Download Dihybrid Problems - Milan Area Schools

Dihybrid Crosses
Name:
Dihybrid crosses are 16 square punnetts that predict the chance of offspring having 2 different traits. Let's
call them trait A and trait B.
Hints for solving dihybrid crosses:
* Use the 1,3 1,4 2,3 2,4 Rule
* Number the alleles in the parent genotype 1 -4
* Match 1 and 3 together, 1 and 4 together, 2 and 3 together, 2 and 4 together
* You should have 2 letters in each square along the top and on the side (the parent genotypes)
* Place the letters in the inner squares just like you did for the shorter Punnett squares
* There should be 4 letters in each of the inner squares
* Put the 2 similar letters next to each and put capital letters first
Finding Genotype Probabilities:
* Find all the different letter combinations
* (There will never be more 9 different combinations)
* Count how many you have of each kind
* Your probability will be the number of that combination / 16
Finding Phenotype Probabilities:
* There will never be more than 4 different phenotype combinations
* Count the number of offspring with dominant trait A and dominant trait B
* find those with dominant A, recessive B
* find those with recessive B, dominant A
* find those with recessive A and recessive B
* your probability will be the number of each combination / 16
Problems - 1st one is partially done for you – we will finish it together in class.
1.
G-Normal, g – Green N- Long Nosed, n – Short Nosed
Cross GgNn x GgNn. Give the genotype and phenotypes of the offspring. Be sure to
list expected probabilities of each phenotype (ie. 3/16).
Parent 1: GgNn (normal, long-nose)
1234
GN (1,3)
Gn (1,4)
gN (2,3)
GN
GGNN
Gn
GGNn
GGnn
gN
GgNN
GgNn
ggNN
gn
GgNn
Ggnn
ggNn
Parent 2: GgNn (normal, long-nose)
12 34
gn (2,4)
ggnn
Genotypes
GGNN = /16
GGNn = /16
GGnn = /16
GgNN = /16
GgNn = /16
Ggnn = /16
ggNN = /16
ggNn = /16
ggnn = / 16
Phenotypes
normal, long nose =
normal, short nose =
green, long-nose =
green, short-nose =
/16
/16
/16
/16
2.
3.
G-Normal, g – Green N- Long Nosed, n – Short Nosed
Cross GGNn x ggnn. Give the genotype and phenotype ratios of the offspring.
Be sure to list expected probabilities of each phenotype (ie. 3/16).
Genotypes
In peas, a gene for tall plants, T, is dominant over its allele for short plants, t. The gene for smooth peas, S, is dominant over
its allele for wrinkled peas, s. Calculate both the genotypes and phenotypes for the following cross: TtSS x ttSs
Genotypes
4.
Phenotypes
Phenotypes
Long noses are recessive to short noses. Attached ear lobes are dominant to detached ear lobes. What would the resultant
offspring be of a cross between a homozygous short nosed, homozygous attached ear person and a heterozygous short nosed,
heterozygous attached ear person? Give the genotypes and phenotypes of the offspring as well as the ratios.
S = short nose
s = long nose
A = attached earlobe
Parent 1:_________
Parent 2: ________
a = detached earlobe
Genotypes
5.
Phenotypes
2 Vietnamese pigs were mated together. If all offspring contained the same genotype RrTt (round, squiggly tailed), what were
the genotypes of the two parent pigs? (there may be more than one correct answer)
Genotypes
RrTt
RrTt
RrTt
RrTt
RrTt
RrTt
RrTt
RrTt
RrTt
RrTt
RrTt
RrTt
RrTt
RrTt
RrTt
RrTt
6.
In the fruit fly Drosophila melanogaster, vestigial (stubby) wings and hairy bodies are produced by two recessive genes
located on different chromosomes. The normal alleles, long wings and hairless bodies, are dominant.
Suppose a vestigial-winged hairy male is mated with a homozygous normal female. What types of offspring would be
expected?
L= long wing
l = stubby (vestigial) wing
H = hairless body
Parent 1(male): _________
Parent 2 (female): ________
h = hairy body
Phenotypes (Offspring types)
7.
Suppose a hairy female with vestigial wings is crossed with a vestigial-winged male heterozygous for the hairless
characteristic. What will be the genotypes and phenotypes in the F1?
Genotypes
8.
Phenotypes
A rare recessive gene in the gerbil (d), is lethal (deadly) in the homozygous condition. Geneticists have noticed that if these
gerbils also contain the dominant allele (N), they will survive (a condition known as epistasis). The alleles D and N are coded
for on different genes. If you mate a ddNn male with a Ddnn female, how many offspring will be viable (live)?
Hint: What combinations are deadly? Remember that dd = deadly unless it is paired
with an N.
Number of Viable (living) offspring = ________________
9.
In hogs, a gene that produces a white belt around the animal’s body is dominant over its allele for a uniformly colored body.
Another gene produces a fusion of the two hoofs on each foot, a condition known as syndactyly; this gene
is dominant over its allele which produces normal hoofs.
Suppose a uniformly colored hog homozygous for syndactyly is mated with a normal-footed hog
homozygous for the belted character. What would be the phenotype of the F1? If the F1 individuals are
allowed to breed freely among themselves, what genotypes and phenotypes would be predicted in the F2?
(Hint: You can use the trick that I taught in class rather than doing the punnet squares if you like)
W = white belt
Parent 1: _______________
Parent 2: ________________
w = uniform color
H = fused hoofs
h = normal hoofs
Syndactyly in a
human
F1 Generation
Genotypes
Phenotypes
Genotypes
Phenotypes
F2 Generation
10. Write your own problem. Be sure to pick two genes with two variants (alleles) for each. BE creative, they do not need to be
real traits. Be sure to solve the problem as well, showing the offspring’s expected genotypes and phenotypes.
Genotypes
Trait A= ___________________
Trait B = __________________
Ex. Cred. Draw your variations
Phenotypes