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bio-Module 6 questions




Question;Experiment 1: Punnett square crossesMaterialsRed beadsBlue beadsGreen beadsYellow beads2 100mL BeakersProcedure1. Set up and complete Punnett squares for each of the following crosses:(remember Y = yellow, and y = blue)Y Y and Y yY Y and y y?????YYYYYYYyYyYy?????YYyYyYyyYyYya) What are the resulting phenotypes?? 2010 eScience Labs, Inc.All Rights ReservedThe punnet square shows all of the possible combinations that would resultfrom the mating of the given genotypes. In this example the dominant allele isrepresented by Y, and determines the color of the corn to be yellow. Theresults from the square is: 100 percent of the offspring will have a yellowphenotype, because every combination expresses the dominant allele Y.b) Are there any blue kernels? How can you tell?For a recessive allele to be expressed there must be two recessive allelespresent.In this particular example, there is no combination that leads to ahomozygous recessive genotype or yy. Meaning there is no blue cornexpressed.2. Set up and complete a Punnett squares for a cross of two of the F1 from 1babove:a) What are the genotypes of the F2 generation?The results of the completed punnet square shows the probability of producinga homozygous recessive offspring is 25%, a 50% chance of producing aheterozygous offspring and a 25% chance the offspring will be homozygousdominant.b) What are their phenotypes?The results show a 75% chance the offspring will be yellow and 25% chancethe offspring will be blue.? 2010 eScience Labs, Inc.All Rights Reservedc) Are there more or less blue kernels than in the F1 generation?In F2 generation there is a 25% probability that a blue phenotype will beexpressed. However, all the genotypes in the F1 generation contain adominant allele supressing the recessive allele leaving a 0% chance of havinga blue phenotype. In other words, F1 has less blue kernels than the F2generation.3. Identify the four possible gametes produced by the following individuals:a) YY Ss:YSYSb) Yy Ss:YSySYsYsYsysc) Create a Punnett square using these gametes as P and determine thegenotypes of the F1:There are 6 distinct genotypes expressed in F1. YYSS: expressed two timesYYSs: expressed four times YYss: expressed two times YySS: expressed twotimes YysS: expressed four times Yyss: expressed two times.d) What are the phenotypes? What is the ratio of those phenotypes?When determining the phenotype of a gene the dominant allele is alwaysexpressed as the phenotype. For Example in this punnet Square Ss=SS andYY=Yy, so any time Y is expressed the phenotype is yellow and when S isexpressed the phenotype is smooth. There are two distinct phenotypes: 1.yellow and smooth 2. yellow and wrinkled. Therefore, 75 percent of theoffspring will be yellow and smooth, and the other 25 percent will be yellowand wrinkled. The ratio for the phenotype is 12:4. 12 combinations of Yellowand smooth and 4 combinations of yellow and wrinkled.4. You have been provided with 4 bags of different colored beads. Pour 50 of theblue beads and yellow beads into beaker #1 and mix them around. Pour 50 ofthe red beads and green beads in beaker #2 and mix them.Attention! Do not pour the beakers together.? 2010 eScience Labs, Inc.All Rights Reserved?#1 contains beads that are either yellow or blue.?#2 contains beads that are either green or red.?Both contain approximately the same number of each colored bead.?These colors correspond to the following traits (remember that Y/y isfor kernel color and S/s is for smooth/wrinkled):?Yellow (Y) vs. Blue (y)Green (S) vs. Red (s).A. Monohybrid Cross: Randomly (without looking) take 2 beads out of #1.?This is the genotype of individual #1, record this information. Do notput those beads back into the beaker.?Repeat this for individual #2. These two genotypes are your parentsfor the next generation. Set up a Punnett square and determine thegenotypes and phenotypes for this cross.?Repeat this process 4 times (5 total). Put the beads back in theirrespective beakers when finished.a) How much genotypic variation do you find in the randomly pickedparents of your crosses?When I randomly picked the parents from the beaker I found six distinctgenotypes. Within all of the randomly selected parents, I chose aheterozygous genotype five out of the ten times, homozygous dominantgenotype two out of the ten times, and a homozygous recessive genotypethree out of the ten times. The genotypes that were chosen: Yy, Ss, yy, SS,ss, and YY.b) How much in the offspring?There was five homozygous dominant genotypes, seven homozygousrecessive genotypes, and eight heterozygous genotypes chosen out of thetwenty total offspring.? 2010 eScience Labs, Inc.All Rights Reserved? 2010 eScience Labs, Inc.All Rights Reservedc) How much phenotypic variation?There were four different phenotypes for this exercise: 1. Yellow and Smooth2. Yellow and Wrinkled 3. Blue and Smooth 4. Blue and Wrinkled. Five wereyellow and smooth phenotype, three were yellow and wrinkled phenotype, fivewere blue and smooth phenotype, seven were blue and wrinkled phenotype ofthe twenty total offspring.d) Is the ratio of observed phenotypes the same as the ratio of predictedphenotypes? Why or why not??????e) Pool all of the offspring from your 5 replicates. How much phenotypicvariation do you find??????f)What is the difference between genes and alleles?A gene is a section of DNA that determines a specific trait such as hair color orblood type. Genes are functional units of heredity. Alleles are variations of thesame gene. For Example: One single gene determines a persons height andwhether or not you are tall or short would be determined through thecombinations of alleles within the height gene. Genes determine the genetictraits a person will have and the different sequences of alleles within the genewill determine a single characterisitc of the individual.g) How might protein synthesis execute differently if there a mutation? 2010 eScience Labs, Inc.All Rights Reservedoccurs?A gene mutation is an alteration in the sequence of nucleotides in DNA. DNAconsists of nucleotides. During protein synthesis, DNA is transcribed into RNAand then translated to produce proteins. Altering nucleotide sequences mostlikely will result in nonfunctioning proteins. For example: if a nonsensemutation occurs it will alter the nucleotide sequence coding a stopcodon inplace of a amino acid, signaling the end of the translation process, stoppingprotein production.h) Organisms heterozygous for a recessive trait are often called carriersof that trait. What does that mean?Carriers contain a recessive allele and are capable of passing on therecessive allele to their offspring, even though they do not express therecessive allele, because of the presence of a dominant allele. A person thatinherits a genetic trait but does not display or show symptoms of that trait, butare able to pass the gene on to their offspring.i)In peas, green pods (G) are dominant over yellow pods. If ahomozygous dominant plant is crossed with a homozygous recessiveplant, what will be the phenotype of the F1 generation? If two plantsfrom the F1 generation are crossed, what will the phenotype of theiroffspring be?If a homozygous dominant plant is crossed with a homozygous recessiveplant, 100% of the offspring will be green. If two plant from the F1 generationare crossed,since they are all heterozygous plants, 3/4 (75%) of the plant'soffspring will be green and 1/4 (25%) of the plant's offspring will be yellow.? 2010 eScience Labs, Inc.All Rights ReservedB. Dihybrid Cross: Randomly (without looking) take 2 beads out of beaker #1 AND 2beads out of beaker #2.??These four beads represent the genotype of individual #1, record thisinformation.Repeat this process to obtain the genotype of individual #2.a) What are their phenotypes??????b) What is the genotype of the gametes they can produce???????Set up a Punnett square and determine the genotypes andphenotypes for this cross.c) What is your predicted ratio of genotypes? Hint: think back to ourexample dihybrid cross??????Repeat this process 4 times (for a total of 5 trials).? 2010 eScience Labs, Inc.All Rights Reservedd) How similar are the observed phenotypes in each replicate??????e) How similar are they if you pool your data from each of the 5replicates??????f)Is it closer or further from your prediction??????? 2010 eScience Labs, Inc.All Rights Reservedg) Did the results from the monohybrid or dihybrid cross most closelymatch your predicted ratio of phenotypes??????h) Based on these results, what would you expect if you were lookingat a cross of 5, 10, 20 independently sorted genes??????i)Why is it so expensive to produce a hybrid plant seed??????? 2010 eScience Labs, Inc.All Rights Reservedj)In certain bacteria, an oval shape (S) is dominant over round andthick cell walls (T) are dominant over thin. Show a cross betweena heterozygous oval, thick cell walled bacteria with a round, thincell walled bacteria. What are the phenotype of the F1 and F2offspring??????5. The law of independent assortment allows for genetic recombination. Thefollowing equation can be used to determine the total number of possiblegenotype combinations for any particular number of genes:2g= Number of possible genotype combinations (where ?g? is the number ofgenes)1 gene:21= 2 genotypes2 genes: 22= 4 genotypes3 genes: 23 = 8 genotypesConsider the following genotype:Yy Ss TtWe have now added the gene for height: Tall (T) or Short (t).a) How many different gamete combinations can be produced??????? 2010 eScience Labs, Inc.All Rights Reservedb) Many traits (phenotypes), like eye color, are controlled by multiplegenes. If eye color were controlled by the number of genesindicated below, how many possible genotype combinations wouldthere be?5:10:20:??????????


Paper#63040 | Written in 18-Jul-2015

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