Download Answer Key Lab 10 Human Inheritance

Lab 10: Human Inheritance
Part 1: Human Traits Governed by Mendelian Genetics
BREAKING NEWS! THESE ARE ALL MYTHS! Only ear wax is inherited in Mendelian ratios!
For more information, see http://udel.edu/~mcdonald/mythintro.html
Materials
Small piece of PTC paper
Procedure
1. Record your phenotype and genotype for the following Mendelian traits:
Trait
Tongue roller (R/r)- dominant allele allows you to roll
your tongue
Your Phenotype
Your Genotype
Tongue roller
RR or Rr
Free ear lobes (E/e)- dominant allele leads to free ear
lobes
Free ear lobe
EE or Ee
Mid-digital hair (M/m)- dominant allele results in hair
on middle joint of finger
No hair on middle joint of
finger
mm
Widows peak (W/w)- dominant allele results in widows No widows peak
peak
Hitch-hikers thumb (H/h)- recessive allele results in
ability to bend thumb more than 50°
Hitch-hikers thumb
Bent little finger (B/b)- dominant allele causes first joint Little finger points inward
to point inwards
ww
hh
BB or Bb
Dimples (D/d)- dominant allele results in dimples
No dimples
dd
Taste PTC (T/t)- dominant allele allows you to taste
PTC
Taster
Tt or TT
Polydactyly (P/p)- dominant allele results in more than
five fingers
Five fingers
pp
2. Two people who are heterozygous for tasting the chemical PTC marry. List the genotypes
possible for their children regarding the tasting of PTC. Show your work.
T
t
T
TT
Tt
t
Tt
tt
TT, Tt, and tt
3. Alfred is heterozygous for tongue rolling and has five fingers. Alfreda, his wife, cannot roll her
tongue and is heterozygous for polydactyly.
A. What is Alfred’s genotype? Rrpp
B. What is Alfreda’s genotype? rrPp
4. What are Alfred’s possible sperm with regard to these two traits?
Rp and rp
5. What are Alfreda’s possible eggs with regard to these two traits?
rP and rp
6. What is the probability that their first child will not roll its tongue and will have the normal
number of fingers?
1 in 4
Rp
rp
rP
RprP
rPrp
rp
Rprp
rprp
Part 2: Human Traits not Governed by Mendelian Genetics
There are many exceptions to Mendel’s Rules. For example, blood types in humans exhibit two exceptions:
codominance and multiple alleles.
1. Codominance- both alleles are expressed simultaneously (A and B alleles)
2. Multiple Alleles- more than two alleles exist in a population (ABO)
Background Information
Monoclonal antibodies are used to detect the blood surface markers governed by the ABO and rh factor blood
determinants. A sample of blood is combined with each specific antibody: anti-A, anti-B, anti-O or anti-rh+
antibodies. A positive reaction, usually indicated by blood coagulation, indicates the presence of that particular
blood surface marker. Antigens are molecules that trigger an immune response, causing the production of
antibodies specific to that antigen.
Image accessed 6/13/14 from https://commons.wikimedia.org/wiki/File:ABO_blood_type.svg. Image in the public domain.
Blood type A
Antibody: B
Antigen: A
Blood type B
Antibody: A
Antigen: B
Blood type AB
Antibody: Neither
Antigen: AB
Blood type O
Antibody: Both
Antigen: Neither
Materials
Synthetic blood
Synthetic antibodies
Blood typing tray
Toothpicks
Procedure
1. Record your unknown blood type here: See Chart at the end
2. Place one drop of the unknown blood into each of the three wells on the plastic plate.
3. Apply one drop of anti-A antibody to one of the three wells and mix with a toothpick. If clumping
occurs, the red blood cells have the A antigen embedded in their surface.
4. Repeat the procedure using anti-B antibody in the second well.
5. Finally repeat the procedure for the third well using anti-Rh factor antibody.
Results
Unknown blood type
Rh+ or Rh-
1. Can two people with type A blood have a baby with type O blood? Explain your answer using
a Punnett Square.
Yes. Both parents can be heterozygous with the genotype AO. They have a 1 in 4 chance of
having a baby with O blood type.
A
O
A
AA
AO
O
AO
OO
2. A mother is Rh negative and her husband is Rh positive. If she becomes pregnant, what are the
implications from an immunological standpoint for both fetus and mother, as well as subsequent
fetuses?
If during the birth of the first baby, the baby’s blood mixes with the mother’s blood, she will
produce antibodies against the Rh antigens on the baby’s red blood cells. If a subsequent baby is Rh
positive, the antibodies in the mother’s blood will attack the baby’s red blood cells. This will cause severe
anemia in the baby. Good thing there is a drug today that will stop the mother from producing the
antibodies against the Rh positive blood cells.
Part 3: Human Genetics Problems
Genetic counselors are trained to detect inheritance patterns of genetic diseases based on information they
obtain from the family. Imagine that you are a genetic counselor and you must solve the following cases based
on the information provided. Use the following steps to solve each problem:
i. Create a legend which indicates the gene pairs (alleles) involved. Use a capital letter to denote the
dominant allele and lowercase letter to denote the recessive allele.
Example: D= dimples d= no dimples
ii. Write the genotype and phenotype of the parents. Example: DD → dimples
iii. Use a Punnet Square to cross the potential gametes of the parents.
iv. Determine the probability based on the Punnet Square.
1. Autosomal Recessive Inheritance (trait only expressed if homozygous recessive)
An albino man (nn) marries a normally pigmented woman (N_) who has an albino mother. What is the
chance that their children will be albino?
Since this trait is expressed only if the genotype is homozygous recessive, the mother with albinism
must have the genotype (nn). Since the woman is normally pigmented she must have gotten the (N) allele
from her father. The other mystery allele must be (n) from her mother, making her a heterogygous for
albinism. The chance that their children will albino is 50%.
N
n
n
Nn
nn
n
Nn
nn
2. Autosomal Dominant Inheritance (trait expressed as long as one dominant allele is present)
A daughter wants to know what the chances are that she will develop Huntington’s disease, a degenerative
disorder of the nervous system which appears during the ages of 30 to 40’s. Her mother has Huntington’s
while her father does not have the disease. Try to determine the possibilities from the information you have
at hand. What further information do you need in order to more accurately determine the probability?
H
h
h
h
Hh
hh
Hh
hh
H
h
H
Hh
Hh
Hh
Hh
h
50%
You need to know if the mother is heterozygous or homozygous dominant.
100%
3. X-Linked or Sex-Linked Recessive Inheritance (males more likely to express trait than females)
Hemophilia is a blood disorder which is sex-linked. A woman carrier has children with a normal man.
Determine the chances for girls and boys with hemophilia. [Remember that females have the XX genotype
and males have the XY genotype. Do not place an allele on the Y chromosome. Example: XN Xn for
female; Xn Y for male]
Girls have no chance to get hemophilia.
Boys have 50% chance of getting hemophilia and 50% chance of not getting the blood disorder.
𝑋𝑁
𝑋𝑛
𝑋𝑁
𝑋𝑁 𝑋𝑁
𝑋𝑁 𝑋𝑛
Y
𝑋𝑁 Y
𝑋𝑛 Y
4. Multiple Alleles (more than two alleles in the gene pool of the population)
In the population as a whole there are three alleles (A, B and O) determining blood type, although any one
individual can have at most only two of the three alleles. A woman and her son are both blood type O.
The woman claims that a man with blood type A is the father of the boy. Is this possible? Explain your
answer.
The father could be heterozygous for the blood antigen A with the genotype (AO). If the woman
is homozygous recessive for blood type O and the father is heterozygous for blood type A, there is
a 50% chance that their child will have blood type O.
O
A
O
O
AO
AO
OO
OO
5. Sex Determination (female genotype XX and male genotype XY)
A. What fraction of human offspring receive an X chromosome from the mother?
100%
B. What fraction receive a Y chromosome from the father?
50%
C. Mr. and Mrs. Jackson have seven boys and one girl. What is the probability that their next child will be
a girl?
50%
Part 4: Pedigree Analysis
We will trace the inheritance pattern of the autosomal recessive trait albinism through four generations. The
legend is as follows:
In the pedigree chart below determine the genotypes of each individual. Use a Punnet Square analysis to help
you. Remember that the genotype of affected individuals is nn. If you cannot determine both gene pairs of a
normal individual, indicate the genotype as N_. Put the genotype next to each symbol.
Nn
Nn
nn
Nn
Nn
nn
Nn
Nn
Nn
nn
N_
Nn
nn
nn
nn
Nn
Nn
Part 5: Practice Problems
1. A certain disease is linked to the Y chromosome. What is the chance that a female could get this
disease?
There is no chance that a female will get the disease.
2. Polydactyly is an autosomal dominant condition. Assuming two non-polydactyl individuals
make a baby, what is the chance of them having a polydactyl child?
Since polydactyl is an autosomal dominant condition, 2 parents who are homozygous recessive
would have no chance of having a polydactyl child.
3. If one of the parents described in question 2 is heterozygous for polydactyly, what is the chance
of polydactyl offspring?
p
p
P
Pp
pp
p
Pp
pp
50% chance of polydactyl offspring.
4. Two plants, which are heterozygous for a gene that describes plant height, are crossed. What
percentage of their offspring will have the same phenotype as the parents if the trait is inherited
via complete dominance? Incomplete dominance?
T
t
T
TT
t
Tt
Tt
tt
If the height trait is inherited via complete dominance the parents and 75% of the offspring will be tall.
If the height trait is inherited via incomplete dominance the parents and 50% of the offspring will be
medium height.
5. In dogs, wire hair (S) is dominant to smooth (s). In a cross of a homozygous wire-haired dog with
a smooth-haired dog, what will be the phenotype of the F1 generation? Genotypes?
S
Ss
S
s
Ss
S
Ss
Ss
100% of the F1 generation has the wire-haired phenotype.
100% of the F1 generation has the genotype (Ss).
6. A new disease was found to be sex-linked recessive. A man and a woman, wishing to have
children together would like to know which percentage of their offspring could possibly have
the disease. Neither parent has this disease. What is the highest possible percentage of their
offspring that will have this disease? What proportion of their female offspring could get it?
Males?
𝑋𝑁
𝑋𝑛
𝑋𝑁
𝑋𝑁 𝑋𝑁
𝑋𝑁 𝑋𝑛
Y
𝑋𝑁 Y
𝑋𝑛 Y
The woman could be a carrier for the disease, but the man could not be a carrier. 25 % of
the offspring could get the disease. No female offspring will have the disease. 50% of the of
the males could get the disease.
7. A boy, whose parents and grandparents have normal vision, is color-blind. What are the
genotypes for his mother and his maternal grandparents?
𝑋𝑁
𝑁
𝑋
Y
𝑋𝑛
𝑋𝑁 𝑋𝑁
𝑋𝑁 𝑋𝑛
𝑋𝑁 Y
𝑋𝑛Y
Mother’s and grandmother’s genotype: 𝑋 𝑁 𝑋 𝑛
Grandfather’s genotype:𝑋 𝑁 𝑌
8. Which of the following genotypes is not possible for the offspring of the following cross? AABb x
aaBB
AaBB, AaBb, AABB, AABb
AABB and AABb are not possible
AB
Ab
aB
ABaB
AbaB
BB
AaBB
aAbB
9. In a recent case in Spokane, Washington, a woman claimed a man was the father of her child.
The man denied it. The man’s lawyer demanded that blood types be taken to prove the
innocence of his client. The following results were obtained: Alleged father, Type O. Mother, Type
A. Child, Type AB. The court ruled that he was indeed the father.
A. What are the possible genotypes for these three people?
Childs’s genotype: AB
Mother’s genotype: AA or AO
Father’s genotype: OO
B. Do you agree with the court’s decision? Why or why not?
No. Neither the mother nor the alleged father have the B allele. The B allele must have come from a
different father.
10. It was suspected that two babies had been exchanged in a hospital. Mr. and Mrs. Jones
received baby #1 and Mr. and Mrs. Simon received baby #2. Blood typing tests on the parents
and the babies showed the following:
Mr. Jones: Type A
Mr. Simon: Type AB
Mrs. Jones: Type O
Mrs. Simons: Type O
Baby #1: Type A
Baby #2 Type O
Were the babies switched? Do you know this for sure?
Baby #1 could be the baby of Mr. and Mrs. Jones or it could be the baby of Mr. and Mrs. Simons. Baby #2
could not be the baby of Mr.and Mrs. Simons, but it could be the baby of Mr. and Mrs. Jones. The babies
could have been switches.
O
O
A
O
AO
AO
OO
OO
A
B
O
AO
BO
O
AO
BO
Unknown Identification Chart
Sample 1
Sample 2
Sample 3
Sample 4
Anti A
+
-
+
-
Anti B
-
+
+
-
Anti RH
+
+
-
+
Blood Type
A+
B+
AB-
O+