Christopher Long Biology I Mr. Snyder October 20th, 2008
Activity One
Mendel 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”: “Tails”/”Tails”: “Tails”/“Heads” and “Heads”/“Tails”:
20 25 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
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The probability that the A phenotype will occur? #
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The probability that a zygote can be heterozygous (i.e. Aa or aA) !
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The probability that a zygote can be either AA, Aa, or aA? #
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The probability that a zygote can be either AA, Aa, or aA? #
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The probability that a zygote can be either AA, Aa, or aA? 1
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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
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The probability that the A phenotype will occur? #
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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 Two
A 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 Three
Principle 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 Four
Understanding 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: !
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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
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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
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Phenotype: RY
3. Complete this Punnet Square for self fertilized F1 plant from the above cross to predict future phenotypes: !
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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: round green: wrinkled yellow: wrinkled green:
45 19 11 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 Transcription RNA 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
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Codon: a three-base sequence located on mRNA that determines which amino acids will be created during protein synthesis.
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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.
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Ribosome: an organelle found within the cytosol of a cell, which binds to the mRNA and facilitates the making of proteins.
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Transcription: the process by which mRNA is constructed from a DNA template.
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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.” !
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“The process in which a ribosome controls protein synthesis is called transcription.” !
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False. Thymine is always paired with adenine in double strand DNA.
False. This process is named translation.
“Adenine and guanine are purines; cytosine and thymine are pyrimidines.” !
True
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“Protein Synthesis occurs within the nucleus of a cell.” !
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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