Solved by a verified expert:Experiment 1: Punnett Square CrossesMaterials-Red Beads-Blue Beads-Green Beads-Yellow Beads-2 100mL BeakersProcedure1. Set up and complete Punnett squares for each of the following corn crosses(remember Y = yellow and y = blue). Please use the following example of a Bb and Bbcross as a guide for filling in your Punnett squares:Bb/BbBbBBBBbbBbbbCross #1:YY/YyYyYYYYyYYYYyCross #2:YY/yyyyYYyYyYYyYya. What are the resulting phenotypes?Cross #1: YellowCross #2: Yellowb. Are there any blue kernels? How can you tell?No, because yellow is the dominant Allele.2. Set up and complete a Punnett square for the F1 generation from Cross #2 (above):YY/yyyyYYyYyYYyYya. What are the genotypes of the F2 generation?Click here to enter text.b. What are their phenotypes?Yellowc. Are there more or less blue kernels than in the F1 generation?Click here to enter text.3. Identify the four possible gametes produced by the following individuals:IndividualGamete#1Gamete#2Gamete#3Gamete#4YY SsYy Ssa. List the genotypes of the F1 generation that result from a cross of these twoindividuals.Click here to enter text.b. What are the phenotypes of the F1 generation? What is the ratio of thosephenotypes?Click here to enter text.4. You have been provided with 4 bags of different colored beads. Pour 50 of the blue beadsand 50 of the yellow beads into beaker #1 and mix them around. Pour 50 of the red beads and50 of the green beads into beaker #2 and mix them around. DO NOT POUR THE BEAKERSTOGETHER. These colors correspond to the following traits:MONOHYBRID CROSS: Randomly (without looking) take 2 beads out of Beaker #1.-This is the genotype of parent #1. Record the genotype and place the beads tothe side. Do NOT put those beads back into the beaker. For example, if you pulla yellow and a blue bead out of Beaker #1, the resulting genotype will be Yy.-Repeat this process for parent #2. These two genotypes are your parents forthe next generation. Set up a Punnett square and determine the genotypes andphenotypes of the offspring.-Repeat this process 4 times (for a total of 5 crosses). Return the beads back toBeaker #1 when finished.Trial#12Parent #1GenotypeClick hereto entertext.Click hereto enterParent #2GenotypeClick hereto entertext.Click hereto enterOffspringGenotypesOffspringPhenotypesa. How much genotypic variation do you find in the randomly picked parents of yourcrosses?Homozygous Dominant: Click here to enter text.Heterozygous: Click here to enter text.Homozygous Recessive: Click here to enter text.b. How much genotypic variation do you find in the offspring?Homozygous Dominant: Click here to enter text.Heterozygous: Click here to enter text.Homozygous Recessive: Click here to enter text.c. How much phenotypic variation do you find in the randomly picked parents of yourcrosses?Yellow: Click here to enter text.Blue: Click here to enter text.d. How much phenotypic variation do you find in the offspring?Yellow: Click here to enter text.Blue: Click here to enter text.e. The predicted phenotypic ratio for a heterozygous cross is 3:1 yellow:blue. Would youexpect the phenotypic ratio for your offspring to be similar? Why or why not?Click here to enter text.f. What is the difference between genes and alleles?Click here to enter text.g. If a mutation occurs, how might the resulting protein be effected?Click here to enter text.h. Organisms heterozygous for a recessive trait are often called carriers of that trait.What does that mean?Click here to enter text.i. In peas, green pods (G) are dominant over yellow pods (g). If a homozygousdominant plant is crossed with a homozygous recessive plant, what will be thephenotype of the F1 generation? If two plants from the F1 generation are crossed, whatwill the phenotype of their offspring be?Click here to enter text.DIHYBRID CROSS: Randomly (without looking) take 2 beads out of Beaker #1 AND 2beads out of Beaker #2.-This is the genotype of parent #1. Record the genotype and place the beads tothe side. Do NOT put those beads back into the beakers. For example, if youpull a yellow and a blue bead out of Beaker #1 and a red and a green bead out ofBeaker #2, the resulting genotype will be YySs.-Repeat this process to obtain the genotype of parent #2.a. What are the parental phenotypes? What are the genotypes of the gametes they canproduce?Parent #12GenotypePhenotypeGamete #1Gamete #2Gamete #3Gamete #4b. Set up a Punnett square for this cross. What are the resulting genotypes andphenotypes?Click here to enter text.c. Repeat the bead picking process 4 more times to complete 4 additional Punnettsquares. How similar are the observed phenotypes in each replicate?Click here to enter text.d. The predicted phenotypic ratio for a dihybrid cross of heterozygous parents is 9:3:3:1.How similar is the combined phenotypic ratio from your five trials to this ratio? Why doyou think this is?Click here to enter text.e. Did the results of your monohybrid or dihybrid cross more closely match the predictedphenotypic ratio for heterozygous parents?Click here to enter text.f. Based on these results, what would you expect if you were looking at a cross of 5, 10,and 20 independently sorted genes?Click here to enter text.g. Why do you think it is so expensive to produce a hybrid plant seed?Click here to enter text.h. In certain bacteria, an oval shape (S) is dominant over a round shape (s), and thickcell walls (T) are dominant over thin cell walls (t). Show a cross between aheterozygous oval, thick cell walled bacteria with a round, thin cell walled bacteria. Whatare the phenotypes and phenotypic ratio of the offspring?Click here to enter text.5. The law of independent assortment allows for genetic recombination. The following equationcan be used to determine the total number of possible genotype combinations for any particularnumber of genes:2g = Number of possible genotype combinations (where g is the number of genes)Consider the following genotype: YySsTta. How many different gamete combinations can be produced?Click here to enter text.b. Many traits (phenotypes), like eye color, are controlled by multiple genes. If eye colorwere controlled by the following number of genes, how many possible genotype combinationswould there be?NumberofGenes51020Number of PossibleGenotypes
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