Biology TEACHER
EQ: Do I carry deadly genes?
Enduring Understanding
Nucleic acids transfer genetic information from generation to
generation.
Broad Brush Knowledge
Mendelian genetics, probability, sex-linked, multiple alleles
Advanced Genetics
Problems
Targeted Skills
analysis, probability
Target
3. Apply Punnett squares to determine the probability of two
factor crosses, co-dominance ,sex-linked traits, and multiple
alleles
Concepts Important to Know and Understand
Heredity
Core Objectives
7. Interpret the role of genetics in determining heredity as it applies to
biotechnology.
TEKS
(6D) compare genetic variations observed in plants and animals.
PURPOSE: To predict and compare the genetic variations that will result from a genetic cross.
TEACHER MANAGEMENT
Estimated Time:
45 minutes
Materials: student handout
Teacher Prep: make copies of handouts
PRE-ACTIVITY: Students should have a working understanding of monohybrid and dihybrid crosses.
They should also be familiar with the rules of probability and the terms genotype and phenotype.
Advanced Genetic Problems – Biology- TEACHER (Revised January 22, 2009)
(printed 2/16/2016) p. 1
Biology TEACHER
EQ: Do I carry deadly genes?
Advanced Genetics
Problems
Target
3. Apply Punnett squares to determine the probability of two factor
crosses, co-dominance ,sex-linked traits, and multiple alleles
Enduring Understanding
Nucleic acids transfer genetic information from generation to generation.
Broad Brush Knowledge
Mendelian genetics, probability, sex-linked, multiple alleles
Concepts Important to Know and Understand
Heredity
Core Objectives
7. Interpret the role of genetics in determining heredity as it applies to
biotechnology.
PURPOSE: To predict and compare the genetic variations that will result from a genetic cross.
Codominance occurs when both alleles contribute to the phenotype, leaving a phenotype that is
intermediate of the parents.
Example: Red flowers are fertilized by white flowers. The resulting offspring have red and white flowers.
PRACTICE:
1. Sickle cell anemia is a codominant disorder, where NA
stands for normal red blood cells and NS stands for sickleshaped red blood cells. Heterozygous individuals have a
phenotype showing both sickle-celled and normal-shaped
red blood cells. Cross two people who are heterozygous for
this trait.
NA
NS
NA
NANA
NANS
NS
NANS
NSNS
NANS x NANS
**Remember to express probability in all acceptable forms: fraction, decimal and percentage.
a. What is the probability their offspring are likely to have sickle cell anemia? ¼ = .25 = 25% __________
b. What is the probability their offspring will have both normal and sickle cell-shaped red blood cells?
2/4 = ½ = .50 = 50% _______________________________________________________________
c. What is the probability their offspring will be able to pass along the sickle trait to their children?
3/4 = .75 = 75% ___________________________________________________________________
2. Multiple alleles occur when a trait has more than just a dominant and recessive allele. For example,
coat color in rabbits is determined by a single gene with four different alleles: C, cch, ch, c . The
combination of two of these alleles determines the rabbit’s phenotype.
Full color (C) - brown
(dominant to all other alleles)
CC, Ccch, Ccch, Cc
Chinchilla (cch) – gray
(dominant to ch and c)
cchch, cchcch, cchc
Himalayan (ch) – mostly white
(dominant to c allele)
chch, chc
Albino (c) – no color
(recessive to all other alleles)
cc
a. If a homozygous Himalayan rabbit and an albino rabbit mate, what are the possible phenotypes of their
offspring? all Himalayan ______________________________________________________________
Advanced Genetic Problems – Biology- TEACHER (Revised January 22, 2009)
(printed 2/16/2016) p. 2
3. Blood types are controlled by multiple alleles. Look at the chart below containing blood types for the
human population. Type A and B are both dominant, while type O is recessive.
BLOOD TYPE
A
B
AB
O
GENOTYPES
IAIA, IAi
IBIB, IBi
IAIB
ii
b. A man with type O blood marries a woman with type AB blood.
What is the probability that they will have a child with type A
blood? 2/4 = ½ = .50 = 50% ____________________________
(Show your work in a Punnett Square.)
ii x IAIB
IA
IB
i
IAi
IBi
i
IAi
IBi
b. If child X has type A blood and father Y has type AB blood, what are the child’s possible genotypes?
Child X genotype = IAi or IAIA (father type AB = IAIB) _____________________________________
c. If child X has type AB blood and mother Y has type B blood, what blood type does the father have?
father is Type A to give IA gene to child (child X type AB = IAIB, mother Y type B = IBIB or IBi) __
d. Vincent has type A blood and his mother has type O blood. What is his genotype? IAi _____________
(Since Vincent is type A, he must have an IA gene. His mother is type O (ii) and gave him an i gene) __
e. Christy has type B blood and her father has type O blood. What is her genotype? IBi ______________
(Since Christy is type B, she must have an IB gene. Her father is type O (ii) and gave her an i gene) __
f. If Vincent from problem 3 and Christy from problem 4 have
IB
i
children, what are the children’s likely genotypes?
A B
A
B
I I , I i, I i, ii ______________________________________
IAi
IA
IAIB
(Show your work in a Punnett Square.)
IAi x IBi
i
IBi
ii
4. Sex-linked genes are so-called because they are located on one of the sex chromosomes, either the X
or the Y. Since the X chromosome contains more genes than the smaller Y chromosome, mutations on
the X appear more frequently in the population. In addition, sex-linked disorders are found more
commonly in males than females because they have only one copy of the X chromosome. The
genotypes for sex-linked traits are also written with superscripts: XBY or XbY for males or XBXB, XBXb, or
XbXb for females.
Complete the following chart by giving the phenotypes and genders for the following genotypes, X N gene
for normal vision and Xn gene for color blindness.
GENOTYPE
PHENOTYPE
GENDER
XNXN
Normal vision
Female
XNY
Normal vision
Male
XNXn
Normal vision (carrier)
Female
XnY
Color blind vision
Male
XnXn
Color blind vision
Female
Advanced Genetic Problems – Biology- TEACHER (Revised January 22, 2009)
(printed 2/16/2016) p. 3
a. What are the possible genotypes and phenotypes resulting from
a cross between a carrier female for color blindness and a
normal vision male? Genotypes: XNXN, XNXn, XNY, XnY _______
Phenotypes: normal female (includes carrier), normal
male,colorblind male
XN
Y
XN
XNXN
XNY
Xn
XNXn
XnY
XD
Y
XD
XDXD
XDY
Xd
XDXd
XdY
(Show your work in a Punnett Square.)
XNXn x XNY
b. Duchenne Muscular Dystrophy is caused by a recessive allele
located on the X chromosome. Affected people experience
progressive weakening and loss of skeletal muscle. How can two
unaffected people have a child with Duchenne MD?
mother is a carrier XDXd and father is normal XDY __________
(Show your work in a Punnett Square.)
XDXd x XDY
a) Is the child with MD a male or female? Male ______________________________________________
b) Who did this child inherit the disorder from? Why? Mother, he got the Y chromosome from his father
and the only Xd chromosome/gene from his mother.
c) What is the probability that the couple’s next child will have MD? ¼ = .25 = 25% _________________
d) What percentage of the couple’s female children will probably have MD? 0/4 = 0 = 0% ____________
e) What percentage of the couple’s male children will probably have MD? ¼ =.25 = 25% _____________
d. Hemophilia is caused by a recessive allele on the X chromosome
that causes abnormal clotting of blood when cut or bruised. How
can a mother whose blood clots normally can have a daughter
with hemophilia.? daughter with hemophilia (XhXh) would need
both parents to have the Xh gene; a mother who was normal but
a carrier (XHXh) and the father would have hemophilia (XhY) ___
Xh
Y
XH
XHXh
XHY
Xh
XhXh
XhY
(Show your work in a Punnett Square.)
XHXh x XhY
a) What is the father’s genotype? XhY _____________________________________________________
b) Why is it extremely rare for a female to have hemophilia? Females must have two recessive Xh gene,
one Xh from each parent which is rare. Usually they have XHXh so that they are normal but carriers.
c) What is the probability that their offspring will have hemophilia? 2/4 = ½ = 50% __________________
d) What is the probability that their offspring can pass hemophilia on to their children? ¾ = 75% ________
e) In this case, what do we call the mother if she is heterozygous for this trait? A carrier _____________
e. Not all sex-linked genes are recessive. For example,
hypophosphatemia, which causes defective bone development is
caused by a dominant allele on the X chromosome. Work a cross
showing a woman with a genotype of XAXa and a man with a
genotype of XaY.
a) What are the phenotypes of the mother and father? Mother has
defective bone development, father has normal bone
development ________________________________________
Xa
Y
XA
XAXa
XAY
Xa
XaXa
XaY
b) What is the probability that the children will have hypophosphotemia? 2/4 = ½ = .50 = 50% ___
Advanced Genetic Problems – Biology- TEACHER (Revised January 22, 2009)
(printed 2/16/2016) p. 4