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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

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Diploid

Haploid

Allele

Dominant

Co-Dominant

Recessive

Genotype

Phenotype

Homozygous

Heterozygous

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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

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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?

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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?

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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?

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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____ __ __ /__ __ _____________

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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

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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

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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 _________________________________________

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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.

_____________________________________________________________

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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.

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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

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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:__________________________________________________

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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).

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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)

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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?

____________________________________________________________

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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: __________