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Punnett square crosses




I am having problems with this homework. The instructions are not clear to me and its hard for me to understand. Class is online so teacher is not readily available for questions or to show me how to do it.;Attachment Preview;Module 6.doc Download Attachment;Experiment 1: Punnett square crosses;Materials;Red beads;Blue beads;Green beads;Yellow beads;2 100mL Beakers;Procedure;1. Set up and complete Punnett squares for each of the following crosses;(remember Y = yellow, and y = blue);Y Y and Y y;Y Y and y y;Y;Y;Y;YY;YY;y;Yy;Yy;Y;Y;y;Yy;Yy;y;Yy;Yy;a) What are the resulting phenotypes?;2010 eScience Labs, Inc.;All Rights Reserved;The punnet square shows all of the possible combinations that would result;from the mating of the given genotypes. In this example the dominant allele is;represented by Y, and determines the color of the corn to be yellow. The;results from the square is: 100 percent of the offspring will have a yellow;phenotype, 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 alleles;present.In this particular example, there is no combination that leads to a;homozygous recessive genotype or yy. Meaning there is no blue corn;expressed.;2. Set up and complete a Punnett squares for a cross of two of the F1 from 1b;above;a) What are the genotypes of the F2 generation?;The results of the completed punnet square shows the probability of producing;a homozygous recessive offspring is 25%, a 50% chance of producing a;heterozygous offspring and a 25% chance the offspring will be homozygous;dominant.;b) What are their phenotypes?;The results show a 75% chance the offspring will be yellow and 25% chance;the offspring will be blue.;2010 eScience Labs, Inc.;All Rights Reserved;c) 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 be;expressed. However, all the genotypes in the F1 generation contain a;dominant allele supressing the recessive allele leaving a 0% chance of having;a blue phenotype. In other words, F1 has less blue kernels than the F2;generation.;3. Identify the four possible gametes produced by the following individuals;a) YY Ss;YS;YS;b) Yy Ss;YS;yS;Ys;Ys;Ys;ys;c) Create a Punnett square using these gametes as P and determine the;genotypes of the F1;There are 6 distinct genotypes expressed in F1. YYSS: expressed two times;YYSs: expressed four times YYss: expressed two times YySS: expressed two;times 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 always;expressed as the phenotype. For Example in this punnet Square Ss=SS and;YY=Yy, so any time Y is expressed the phenotype is yellow and when S is;expressed the phenotype is smooth. There are two distinct phenotypes: 1.;yellow and smooth 2. yellow and wrinkled. Therefore, 75 percent of the;offspring will be yellow and smooth, and the other 25 percent will be yellow;and wrinkled. The ratio for the phenotype is 12:4. 12 combinations of Yellow;and smooth and 4 combinations of yellow and wrinkled.;4. You have been provided with 4 bags of different colored beads. Pour 50 of the;blue beads and yellow beads into beaker #1 and mix them around. Pour 50 of;the red beads and green beads in beaker #2 and mix them.;Attention!Donotpourthebeakerstogether.;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 is;for 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 not;put those beads back into the beaker.;Repeat this for individual #2. These two genotypes are your parents;for the next generation. Set up a Punnett square and determine the;genotypes and phenotypes for this cross.;Repeat this process 4 times (5 total). Put the beads back in their;respective beakers when finished.;a) How much genotypic variation do you find in the randomly picked;parents of your crosses?;When I randomly picked the parents from the beaker I found six distinct;genotypes. Within all of the randomly selected parents, I chose a;heterozygous genotype five out of the ten times, homozygous dominant;genotype two out of the ten times, and a homozygous recessive genotype;three 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 homozygous;recessive genotypes, and eight heterozygous genotypes chosen out of the;twenty total offspring.;2010 eScience Labs, Inc.;All Rights Reserved;2010 eScience Labs, Inc.;All Rights Reserved;c) How much phenotypic variation?;There were four different phenotypes for this exercise: 1. Yellow and Smooth;2. Yellow and Wrinkled 3. Blue and Smooth 4. Blue and Wrinkled. Five were;yellow and smooth phenotype, three were yellow and wrinkled phenotype, five;were blue and smooth phenotype, seven were blue and wrinkled phenotype of;the twenty total offspring.;d) Is the ratio of observed phenotypes the same as the ratio of predicted;phenotypes? Why or why not?;e) Pool all of the offspring from your 5 replicates. How much phenotypic;variation 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 or;blood type. Genes are functional units of heredity. Alleles are variations of the;same gene. For Example: One single gene determines a persons height and;whether or not you are tall or short would be determined through the;combinations of alleles within the height gene. Genes determine the genetic;traits a person will have and the different sequences of alleles within the gene;will determine a single characterisitc of the individual.;g) How might protein synthesis execute differently if there a mutation;2010 eScience Labs, Inc.;All Rights Reserved;occurs?;A gene mutation is an alteration in the sequence of nucleotides in DNA. DNA;consists of nucleotides. During protein synthesis, DNA is transcribed into RNA;and then translated to produce proteins. Altering nucleotide sequences most;likely will result in nonfunctioning proteins. For example: if a nonsense;mutation occurs it will alter the nucleotide sequence coding a stopcodon in;place of a amino acid, signaling the end of the translation process, stopping;protein production.;h) Organisms heterozygous for a recessive trait are often called carriers;of that trait. What does that mean?;Carriers contain a recessive allele and are capable of passing on the;recessive allele to their offspring, even though they do not express the;recessive allele, because of the presence of a dominant allele. A person that;inherits a genetic trait but does not display or show symptoms of that trait, but;are able to pass the gene on to their offspring.;i);In peas, green pods (G) are dominant over yellow pods. If a;homozygous dominant plant is crossed with a homozygous recessive;plant, what will be the phenotype of the F1 generation? If two plants;from the F1 generation are crossed, what will the phenotype of their;offspring be?;If a homozygous dominant plant is crossed with a homozygous recessive;plant, 100% of the offspring will be green. If two plant from the F1 generation;are crossed,since they are all heterozygous plants, 3/4 (75%) of the plant's;offspring will be green and 1/4 (25%) of the plant's offspring will be yellow.;2010 eScience Labs, Inc.;All Rights Reserved;B. Dihybrid Cross: Randomly (without looking) take 2 beads out of beaker #1 AND 2;beads out of beaker #2.;These four beads represent the genotype of individual #1, record this;information.;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 and;phenotypes for this cross.;c) What is your predicted ratio of genotypes? Hint: think back to our;example dihybrid cross;Repeat this process 4 times (for a total of 5 trials).;2010 eScience Labs, Inc.;All Rights Reserved;d) How similar are the observed phenotypes in each replicate?;e) How similar are they if you pool your data from each of the 5;replicates?;f);Is it closer or further from your prediction?;2010 eScience Labs, Inc.;All Rights Reserved;g) Did the results from the monohybrid or dihybrid cross most closely;match your predicted ratio of phenotypes?;h) Based on these results, what would you expect if you were looking;at 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 Reserved;j);In certain bacteria, an oval shape (S) is dominant over round and;thick cell walls (T) are dominant over thin. Show a cross between;a heterozygous oval, thick cell walled bacteria with a round, thin;cell walled bacteria. What are the phenotype of the F1 and F2;offspring?;5. The law of independent assortment allows for genetic recombination. The;following equation can be used to determine the total number of possible;genotype combinations for any particular number of genes;2g= Number of possible genotype combinations (where g is the number of;genes);1 gene;21= 2 genotypes;2 genes: 22= 4 genotypes;3 genes: 23 = 8 genotypes;Consider the following genotype;Yy Ss Tt;We 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 Reserved;b) Many traits (phenotypes), like eye color, are controlled by multiple;genes. If eye color were controlled by the number of genes;indicated below, how many possible genotype combinations would;there be?;5;10;20


Paper#16642 | Written in 18-Jul-2015

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