lab #13 biology 10 bcc topic: mendelian genetics part1...

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Lab #13 Biology 10 BCC Topic: Mendelian Genetics

Part1: 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 2 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:

Diploid

Haploid

Allele

Dominant

Recessive

Genotype

Phenotype

Heterozygous

Homozygous

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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.

1. What are the phenotypes of the parent plants? The parents are considered the P generation.

2. Determine the genotypes of each parent plant.

3. Imagine each parent goes through MEIOSIS to produce gametes. List the genotype(s) of the possible gametes that each parent would produce.

4. 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.

5. Now allow the F1 generation to self-pollinate. What are the possible gametes that each F1 parent can produce?

6. 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 F2 (second filial) generation.

Note: Always reduce the phenotypic and genotypic ratios to their lowest terms.

7. What are the phenotypes and what is the phenotypic ratio of the F1 generation?

8. What are the genotypes and what is the genotypic ratio of the F1 generation?

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9. What is the phenotypes and the phenotypic ratio of the F2 generation?

10. What are the genotypes and the genotypic ratio of the F2 generation?

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 classroom board and when all results are recorded, you will need to tally the numbers for

the “Class Totals” column in chart above.

5. What is the expected genotypic ratio for a cross between two Aa coins?

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6. Did the observed and expected genotypic ratios match? Why or why not?

Part 4: Probability Continued

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 _________________________________________

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)

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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. 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.

6.

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.

___________________________________________________________________________________________

7. 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.

___________________________________________________________________________________________

____________________________________________________________________________________________

8. 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?

______________________________________________________________________________

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Part 5: The Law of Independent Assortment

The Law of Independent Assortment states that genes located on different chromosomes assort independently from one another. To see the effects of this law, you must examine two different genes that are carried on two different chromosomes. We can investigate this phenomenon by looking at “babies”.

For this experiment, each group will examine a special ear of corn. These ears were created when a mama and papa corn plant, both heterozygous for SEED COLOR (P = purple, p= yellow) and SEED SHAPE (S = smooth, s = wrinkled), made baby corns. Corn is cool, because an EAR of corn is just a WHOLE BUNCH OF BABIES held all in one place! By counting the corn babies (each kernel is a baby), you can investigate the principle of Independent Assortment.

Materials-1 ear of corn/group

Procedure

1. What were the phenotypes of the mama and papa corn plants that gave rise to your cob of babes? (Read the

previous paragraph to answer this question!)

2. What were the genotypes of the mama and papa corn plants that gave rise to your cob of babes?

3. What were all the possible gametes each parent corn could produce?

Mama corn: Papa corn:

4. Make a dihybrid cross illustrating ALL the POSSIBLE baby corns produced by these parents. Then calculate the

EXPECTED phenotypic ratios of the babies.

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5. Count and record the phenotypes of 100 kernels on your cob. Record your results below.

Phenotypes of possible offspring

Smooth Purple Smooth Yellow Wrinkled Purple Wrinkled Yellow

Tally Total Tally Total Tally Total Tally Total

Expected ratio

Total # counted

Observed Ratio (16*count) total

flips

Observed Ratio (Class average)

Calculate the ratios using this formula:

Phenotypic Ratio = # of possible combinations (16) x # of kernels of a given phenotype

total number of kernels counted (100)

Note: If calculating class totals, the denominator in this equation is equal to the total of all kernels

counted by all students in the class.

1. Do the observed phenotypes agree with the expected phenotypes? Why or why not?

2. Can you determine the genotypes of the purple kernels or the smooth kernels in this lab

exercise? Why or why not?

3. Can you determine the genotypes of the yellow kernels or the wrinkled kernels in this lab

exercise? Why or why not?

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Part 6: Pedigrees

1. 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?

________________________________________________________

2. Pedigree analysis is often used to determine the mode of inheritance (dominant or recessive, for

example). Complete the unlabeled pedigree by indicating the genotypes for all involved. What is the

mode of inheritance for this pedigree?

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3. Explain why you know the genotype of one female in the third generation, but are unsure of the other.

__________________________________________________________________________________

__________________________________________________________________________________

4. Describe what you think is important to know medically about the behavior of recessive alleles.

__________________________________________________________________________________

__________________________________________________________________________________

5. 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).

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.

_______________________________________________________________________

_______________________________________________________________________

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6. 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)

7. 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

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I 1 III 5

II 2 III 9

II 6 IV 4

II 7 IV 12

8. 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?

_______________________________________________________________________

9. Identify the inherited disorder described below.

a. Recessively inherited disorder in which a defective protein causes the buildup of

chloride ions in cells, the osmotic uptake of water from the surrounding mucus, and

results in very thick, viscous mucus. This thick mucus accumulates in the pancreas,

lungs, & digestive tract.

Disorder:_________________________________________________

b. Recessively inherited disorder where a missing enzyme results in the accumulation of

lipids in the brain. The accumulation of lipids in the brain causes seizures, blindness,

and degeneration of motor & mental performance.

Disorder:_________________________________________________

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c. Recessively inherited disorder that results in a single amino acid substitution in hemoglobin. This substitution results in abnormally shaped hemoglobin molecules and as a result abnormally shaped red blood cells.

Disorder:_________________________________________________

d. This disorder is caused by a late-acting lethal dominant allele. The phenotype does not appear until 35 to 40 years of age and causes degeneration of the nervous system and eventually death.

Disorder:_________________________________________________

10. Explain how a lethal recessive gene can be maintained in a population.

______________________________________________________________________________

______________________________________________________________________________

11. Identify each pedigree as autosomal recessive, autosomal dominant, X-linked, or Y-linked

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Part 7: Polygenic inheritance, pleiotropy, codominance, incomplete dominance

1. Define each of the following:

Complete

Dominance

Incomplete

Dominance

Codominance

2. 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?

___________________________________________________________

3. 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?

___________________________________________________________

4. The chart shows the results from several matings between different tribbles.

Parental Cross Offspring Parental Cross Offspring

Blue x red All purple Yellow x white All yellow

Blue x yellow All green Blue x black All blue

Yellow x red All orange Red x black All red

Blue x white All blue Yellow x black All yellow

Red x white All red Black x white All gray

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a. Which crosses are examples of complete dominance?

____________________________________________________

____________________________________________________

b. Which crosses are examples of incomplete dominance?

______________________________________________________

c. Give the genotypes for each of the following tribble colors. Remember to use a capital letter

to indicate a dominant allele and a lower case letter to indicate a recessive letter. If the color is

the result of incomplete dominance, two capital letters (or two lower case letters) should be

used. For example, in some flowers when red flowers (RR) are crossed with white flowers

(WW), pink (RW) flowers are produced.

Blue ____________ Red ____________

Purple ____________ Yellow ____________

Green ____________ White ____________

Black ____________ Grey ____________

d. Two blue tribbles mate and produce offspring that include white tribbles. What are the

genotypes of the parents?

______________________________________________________

e. If two orange tribbles mate, how many of the 852 offspring would you expect to be yellow?

Show your work.

______________________________________________________

f. A special investigation was conducted to determine the identity of the parents of an abandoned litter of tribbles. The litter included blue, purple, green, and orange tribbles. What are the phenotypes and genotypes of the parents?

______________________________________________________

5. Tribbles can be spotted (color against a white background) or solid color. The spotted allele is

dominant to the solid color allele.

a. When two spotted tribbles were mated, 45 spotted tribbles and 15 solid-colored tribbles were

produced. How many of the spotted tribble offspring would you expect to be

heterozygous? Show your work.

______________________________________________________

________________________________________________________

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b. Two red spotted tribbles were mated. Each tribble had a black solid colored parent. How many

of the 5,280 offspring would you expect to be black, solid-colored? Show your work.

______________________________________________________

______________________________________________________

6. Define multiple alleles: __________________________________________________________

___________________________________________________________

Example:___________________________________________________

7. Explain why a type O person can donate blood to all other blood types but can only receive type O

blood.

___________________________________________________________

___________________________________________________________

8. 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.

Blood Group of Mother Blood Group of Child Blood Group that would

Exonerate Man

AB A

O B

A AB

O O

B A

9. 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.

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10. 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?

11. 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?

12. 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.

________________________________________________________________________________

________________________________________________________________________________

13. Match the description/example with the correct pattern of inheritance.

A. Epistasis B. Pleiotropy C. Polygenic Inheritance

______ Single gene with multiple effects

______ Gene at 1 locus alters the phenotypic expression of a second gene

______ Several genes determine one phenotype

______ Sickle-celled anemia

______ Coat color in mice and rodents

______ Skin color in humans

______ Height in humans

14. 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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Part 8: Sex Linked Traits

Fruit Fly Linkage

**In fruit flies, eye color is a sex-linked trait. Red is dominant to white.**

1. What are the sexes and eye colors of flies with the following genotypes?

X R X r _________ X R Y _________ X r X r _______

X R X R _________ X r Y _________

2. What are the genotypes of these flies:

white eyed, male ______

red eyed female (heterozygous) ________

white eyed, female ________ red eyed, male _________

3. Show the cross of a white eyed female Xr Xr and a red-eyed male XR Y .

How many are:

white eyed, male ____ white eyed, female ____ red eyed, male____ red eyed, female ____

4. Show a cross between a pure red eyed female and a white eyed male.

What are the genotypes of the parents: _________ and ____________

How many are:

white eyed, male ____ white eyed, female ____ red eyed, male ____ red eyed, female ____

5. Show the cross of a red eyed female (heterozygous) and a red eyed male.

What are the genotypes of the parents? _________ & ____________

How many are:

white eyed, male ____ white eyed, female ____ red eyed, male ____ red eyed, female ____

Math: What if in the above cross, 100 males were produced and 200 females.

How many total red-eyed flies would there be? ________

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Human Sex Linkage

6. In humans, hemophilia is a sex-linked trait. Females can be normal, carriers, or have the disease. Males will

either have the disease or not (but they won’t ever be carriers)

X H X H = female, normal

X H X h = female, carrier

X h X h = female, hemophiliac

X H Y = male, normal

X h Y= male, hemophiliac

Show the cross of a man who has hemophilia with a woman who is a carrier.

How many children will have the disease? _____

7. A woman who is a carrier marries a normal man. Show the cross.

How many children will have the disease? _____

What is the sex of the child with the disease? _____

8. A woman who has hemophilia marries a normal man. How many of their children will have hemophilia, and

what is their sex?

How many children will have the disease? _____ What is the sex of the child with the disease? _____

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Calico Cat Genetics

9. In cats, the gene for calico (multicolored) cats is codominant. Females that receive a B and an R gene have

Black and oRange splotches on white coats. Males can only be black or orange, but never calico. Show the

cross of a female calico cat with a black male?

Female, Calico = X B X R Male, Black = X B Y

How many offspring will be:

Female and calico ____ Female and black ____

Male and black ____ Male and orange ____

Male and calico ____

10. Show the cross of a female black cat and a male orange cat.

What percentage of the kittens will be calico and female? _____

What color will all the male cats be? _____

lab #13 biology 10 bcc topic: mendelian genetics part1...

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