Transcript Punnett Square Problems 2_2

Name ______________________
Name ______________________
One flower is heterozygous red (Rr) and it is crossed with a
homozygous white (rr) plant. Use a Punnett square to
determine the probability of one of their offspring having a
red color.
R
50%
probability of
being red
r
r
Rr
rr
r
Rr
rr
F.O.I.L
• How to list all possible combinations of two alleles.
• First-Outside-Inside-Last
• RrYy
RY
First:
______
Ry
Outside: ______
Inside: ______
rY
ry
Last:
______
1. In mice, black fur (B) is dominant over brown (b) and short tails (S) are dominant
over long (s).
• Write the genotype for a heterozygous black, short-tailed mouse: ______________
• Cross two of these individuals.
• From the Punnett Square, describe the phenotype ratios of the offspring.
BbSs
BS
Bs
BS
BBSS
BBSs
BbSS
BbSs
Bs
BBSs
BBss
BbSs
Bbss
bS
bs
BbSS
BbSs
bbSS
bbSs
BbSs
Bbss
bbSs
bbss
bS
bs
9 Black Short; 3 Black Long; 3 Brown Short; 1 Brown Long
2. What proportion of the mice offspring of the cross BbSs x BBss will have black fur and
long tails?
Bs
BS
Bs
BBSs
BBss
bS
BbSs
bs
Bbss
Bs
Bs
Bs
1/2 will have black fur, long tails (50%)
3. A couple has three children, all of whom have brown eyes and blond hair. Both parents
are homozygous for brown eyes (BB), and one is blond (rr) while the other is a redhead
(Rr). What is the probability that the next child will be a brown-eyed redhead?
BB rr
Br
BR
BBRr
Br
BBrr
Br
BB Rr
Br
Br
BR
Br
Brown-eyed redhead child = BBRr = 50% chance
Review
•
•
•
•
Same
homozygous = __________
combination of genes,
e.g. BB, bb
Different combination of genes,
heterozygous = __________
e.g. Bb
genes
genotype = What combination of __________
it has,
e.g. Bb, BB
looks like, e.g. brown eyes
phenotype = what it ________
4. In guinea pigs, black fur color (B) is dominant over white (b), and short hair (S) is dominant
over long (s).. Show the Punnett Square for a cross between a homozygous black, short-haired
guinea pig and a homozygous white, long-haired guinea pig. What do the offspring look like?
Homozygous black Short hair = BBSS
Homozygous white, long haired = bbss
BS
bs
BS
BS
BbSs
bs
bs
bs
All BbSs, black fur, short hair
BS
5. In humans, the presence of freckles is due to a dominant gene (F) and the non-freckled
condition is due to its recessive allele (f). Dimpled cheeks (D) is dominant to non-dimpled
cheeks (d). Two persons with freckles and dimpled cheeks have two children. One has
freckles but no dimples and one has dimples but no freckles.
a. What are the genotypes of the parents?
Parents: F__D__ x
F__D__
Children F__ dd ,
Parents must be heterozygous for each gene:
FfDd
x
FfDd
ffD___
b. What are the ratios of the possible phenotypes and genotypes of the children they could
produce?
FfDd
x
FfDd
FD
Fd
FD
FFDD
Fd
fD
fd
F.O.I.L
First-Outside-Inside-Last
fd
FFDd
fD
FfDD
FfDd
FFDd
FFdd
FfDd
Ffdd
FfDD
FfDd
ffDD
ffDd
FfDd
Ffdd
ffDd
ffdd
9 Freckled Dimpled, 3 Freckled Non Dimpled,
3 Non-freckle Dimple; 1 Non Freckle Non Dimple