the
appearance of the F1 generation?
3. If two horses of the F1 generation in #2 were mated, what kinds of
offspring could be produced and in what proportions?
4. If an F1 male from problem #2 was mated to a homozygous female black
pacer, what kinds of offspring could be produced and in what
proportions?
5. In rabbits, black is due to a dominant gene (B), and brown to its
recessive allele (b). Short hair is due to a dominant gene (L), and long
hair to its recessive allele (l). In a cross between a homozygous black
short-haired male and homozygous brown long-haired female, what
would be the genotype and phenotype of the F1 and the F2 generations?
6. In guinea pigs, rough coat is due a dominant gene (R) and smooth to its
recessive allele (r). Short hair is dependant upon a dominant gene (L),
and long hair upon its recessive allele (l). Give the phenotype and
genotype and the ratios for each of the following crosses:
a. smooth long-haired x completely heterozygous
b. homozygous rough long-haired x homozygous smooth shorthaired
c. RrLL x RRLl
7. An anemic condition in humans called thalassemia is controlled by
codominant alleles. Homozygotes for the defective allele, TmTm, have a
severe anemia (thalassemia major), whereas heterozygotes, TmTn, have
mild anemia (thalassemia minor). Normal individuals are homozygous
TnTn. An independent gene in humans determines the presence or
absence of pigmentation, with the allele A (normal pigmentation)
completely dominant over the allele a (albino).
Show what the genotypes and phenotypes would be in the P, F1, and F2
generations from the cross: TmTmAA x TnTnaa.