Christopher Long

Biology I

Mr. Snyder

October 20th, 2008

Activity OneMendel and the Laws of Chance;

Calculation of Genetic Ratios:

A great scientist once wrote… “all science is measurement”. In genetics, much of

that which is measured concerns the ratios of different phenotypes (outward

appearances) and genotypes (genetic makeup). These genetic relationships arise

from probability relationships—the chance–segregation and assortment of genes

in games, and their chance combination to form a zygote (a diploid cell resulting

from the union of two gametes).

In this activity, you explore how the determination of genetic ratios is derived

from two basic laws of chance.

In applying mathematics to the study of genetics, Mendel was stating that the

laws of chance apply to biology as well as they do to the physical sciences—a

radical concept for the times!

Laws of Chance:

Materials needed per group:

! Two double-sided plastic “coins”—marked “heads” and “tails”

! Student Study and Analysis Sheet (one per student)

Toss the two–sided coin. The chance that it will turn up “heads” is !. If two coins

are tossed, the chance that both will turn up “heads” is again !. The chance that

the second will also turn up “heads” is !. The chance that both will turn up

“heads” is ! x ! or ".

Summary:

The probability of two independent events occurring together is simply the

probability of one occurring alone multiplied by the probability of the other

occurring alone.

We can diagram this probability relationship in a checkerboard—a Punnet Square

(as below), which indicates that the combination in each square has an equal,

independent chance of occurring.

!

Note: The Punnet square was named after an english geneticist who first used this sort of diagram for the analysis of genetically determined traits.

1. What would the the probability (chance) be, if there were three plastic

“coins”?

! !

2. Place two plastic “coins” in a shaker cup. Shake and toss the pieces onto

the table top 100 times. Keep track of the results: record totals in the

spaces below. Do the results come close to those predicted by the Punnet

Square.

! “Heads”/“Heads”: 20

! “Tails”/”Tails”: 25

! “Tails”/“Heads” and “Heads”/“Tails”: 55

3. Suppose you toss both plastic coins 1,000 times instead of 100. What

would this larger sample allow?

! The odds would even out—the ratio of your own findings would draw

nearer to the expected ratio.

Calculation of Genetic Ratios:

On the basis of Mendelian principles, a diploid (double set of chromosomes) adult

of genetic constitution Aa may give rise to two games, A, and a. If the genetic

constitution of the parents is not given, two possible explanations (hypotheses)

exist which can account for the presence of phenotypes. A or a in the offspring.

The answer, in terms of probability is to assume that the parents Aa produce two

types of gametes, A and a, equally well and the aa parent produces only one type

of gamete, a. Any combination of gametes depends upon the frequency or

probability of each type of gamete furnished by the parents. Thus the formation

of a zygote is the result of two independent events (two gametes), each with

their own probabilities, which now occur together.

Summary:

The probability that a particular zygote will be formed is equal to the

product of the probabilities of the gametes that compose it.

4. Complete the two checker boards below to calculate the values:

!

5. Complete the following statement for the Aa x Aa cross:

! How many possible kinds of zygotes can be formed?

4

! The probability that the A phenotype will occur?

#

! The probability that a zygote can be heterozygous (i.e. Aa or aA)

!

! The probability that a zygote can be either AA, Aa, or aA?

#

! The probability that a zygote can be either AA, Aa, or aA?

#

! The probability that a zygote can be either AA, Aa, or aA?

1

! What are the genotype and phenotype ratios?

Genotypic Ratio: 1AA:1Aa:1aA:1aa

Phenotypic Ratio: 3A:1a

6. Complete the following statement for the Aa x aa cross:

! Do both genotypes occur with equal frequency?

Yes

! The probability that the A phenotype will occur?

#

! What are the genotype and phenotype ratios?

Genotypic Ratio: 1Aa:1aa

Phenotypic Ratio: 1A:1a

Christopher Long

Biology I

Mr. Snyder

October 20th, 2008

Activity TwoA Monohybrid Cross

This activity investigates crosses between pea plants that are different in but a

single characteristic (gene difference)—a monohybrid cross.

Materials needed per group:

! Cup shaker

! Four discs representing gamete cells.

! Two Red Allele Discs each having W for the purple flow on each side.

! Two Red Allele Discs each having w for the white flow on each side.

! Student Study and Analysis Sheet

! One wax pen

Note: A pea plant homozygous for purple flower color is represented by WW in genetic shorthand. The gene for purple flow coloring is designate W because of a convention by which geneticists, in

indicating a pair of alleles, use the first letter of the less common form (white). The capital indicated the dominant,, the lowercase the recessive. Use the wax pen provided to write the allele

type (W or w) on Each side of the disc.

Read and become familiar with the information presented in the Student Study

Sheet.

1. Place two W allele disc in the cup, representing the union (fertilization) of

male and female gametes. Shake and toss them onto the table. Repeat nine

more times. Record your results (genotypes, phenotype) below:

! Genotype: 1WW

! Phenotype: 1w

2. Repeat Step 1 (above), but this place two w allele discs in the cup. Shake

and toss. Repeat nine additional times. Record the genotype and phenotype

below:

! Genotype: 1WW

! Phenotype: 1w

3. Cross two purple (W) plants created from zygote unions in Step 7 by

allowing them to self–pollinate as Mendel did. Draw a Punnet square and

record the genotypes and phenotypes.

!

4. Write a statement that explains why these plants would continue to “breed

true”

! The plant continue to “breed true” because each has only one distinct

factor for each characteristic.

5. If white plants (w) were substituted for purple (W), would there be any

change in the expected outcome?

! If white plants were substituted for purple plants, there would be no

change in the expected outcome, besides a different phenotype.

Christopher Long

Biology I

Mr. Snyder

October 20th, 2008

Activity ThreePrinciple of Segregation

In this activity you will study single gene differences in various monohybrid

crosses that led Mendel to the discovery of the Principle of Segregation—the

ability to predict the segregation between two different alleles in a single gene

pair and their behavior in each generation.

Materials needed per group:

! Two yellow–colored plastic discs—representing “purebred” (homozygous)

yellow pod color trait

! One green–colored plastic disc—representing “purebred” (homozygous)

green pod color trait

! One red–colored plastic disc—representing “purebred” (homozygous) purple

flower trait

! One blue colored plastic disc

! Clear sticky tape (not provided)

! Student Study and Analysis Sheet

Mendel conducted experiments on seven traits (see table 1) whose characteristics

(itself and its alternate) were clearly defined. This activity will simulate crosses

among two of these: flower and pod color.

1. Simulate a cross between a plant homozygous (GG) for the green pod color

trait (green–colored plastic disk) by crossing it with another homozygous

(gg) for yellow pods (yellow–colored plastic disc).

2. Conduct another simulated cross, this time using red–colored discs to

represent a plant hozygous (WW) purple–colored flowers with yellow–

colored discs to represent a plant homozygous (ww) white–colored flowers.

Secure plastic discs with sticky tape.

3. What generation do both these “disc crosses” represent?

! They represent the F1 generation

4. Hold each disc (representing a cross result) set up to the light and record

the “dominant” observed color below.

! Cross GG x gg: Green

! Cross WW x ww: Red

5. What phenotypic trait is hidden in each cross?

! Cross GG x gg: Yellow pod color

! Cross WW x ww: White flower color

6. Describe the relationship of “dominant” and “hidden” traits

! Dominant traits are dominant, and conceal hidden traits, which are

recessive

7. Complete the Punnet Squares below, writing in both genotype and

phenotype for each of the above crosses.

!

8. Write a statement that describes each cross.

! The offspring resulting from each cross will be be heterozygous

9. Use the information from Step 6 to allow each plant variety (pod and

flower color) to self pollinate. Can you (like Mendel) predict the results?

Complete another Punnet square for each cross—record phenotype and

genotype.

!

10.In what ratio do the dominant and recessive traits appear?

! 3:1

Is there a similar relationship concerning all of the seven traits studied by

Mendel?

! Yes

11. Use the results of your observations together with Mendel's actual

laboratory to explain how recessives disappear so completely and then

reappear again, always in constant proportions.

! In the F1 generation, heterozygous offspring are always created

because both pairs of alleles for each trait are homozygous. Thus, the

dominant phenotype shows. In the F2 generation, two pairs of

heterozygous alleles are combined, which allows " chance for

recessive traits to re-emerge

12. Write a statement describing how Mendel probed the principle of

segregation.

! Mendel saw the appearance and disappearance of traits and their

constant proportions, and that this could be explained if hereditary

characteristics were determined by discrete factors.

Christopher Long

Biology I

Mr. Snyder

October 22nd, 2008

Activity FourUnderstanding Dominance

This activity investigates crosses between pea plants that are different in two trait

characteristics, (a dihybrid cross) with each gene pair having one dominant and

one recessive allele—and how each gene pair acts independently of the other.

Phenotypes in the resultant generations will be, on average, in a ratio of 9:3:3:1

—testifying to the independence, or independent assortment of these two gene

pairs.

Materials needed per group:

! Two four-sided dice—each numbered side representing two gene pairs

! 1 = (RY) round yellow

! 2 = (Ry) round green

! 3 = (rY) wrinkled yellow

! 4 = (ry) wrinkled green

! Student Study and Analysis Sheet

1. Use table one to determine which traits (seed form and seed color) are

dominant, and which are recessive:

! Dominant: round, yellow

! Recessive: wrinkled, green

2. Predict the genotype and phenotype of the F1 generation that results from

a cross of a plant homozygous for round (RR) and yellow (YY) is crossed

with plant having wrinkled (rr) and green–colored (yy) peas.

! Genotype: RrYy

! Phenotype: RY

3. Complete this Punnet Square for self fertilized F1 plant from the above

cross to predict future phenotypes:

!

! Phenotypes: round–yellow, round–green, wrinkled–yellow,

wrinkled–green

4. Use four–sided dice for Steps 27 through 28:

! Two four-sided dice—each numbered side representing two gene pairs

! 1 = (RY) round yellow

! 2 = (Ry) round green

! 3 = (rY) wrinkled yellow

! 4 = (ry) wrinkled green

Each die is read by matching the number visible on the point, correlated to the

gamete designation above. Thus when two die are cast (simulating fertilization)

the genotype of the organism is established. For example: a cast die indicate “4”

and “1”. The corresponding genotype would be RrYy–round yellow.

5. Cast two dice 80 times (or as many times as your teacher directs) to arrive

at organism genotypes. Tally likely phenotype combinations.

! round yellow: 45

! round green: 19

! wrinkled yellow: 11

! wrinkled green: 6

6. What is the phenotypic ratio?

! 45RY:19Ry:11rY:6:ry

7. Add your data to that of the rest of the class; are the phenotypic ratios

altered much? How do they compare with your Punnet square prediction?

! The phenotypic ratios are not altered much. They compare closely

with my Punnet square prediction.

8. Write a statement that summarizes Mendel's principle of independent

assortment based upon your classroom data and Punnet square prediction:

! The contrasting alleles which control a factor separate during the

formation of gametes, and each is joined to another allele during the

creation of a zygote.

Christopher Long

Biology I

Mr. Snyder

October 20th, 2008

Genetics Lab:DNA Molecule and Replication

RNA TranscriptionRNA Translation

Materials Needed per Group:

! 18 Black, deoxyribose molecule pieces

! 9 Blue, ribose molecule pieces

! 25 White, flexible phosphate molecule tubes

! 10 Red, guanine base tubes

! 4 Blue, thymine base tubes

! 8 Green, adenine base tubes

! 10 Black, cytosine base tubes

! 4 White, uracil base tubes

! 18 Solid, hydrogen bond connectors

! 3 Blue tRNA models

! 3 Black, amino acid models

! 2 White, rigid amino acid bonding tubes

! 1 Blue, ribosome model

! Chart (below)

!

Nucleotide Color

Adenine Green

Cytosine Black

Guanine Red

Thymine Blue

Uracil White

Procedure

Student should already have a basic understanding or organic molecules such as

nucleic acids and proteins. Knowledge of cell structures an their functions may

also save explanation time during the laboratory procedure.

Part I: Making a DNA Chain:

A. Laying down the tracks

1. Obtain eight white phosphate flexible model tubes and nine black

deoxyribose sugar pieces. Connect these in a straight chain, so that the

third, open, bonding site on each sugar is facing the same direction. Obtain

eight more phosphate tubes with nine sugar pieces, and form a second

chain with each extra bonding site facing the same direction. Set these

chains parallel on the table in front of you so that the extra bonding sites

are facing each other, forming a structure that resembles tracks of a

railroad. (Figure 1).

A. Partially Completing the DNA Molecule

1. The open bonding site on each nucleotide's sugar is filled by a nitrogen

base. DNA has four possible nitrogen bases, and they have been color

coded for this investigation. Starting at the top of one of your phosphate

sugar tracks, bond the following nitrogen bases, in order, to the strand:

cytosine, thymine, adenine, cytosine, guanine, guanine, adenine, thymine,

guanine. This strand is marked with an asterisk in figure 4 because it will be

the single strand transcribed by the RNA in Part E.

2. The second chain of phosphate-sugar molecules also contains nitrogen

bases, and they are aligned in a specific arrangement dependent on the

first strand's patter. The nitrogen bases guanine and cytosine must always

line up across from each other. Likewise, adenine and thymine will always

match up.

3. Since the first nitrogen base on the top chain was cytosine the first

nitrogen base on the bottom strand must be guanine. Place a guanine on

the first deoxyribose of the bottom chain (Figure 4). Complete the bottom

strand, placing the appropriate nitrogen base on each sugar that will

correctly complement its opposing base.

4. Nitrogen base pairs join via hydrogen bonding to create the “railroad ties”

that run between the phosphate-sugar “tracks”. Obtain nine hydrogen

bonding plugs, and connect the two strands at their complementary

nitrogen bases. This model is a portion of a DNA molecule.

Part II: Messenger RNA and Transcription:

A. Unzipping the DNA

1. In order for the information stored in the DNA to be used, the strand must

temporarily separate at the hydrogen bonds that connect the

complementary nitrogen bases. This model is a portion of a DNA molecule.

B. Preparing RNA Nucleotides

1. The DNA of a cell remains in the nucleus at all times, yet protein synthesis

occurs in the cytoplasm. The information is transferred from the nucleus to

the cytoplasm by a second type of nucleic acid, RNA. The sugar within RNA

is called ribose and it contains one more hydroxyl group (-OH) than DNA's

sugar, deoxyribose. Obtain nine blue ribose pieces from your kit, and

connect a white flexible phosphate tube to each. Do not link the sugar–

phosphate models in a chain, but leave them as individual sections.

2. Complete the nine RNA nucleotides by adding a nitrogen base to each.

Remember from the introduction that thymine is not a possible nitrogen

base in RNA, but that uracil is used instead to compliment adenine. To

successfully continue with the procedure, you will ned to add two adenines,

two guanines, two uracils, and three cytosines to your sugar phosphate

combinations.

B. Transcribing the mRNA

1. Using the “unzipped”, single strand of DNA, marked with asterisk in figure

2, and the hydrogen bond connectors, pair up the RNA nucleotides with

their complementary DNA nitrogen bases. Recall that cytosine will pair with

guanine, while adenine will pair with thymine of uracil.

2. Connect each nucleotide's phosphate group to the neighboring nucleotide's

sugar.

3. Detach the RNA strand from the DNA strand at the hydrogen bond

locations, allowing RNA to keep the bonds. This strand of ribonucleic acid is

called messenger RNA (mRNA) because it carries the information stored on

a DNA in the nucleus to a ribosome located in the cytoplasm. The mRNA is

like a photographic negative of the original DNA single strand. Compare

your mRNA to the complementary single strand of DNA, from Part B. Aside

from uracil replaing theymine, and ribose replacing deoxyribose, the strands

are identical

4. Using the remaining hydrogen bond connectors, reconnect two single DNA

strands, and save your double stranded DNA model for further observations

later in this investigation.

Part III: Transfer RNA and Translation:

A. Ribosomes and Codons

1. After seperation from the DNA, the mRNA exists the nucleus and parteners

with a ribosome in the cytoplasm of a cell. Place your mRNA model on the

plastice ribosome plate from your kit.

2. A ribosome recognizes three nucleotides at a time on the mRNA. The group

of three nucleotides is called a codon, and codons spell out a message that

will translate to a specifiec amino acid in the protein synthesis sequence.

Your current chain of nine RNA nucleotides is actually considered three

mRNA codons. Compare your codons to the illustration in figure 4.

B. Transfer RNA and Anticodons

1. A second type of RNA awaits the mRNA in the cytoplasm of the cell.

Transfer RNA (tRNA) is a relatively small molecule that bonds to an amino

acid on one end and a mRNA codon on the other (Figure 7). Obtain your

three tRNA models and notice that each has three nitrogen base bonding

sites. These three sites make up an anticodon, which complements a codon

on the messenger RNA.

2. Build your

anticodons b

placing the

following groups

of three nitrogen bases on the tRNA models: cytosine, uracil, and adenine

(CUA); cytosine, guanine, and guanine (CGG) and; and adenine, uracil, and

guanine (AUG).

H. Translation

1. Attach the anticodons to their complementary codons using the hyrdogen

bond connectors. Again, guanine and cytosine must partener, while adenine

and uracil partener. The tRNAs are now ready to receive amino acids. Bond

each amino acid to its respective tRNA. Notice that the amino acid sequence

has ben determined by how the anticodons were arganized.

1. Complete the synthesis of this portions of a protein by bonding the

amino acid models together using the white bonding tubes. (See Figure

6 for comparison).

Each type of amino acid is carried to the ribosome by a particular form of

tRNA, carrying an anticodon that forms a temporary bond with one of the

codons in the mRNA. As shown above, the ribosom moves along the mRNA

chain and “Read off” the codons in sequence.

Christopher Long

Biology I

Mr. Snyder

November 13th, 2008

Assessment

1. Define the Following terms:

! Anticodon: a sequence of three nucleotides located on tRNA that

complements a corresponding mRNA codon

! Codon: a three-base sequence located on mRNA that determines

which amino acids will be created during protein synthesis.

! Nucleotide: the basic building block of DNA and RNA, which is

composed of a sugar, a phosphate group, and a nitrogen base, which

is always adenine guanine cytosine thymine or uracil.

! Ribosome: an organelle found within the cytosol of a cell, which binds

to the mRNA and facilitates the making of proteins.

! Transcription: the process by which mRNA is constructed from a DNA

template.

! Translation: the process in which mRNA attaches to a ribosome for

protein synthesis.

2. Draw a diagram of your DNA molecule. Keep the illustration in a straight

“ladder” form as seen in Figure 2; do not attempt to draw the double helix

shape. Specifically identify each nitrogen base.

!

3. A partial strand of DNA has the nitrogen base pattern shown below. Indicate

what nitrogen bases would be needed for a mRNA to complement this

strand.

!

4. What would be the nitrogen base pattern for the anticodons (tRNAs) that

would bond to the mRNA strand in Question 3.

!

5. How can protein be synthesized in the cytoplasm of a cell when DNA is

contained in the nucleus?

! The genetic information contained in DNA is transcribed into mRNA,

which goes into the cytosol for translation.

6. Complete the following Chart

!

7. Using a venn diagram, compare and contrast codons to anticodons.

Similarities and differences can relate to structure, components, and/or

function.

!

8. Read the following statement and write if each one is true or false. If you

believe the statement is false, explain the reason why below it.

! “Uracil is always paired with adenine in double-stranded DNA.”

! False. Thymine is always paired with adenine in double

strand DNA.

! “The process in which a ribosome controls protein synthesis is called

transcription.”

! False. This process is named translation.

! “Adenine and guanine are purines; cytosine and thymine are

pyrimidines.”

! True

! “Protein Synthesis occurs within the nucleus of a cell.”

! False. Protein synthesis occurs in the cytosol of a cell.

! “Nucleotides contain a sugar, an organic nitrogen base, and a

phosphate group.”

! True

9. The DNA model you constructed was nine base pairs long and yielded three

amino acid protein. Actual human DNA is billions of base pairs in length and

codes for the production of millions of proteins. Some sequences are so

specific that a change in a single base (termed an SNP, or single nucleotide

polymorphism) can result in an entirely new protein structure. Research an

example of an SNP and explain the effects it may have.

! One example of an SNP occurs in the blood disease Sickle Cell

Anemia. In this hereditary condition, the protein Hemoglobin is

mangled as the result of an SNP which occurs when thymine is

substituted for adenine in the three nucleotide sequence which

produces the amino acid glutamine. As a result, the amino acid valine

replaces the amino acid glutamine — which deforms the Hemoglobin

protein so badly that the entire red blood cell is affected, twisting,

clotting, and eventually crystalizing. The combined affects of this

disease result in a drastic shortening of the subject's life.

Christopher Long

Biology I

Mr. Snyder

November 25th, 2008

Classification Lab

What you need per group:

! Eight critter cards

! Key to the Kingdoms of Life

! key to each of the six kingdoms (Archaebacteria, Eubacteria, Protista,

Fungi, Plantae, Animalia)

! Critter Characteristic Sheets (one for each kingdom)

A Closer Look at Critter Cards:

Take a moment to familiarize yourself with the information contained on each

critter card.

!

1. Review each of your group's “critters” and determine to which of the six

kingdoms they belong. Use the information provided on the Critter Card,

the information in the key booklet and the Key to the Kingdoms of life to

make this decision. Record the “organism number” and scientific name(s) in

the left–hand column on the appropriate sheet for each critter.

Share your findings with your teacher before proceeding.

2. Become familiar with using a dichotomous or two–answer key.

! Using a Dichotomous Key:

A “dichotomous” or “two answer” key allow you to compare two

choices for each of an organism's characteristics. You should be able

to read the pair of characteristics, and after examining you organism,

you'll answer “no” to one pair and “yes” to the other. Then, using the

line for “yes” characteristics, follow the instructions on where to go

next in the key. If the characteristic you said “yes” to ends at the

name instead of the directions to the next step, you've finished keying

out the organism! The name at the end of the line is the group to

which your organism belongs.;

To get some practice using the dichotomous key, lets identify this unique

organism:

!

Begin at 1a:

1a Organism shaped like a cube or a sphere

If yes………….………....…..…………Go to 1a

If no………….………....…..………….Go to 1b

1b Organism shaped like a cylinder

If yes, the organism is………….Cylinderous cellous

If no………….………....…..…………Go to 2A

2A Organism is a cube

If yes, the organism is…………Cubous cellous

If no………….………....…..…………Go to 2B

2B organism is a square………….……Squarous cellous

Based on the characteristics of its cubes shape, this “organisms would be

identified as a “cubed cell” or “Cubous cellous.”

A word of caution when using a dichotomous key –if you get to a point in

the key where you think the answer to both statements in a pair is “no”—

you probably made a mistake at an earlier step. Go back and give it another

try. You should always be able to say “yes” to one of the two statement in

each pair.

3. Identify the phylum to which a particular critter belongs by following the

key until it ends in a phylum name:

4. Write down the group characteristic(s) that identify the phylum on the

Critter Characteristic Sheet that matches the kingdom to which the critter

belongs.

For example:

!

This Critter belongs to the Kingdom Animalia (Eumetazoa) and the phylum

Arthropoda. It has these groups of characteristics:

!

5. In the animal key, subphylum groups are also identified. If when keying out

your critter you encounter a subphylum group, draw a line below the

phylum characteristics and write down the subphylum characteristics below

it.

For example :

!

6. In the Kingdom Eubacteria, Kingdom Plantae, and Kingdom Animalia

(Eumetazoa) key class group are also identified in tables. Search the table

for the matching class group under a phylum or subphylum group name.

For example :

!

7. Your teacher will check your

group's summaries of the phylum

andor class characteristics of each

assigned critter.

1. How many cell types are present in your group's set of critters?

! Two cell types; Prokaryotes and Eukaryotes.

2. How many types of body organization, or body plan, are represented in yor

group's set of critters?

! Two; unicellular and multicellular.

3. Which kingdoms represented in your set of critters produce young through

sexual reproduction?

! Animalia, Plantae, Protista, and Fungi.

4. In which kingdoms do critters represented in your group's set reproduce

from only one cell or parent (i.e asexual reproduction)?

! Eubacteria and Archaebacteria.

5. Give an example of asexual reproduction in a common food plant.

! The bud of a potato is an example of sexual reproduction in a

common food plant.

6. Which life characteristics are present only in Eubacteria and Archaebacteria?

! Cells without nuclei

7. What are different ways in which two organism can get energy?

! Photosynthesis, chemicals, ingestion, and absorption