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Heredity Lab
Mendelian Genetics
Part 1: Terminology
Beginning students of biology always learn about Mendelian genetics. Inevitably, the study of
inheritance always leads to additional questions. In fact, Mendelian inheritance patterns are
exceedingly rare, especially in humans. We now know that inheritance is much more complex,
usually involving many genes that interact in varied ways. Nonetheless, a clear understanding
of basic inheritance patterns that follow Mendel’s original observations will provide a
springboard for understanding current scientific exploration.
Inheritance patterns that follow Mendelian rules are as follows:
• Traits are governed by single genes
• There are two alternate forms of a gene, known as alleles
• Alleles are expressed as dominant and recessive
It just so happened that the traits Gregor Mendel observed in his pea plants did indeed
conform to these rules. After collecting and analyzing his data, Gregor Mendel developed
two laws of inheritance: The Law of Segregation and the Law of Independent Assortment.
1. Describe these laws below:
A. The Law of Segregation
B. The Law of Independent Assortment
2. Before you can work with problems involving Mendelian inheritance, you need to be comfortable with the following terms: Define them and give examples.
DNA
Chromosome
Gene
Locus
2
Diploid
Haploid
Allele
Dominant
Co-Dominant
Recessive
Genotype
Phenotype
Homozygous
Heterozygous
3
Part 2: Mendel’s First Law- Law of Segregation
The Law of Segregation states that alternative alleles of a trait segregate independently
during meiosis.
Using a technique known as Punnett Square analysis, we will see how Mendel analyzed his
monohybrid crosses to come up with the Law of Segregation.
Procedure
Carefully follow each step to create a Punnett square analysis. You can use these SAME
general procedures to analyze EVERY Punnett Square you do!
Problem: In pea plants, height is coded for by the “T” gene. The dominant allele (T) codes for the tall phenotype while the recessive allele (t) codes for the short phenotype. Make a cross between a true breeding tall pea plant and a true breeding short pea plant.
Things to consider and how to solve the problem;
1. What are the phenotypes of the parent plants? The parents are considered the P
generation. 1st Parent ____________ 2nd Parent___________
Determine the genotypes of each parent plant. 1st Parent _____ 2nd Parent_____
2. Imagine each parent goes through MEIOSIS to produce gametes. List the genotype(s) of the possible gametes that each parent would produce.
3. Create a Punnett square that displays the GENOTYPES of the possible offspring. Also
label the PHENOTYPES of the possible offspring. These offspring are considered the F1
(first filial) generation. Remember, parents only give up one allele for each trait, so there
is only 1 allele above each box. When you “multiply” alleles the offspring will each have 2;
one from each parent.
Genotypes of parent
4
4. Now allow the F1 generation to self-pollinate. What are the possible gametes that each F1 parent can produce?
5. Create a Punnett square crossing two F1 offspring that displays the Genotypes of the possible offspring. Also label the PHENOTYPES of the possible offspring. These offspring are considered the F2 (second filial) generation. Genotypes of parent
Note: Always reduce the phenotypic and genotypic ratios to their lowest terms.
1. What is the phenotypic ratio of the F1 generation?
2. What is the genotypic ratio of the F1 generation?
3. What is the phenotypic ratio of the F2 generation?
4. What is the genotypic ratio of the F2 generation?
5
Part 3: Probability
Do the expected and observed phenotypic and genotypic ratios always match up in real life? In the case of flipping coins, we would expect to see heads 50% of the time and tails 50% of the time. But, does this always occur? Let’s explore! Materials- 2 coins
Procedure 1. Working with a partner, take two coins and assume that heads represent the dominant allele (A)
and tails represents the recessive allele (a). The genotype for each coin is heterozygous (Aa).
2. Assume that each coin represents one parent. When a single coin is flipped, one gamete is formed (through the process of meiosis). If the flipped coin is on heads, then the gamete has the dominant allele (A). When both coins are flipped simultaneously, there will be two possible gametes which can combine through fertilization to form a zygote. Each time you flip both coins, you will record the “genotype” of the offspring.
3. Flip the coins 100 times and record your results in the chart below.
Expected results
(After 100 flips)
Your results (# of flips
with each outcome)
Class results
Genotype Expected
count
Ratio
(4*count)
total flips
Tally Observed
count
Ratio Observed
count
Ratio **
AA
Aa
aa
Total
flips
100 -- Total flips 100 Total
flips
Calculate the ratios using this formula:
Genotypic Ratio = # of possible combinations (4) x # of flips of a given genotype (from tally)
total number of flips counted (100)
**Note: If calculating class totals, the denominator in this equation is equal to the total of all flips counted by all
students in the class.
4. Record your results on the board and when all groups have entered their totals, calculate the “Class Totals” column.
5. What is the expected genotypic ratio for a cross between two Aa coins?
6. Did the observed and expected genotypic ratios match? Why or why not?
6
Part IV: Make a Baby
Purpose: To demonstrate the principles of Mendelian genetics and sex determination, including the
concepts of allele, phenotype, genotype, dominant, recessive, codominant, homozygous and heterozygous
by creating a simulated baby.
Materials: Two pennies, art supplies, paper.
Procedure:
1) Working with a partner, determine the genotype of the baby by flipping pennies. "Mom" flips one penny
to choose an allele for her egg and "Dad" flips the other to choose an allele for his sperm. (Note that
the gender of the baby is a special case and is determined by dad alone. Boys are XY and girls are XX.
Mom can give only an X but dad can give either an X or a Y.)
2) Record the alleles which resulted from the coin flips, and put "sperm and egg" together. (You cannot
pick the traits you want; life doesn't work that way!) Write down baby's genotype for each trait in
Table 1. Heads represents allele #1 and tails represents allele #2.
3) Record the baby's phenotype in Table 1 by looking up the genotype you got in the Genotype/Phenotype
Reference Sheet. Note: Dominant alleles are written with an uppercase letter and recessive alleles are
written as lowercase letters. Dominant alleles mask the expression of recessive ones. Co-dominant
alleles are written as uppercase letters with a subscript. Co-dominant alleles result in a phenotype that
is blended.
4) Repeat steps 1, 2, and 3 for all traits and then draw, color, and name your creation. Remember that you
are drawing a baby's face that represents the traits you got - not a child's or an adult's (no tattoos, no
mustaches, no piercings, etc., and not too much hair!)
Questions:
1. Why is the coin flip used to represent the selection of alleles?
2. Which parent determines the sex of each child, and why?
3. Compare your child to other children in the class. What similarities do the faces have?
4. Do the children look more like one parent than the other? Explain why?
5. We only looked at 23 traits. Think of the thousands of traits that humans possess. Why is it
unlikely any two non-identical siblings will look the same?
7
Data:
Table 1: Check here indicating whether you are the mom or the dad and fill in the data below.
Mom's Name: ________________ Dad's Name _________________ Baby's Name: ________________
Trait Heads Tails Allele from Mom Allele from Dad Genotype Phenotype
Gender X Y ______X______ _____________ _____________ _____________
Face Shape R r _____________ _____________ _____________ _____________
Chin Shape N n _____________ _____________ _____________ _____________
Chin Dimple A a _____________ _____________ _____________ _____________
Freckles F f _____________ _____________ _____________ _____________
Cheek Dimples D d _____________ _____________ _____________ _____________
Lip Thickness T t _____________ _____________ _____________ _____________
Eye Brows B b _____________ _____________ _____________ _____________
Eye Shape W w _____________ _____________ _____________ _____________
Eyelashes L l _____________ _____________ _____________ _____________
Ear Shape R r _____________ _____________ _____________ _____________
Ear Lobes F f _____________ _____________ _____________ _____________
Widow's Peak W w _____________ _____________ _____________ _____________
Hair Curliness C1 C2 _____________ _____________ _____________ _____________
Eyebrow Color D1 D2 _____________ _____________ _____________ _____________
Eye Width W1 W2 _____________ _____________ _____________ _____________
Eye Size S1 S2 _____________ _____________ _____________ _____________
Mouth Size M1 M2 _____________ _____________ _____________ _____________
Nose Size P1 P2 _____________ _____________ _____________ _____________
Birth Mark B1 B2 _____________ _____________ _____________ _____________
Skin Tone S1 S2 _____________ _____________ _____________ _____________
Polygenic Trait (4 coin flips) Alleles from Mom Alleles from Dad Genotype Phenotype
Hair Color A, B a, b #1____ #2____ #1____ #2____ __ __ /__ __ _____________
Eye Color A, B a, b #1____ #2____ #1____ #2____ __ __ /__ __ _____________
8
Genotype/Phenotype Reference Sheet
Trait Genotype/Phenotype
(Homozygous for Allele 1)
Genotype/Phenotype
(Heterozygous)
Genotype/Phenotype
(Homozygous for Allele
#2)
Face Shape RR
Round
Rr
Round
rr
Square
Chin Shape NN
Noticeable
Nn
Noticeable
nn
Less Noticeable
Chin Dimple AA
Absent
Aa
Absent
aa
Present
Freckles FF
Present
Ff
Present
ff
Absent
Cheek
Dimples
DD
Present
Dd
Present
dd
Absent
Lip
Thickness
TT
Thick
Tt
Thick
tt
Thin
Eye Brows BB
Bushy
Bb
Bushy
bb
Fine
Eye Shape WW
Wide
Ww
Wide
ww
Round
Eyelashes LL
Long
Ll
Long
ll
Short
Ear Shape RR
Long
Rr
Long
rr
Round
9
Ear Lobes FF
Free
Ff
Free
ff
Attached
Widow's
Peak
WW
Present
Ww
Present
ww
Absent
Hair
Curliness
C1C1
Curly
C1C2
Wavy
C2C2
Straight
Eyebrow
Color
D1D1
Darker than hair
D1D2
Same as hair
D2D2
Lighter than hair
Eye Width W1W1
Close Together
W1W2
Average
W2W2
Far apart
Eye Size S1S1
Large
S1S2
Medium
S2S2
Small
Mouth Size M1M1
Wide
M1M2
Medium
M2M2
Narrow
Nose Size P1P1
Small
P1P2
Medium
P2P2
Large
Birth Mark
(mole)
B1B1
Left cheek
B1B2
Absent
B2B2
Right cheek
Skin Tone S1S1
Light
S1S2
Medium
S2S2
Dark
Hair Color
AABB=Black AaBB=Dark Brown aaBB=Blond
AABb=Black AaBb=Light Brown aaBb=Blond
AAbb=Red Aabb=Dark Blond aabb=white (albino)
Eye Color
AABB=Deep Brown AaBB=Greenish Brown aaBB=Green
AABb=Deep Brown AaBb=Light Brown aaBb=Light Blue
AAbb=Brown Aabb=Gray-Blue aabb=Pink
10
Part V: Genetics Problems
Note: Get in the practice from the start of showing all your work and answering all parts of the
question as you will be expected to do so on all assignments and examination.
MENDELIAN GENETICS
1. A student has a penny, a nickel, a dime, and a quarter. She flips them all simultaneously and checks for heads or
tails. Show your work.
What is the probability that all four coins will come up heads? ________________________________________
She again flips all four coins. What is the probability that she will get four heads both times? _______________
What probability rule did you use to determine your answer? ________________________________________
2. For the following crosses, indicate the probability of obtaining the indicated genotype in an offspring.
Remember it is easiest to treat each gene separately as a monohybrid cross and then combine the probabilities.
Cross Offspring Probability
AAbb x AaBb AAbb
AaBB x AaBb aaBB
AABbcc x aabbCC AaBbCc
AaBbCc x AaBbcc aabbcc
3. The genotype of F1 individuals in a tetrahybrid cross is AaBbCcDd. Assuming independent assortment of these
four genes, what are the probabilities that F2 offspring would have the following genotypes? Show your work.
a. aabbccdd _________________________________________
b. AaBbCcDd _________________________________________
c. AABBCCDD _________________________________________
d. AaBBccDd ________________________________________________________________
e. AaBBCCdd _________________________________________
11
4. Flower position, stem length, and seed shape were three characters that Mendel chose to study. Each is
controlled by an independently assorting gene and has dominant and recessive expression as follows:
Trait Dominant Recessive
Flower position Axial (A) Terminal (a)
Stem length Tall (T) Short (t)
Seed shape Round (R) Wrinkled (r)
If a plant that is heterozygous for all three traits were allowed to self-pollinate, what proportion of the offspring
would be expected to be as follows? (NOTE: Use the rules of probability and show your work.)
a) Homozygous for the three dominant traits _______________________________________
b) Homozygous for the three recessive traits _______________________________________
c) Heterozygous for the three traits ____________________________________________
d) Homozygous for axial and tall, heterozygous for round _______________________________
5. In squash, an allele for white color (W) is dominant over the allele for yellow (w). Give the expected genotype and
phenotype ratios for the offspring produced by each of the following crosses. (The crosses now use the correct
letters for the alleles.)
a) WW x ww
b) Ww x Ww
c) Ww x ww
6. In human beings, brown eyes are usually dominant over blue eyes. Suppose a blue-eyed man marries a brown-eyed
woman whose father had blue eyes. What proportion of their children would you predict will have blue eyes?
7. Polydactyly (extra fingers and toes) is due to a dominant gene. A father is polydactyl, the mother has the normal
phenotype, and they have had one normal child. P = polydactyl; p = normal
What is the genotype of the father?_________________________________
What is the genotype of the mother?________________________________
What is the probability that a second child will have the normal number of digits? Show your work.
______________________________________________________________________________
8. A woman with the rare recessive disease phenylketonuria (PKU), who had been treated with a diet having low
levels of the amino acid phenylalanine, was told that it was unlikely her children would inherit PKU because her
husband did not have it. However, her first child had PKU.
What is the most likely explanation? _____________________________________________________________
Assuming this explanation is true, what would be the probability of her second child having PKU? Show your
work.
_____________________________________________________________
12
9. PKU is an inherited disease determined by a recessive allele. If a woman and her husband are both carriers, what is
the probability of each of the following? Show your work.
a. All three of their children will be normal _____________________________________
b. One or more of the three children will have the disease. ________________________
c. All three children will be afflicted with the disease. ____________________________
d. At least one child will be normal. _________________________________________
10. In an examination of Mendel’s principles, strain of light brown mice was crossed with a strain of dark brown mice. All F1 were dark brown. In the F2, 42 were dark brown and 15 were light brown. Is this consistent with the law of segregation? Explain.
_______________________________________________________________________________
_______________________________________________________________________________
11. A black guinea pig crossed with an albino one gave 12 black offspring. When the albino was crossed with a
second black one, 7 blacks and 5 albinos were obtained. What is the genotype for:
a. The first black parent? _____________________________
b. The albino parent? _____________________________
c. The second black parent? _____________________________
d. The first black offspring? _____________________________
e. The second black offspring? _____________________________
Why did the two crosses produce different offspring since both involved a cross between a black parent and
albino parent?
______________________________________________________________________________
12. Karen and Steve each have a sibling with sickle-cell anemia. Karen, Steve, nor any of their parents has the
disease, and none of them has been tested to reveal the sickle-cell trait. Based on this incomplete information,
calculate the probability that if this couple has a child, the child will have sickle-cell anemia. Show your work.
13
BEYOND MENDEL
13. A rooster with blue (actually gray) feathers is mated with a hen of the same phenotype. Among their offspring, 15
chicks are blue, 6 are black, and 8 are white.
What is the simplest explanation for the inheritance of these colors in chickens?
___________________________________________________________
What offspring would you predict from the mating of blue rooster and a black hen?
___________________________________________________________
14. If two medium-tailed pigs were mated and the liter produced included three stub-tailed piglets, six medium-tailed,
and four long-tailed piglets, what would be the simplest explanation of these results?
___________________________________________________________
MULTIPLE ALLELES
Blood typing
15. We know that human blood type is determined by a three-allele system at a single locus. For the following, state
whether the child mentioned can actually be produced from the marriage.
a. An O child from the marriage of two A individuals ___________
b. An O child from the marriage of an A to a B ___________
c. An AB child from the marriage of an A to an O _______________
d. An O child from the marriage of an AB to an A _____________
e. An A child from the marriage of an AB to a B ______________
16. Blood typing has often been used as evidence in paternity cases, when the blood type of the mother and child
may indicate that a man alleged to be the father could not possibly have fathered the child. For the following
mother and child combinations, indicate which blood groups of potential fathers would be exonerated (not the
daddy).
Blood Group of Mother Blood Group of Child Blood Group that would
Exonerate Man
AB A
O B
A AB
O O
B A
14
17. Fred has type AB blood, Wilma has type B blood, and Pebbles, their daughter has type A blood. Betty has type
B blood, Barney has type A blood, and their some BamBam has type O blood. In the bloodiest fight ever
witnessed in Bedrock, BCE, Barney accused Betty of having an affair with Fred. Barney also claimed that Fred
is BamBam’s father, sighting evidence from the new field of Geneticsrock. Could Barney be right? Could Fred
be BamBam’s father? Support your answer.
18. A man with group B blood marries a woman with group B blood. Their child has group O blood. What are the
genotypes of these individuals? What other genotypes, and in what frequencies, would you expect in offspring
form this marriage?
19. Imagine that a newly discovered, recessively inherited disease is only expressed in individuals with group O
blood, although the disease and blood group are independently inherited. A normal man with A blood and a
normal woman with B blood have already had one child with the disease. The woman is now pregnant for a
second time. What is the probability that the second child will also have the disease? Assume the parents are
heterozygous for the “disease” gene. Show your work.
________________________________________________________________________________________
Color Patterns
20. Color pattern in a species of duck is determined by a single pair of genes with three alleles. Alleles H and I are
codominant, and allele i is recessive to both. How many phenotypes are possible in a flock of ducks that
contains all the possible combinations of these three alleles?
Epistasis
21. In guinea pigs, the gene for production of melanin is epistatic to the gene for the deposition of melanin. The
dominant allele M causes melanin to be produced; mm individuals cannot produce the pigment. The dominant
allele B causes the deposition of a lot of pigment and produces a black guinea pig, whereas only a small amount
of pigment is laid down in bb animals, producing a light-brown color. Without an M allele, no pigment is
produced so the allele B has no affect and the guinea pig is white. A homozygous black guinea pig is crossed
with a homozygous recessive white: MMBB x mmbb. Give the phenotypes of the F1 and F2 generations.
F1 generation:__________________________________________________
F2 generation:__________________________________________________
15
PEDIGREES
22. Use the information provided below to answer the questions that follow.
a. List the sex of the children, in order of birth, produced by the parents in Generation I.
________________________________________________________
b. Is the trait being tracked in this pedigree dominant or recessive? How do you know?
________________________________________________________________________________
_______________________________________________________________________________
c. How many children did individuals 2 and 3 in Generation II produce?
_______________________________________________________________________________
d. What is the relationship of individual 6 in Generation II to the couple in Generation I?
_______________________________________________________________________________
23. Describe what you think is important to know medically about the behavior of recessive alleles.
______________________________________________________________________
_____________________________________________________________________
24. Albinism (lack of skin pigmentation) is caused by a recessive allele. Consider the following human
pedigree for this trait (solid symbols represent individuals who are albinos).
16
a. What are the genotypes of father and mother in Generation I?
Father:______________ Mother:_____________
b. In Generation II, what is the genotype of:
Mate 1______________ Mate 2:_____________
c. In Generation III, what is the genotype of son 4?_________________
d. Can you predict the genotype of son 3? Explain.
________________________________________________________
________________________________________________________
25. The pedigree below shows the ABO blood group for a family.
What is the genotype for the following individuals?
Individual Genotype Individual Genotype
(a) (f)
(b) (g)
(c) (h)
(d) (i)
(e)
17
26. The following pedigree traces a form of deafness in a family. This deafness is a recessive trait. Using the letters N for normal and n for deafness, provide the genotypes for the individual indicated in the chart that follows.
Individual Genotype Individual Genotype
I 1 III 5
II 2 III 9
II 6 IV 4
II 7 IV 12
27. The pedigree below traces brachydactyly, a condition in which fingers are abnormally short, through several
generations of a family. Those individuals afflicted with brachydactyly are shaded. Use this pedigree to
answer the questions that follow.
a. Examine the children produced by individuals 6 & 7 in Generation II. How do you explain the fact
that 9 is not brachydactyly?
_____________________________________________________________
_____________________________________________________________
b. Is brachydactyly a dominant or recessive disorder?_____________________
c. What is the relationship between individuals 2 & 3 in Generation II?
____________________________________________________________
18
SEX-LINKED TRAITS
28. We know that the most common form of color blindness results from an X-linked recessive gene. A
couple with normal color vision has a daughter with normal vision and a son who is color-blind. What
is the probability that the daughter is a carrier for the color-blindness allele? In other words, what is the
probability that the daughter is heterozygous for the trait?
29. Both John and Cathy have normal colored vision. After 10 years of marriage, Cathy gives birth to a
colorblind son. John filed for divorce, claiming he’s not the father of the child.
(i) Is John justified in his claim for nonpaternity? Explain why
(ii) From which grandparent could the son have inherited his X chromosome?
Cathy’s Mom: _________ Cathy’s Dad: _________
John’s Mom: __________ John’s Dad: __________