Objective: To review the phases of mitosis with the understanding that in a real tissue or a non-synchronized cell culture, different mitotic phases and cells in interphase will be observed at a certain time.Interphase is the period between mitotic events; all other phases are part of mitosis and their individual probabilities add up to the total probability of seeing cells undergoing mitosis.Pay attention to the individual probabilities for cells in each phase of the cell cycle. Problem N° 10For a certain type of animal tissue, the probability of a randomly selected cell being in a particular stage of the interphase/mitosis cycle is given by the following table:Stage ProbabilityInterphase0.5 Prophase0.1 Metaphase0.05Anaphase0.2 Telophase0.15Suppose you examine 100 cells at random from this tissue.A) How many cells do you expect to see in metaphase? _____ cellsB) You have selected a high magnification (2000 X) in your microscope and you see only one cell at a time (50 µm in diameter). How frequently you would see a cell undergoing mitosis? _____ %C) What is the probability of seeing no metaphases in 100 cells? _____ % Background and/or InstructionsProblem N° 9Objective: To apply the concept ofcodominance in the prediction of horse coatcolor.In codominance, both alleles are expressed;both gene products are measurable orvisible as an additional phenotype that isneither dominant nor recessive norintermediate. In incomplete (partial)dominance an intermediate trait isCremello:Palomino:Chestnut:produced.almost whitegolden with lighter mane and tailreddish brownCoat color of horses is a very complex trait.The following table gives ratios obtained in matings of the above horse (Equus caballus) varieties:The colors cremello, chestnut and palominocan be explained using a codominant modelCrossParentsOffspringof inheritance and are known to becremello x cremelloall cremellodetermined by alleles at a single locus inhorses.IIchestnut x chestnutall chestnutFor this problem, use the genotypes:cremello x chestnutall palominoIIC'C' = cremelloIVpalomino x palomino% chestnutC?C? = chestnut½ palomino% cremelloC'c? = palomino.If cross IV (palomino X palomino) is represented by C'C? x c'c², what are the genotypic proportions ofthe offspring?а.Locus = is the physical position of an alleleon a chromosome (plural loci). As theposition of a gene (or allelic form of one),the concept of locus becomes importantwhen mapping chromosomes, i.e., locatingthe positions of loci.Cross III (cremello X chestnut) can be represented by C'C' x C?c?, what is the genotype of the palominooffspring?b.Although mass sequencing projects mayc. Why do crosses I (cremello X cremello) and II (chesnut X chesnut) produce all cremello and all chestnut,respectively, if cremello X chestnut produces all palomino?have obviated the need for chromosomemapping, the association (linkage) betweenor among alleles led to importantdiscoveries and applications in genetics. BackgroundProblem N° 8Objective: To understand the role ofprobabilities in the transmission of traits in8.1. An individual of genotype AA Bb Cc DD Ee is test-crossed. Assuming that the loci undergo independentassortment, what fraction of the progeny are expected to have the genotype Aa Bb Cc Dd Ee?living organisms.This problem is about probabilities inclassical geneticsSYou may express your results as fractions,decimals, or, for large populations, %.When necessary, always write a zero beforethe decimal point. It increases clarity and8.2. Found on chromosome 4 in humans, two blood group alleles, LM and LN, are equally expressed inheterozygous individuals. Such heterozygous individuals are said to be in the MN blood group. Assume that aman and woman, both in the MN blood group, have a child.precision.a. What is the probability that this child will be a girl and be in the MN blood group?For part 8.3., think of the family as aformula that goes like this:Family = Parents + Sibshipb. What is the probability that the next two children born to this couple will be in the MM blood group?.. Sibship = Family - Parents8.3. Red-green colorblindness in humans is caused by a recessive gene on the X chromosome. A husband andwife both have normal vision, although both of their fathers are red-green colorblind. What is the probabilitythat their first child will be:а.a son with normal vision?b. a daughter with normal vision?с.a colorblind son?d.a colorblind daughter?

Question

Objective: To review the phases of mitosis with the understanding that in a real tissue or a non-synchronized cell culture, different mitotic phases and cells in interphase will be observed at a certain time.Interphase is the period between mitotic events; all other phases are part of mitosis and their individual probabilities add up to the total probability of seeing cells undergoing mitosis.Pay attention to the individual probabilities for cells in each phase of the cell cycle. Problem N° 10For a certain type of animal tissue, the probability of a randomly selected cell being in a particular stage of the interphase/mitosis cycle is given by the following table:Stage ProbabilityInterphase0.5  Prophase0.1  Metaphase0.05Anaphase0.2  Telophase0.15Suppose you examine 100 cells at random from this tissue.A)  How many cells do you expect to see in metaphase? _____ cellsB) You have selected a high magnification (2000 X) in your microscope and you see only one cell at a time (50 µm in diameter). How frequently you would see a cell undergoing mitosis? _____ %C) What is the probability of seeing no metaphases in 100 cells? _____ % Background and/or InstructionsProblem N° 9Objective: To apply the concept ofcodominance in the prediction of horse coatcolor.In codominance, both alleles are expressed;both gene products are measurable orvisible as an additional phenotype that isneither dominant nor recessive norintermediate. In incomplete (partial)dominance an intermediate trait isCremello:Palomino:Chestnut:produced.almost whitegolden with lighter mane and tailreddish brownCoat color of horses is a very complex trait.The following table gives ratios obtained in matings of the above horse (Equus caballus) varieties:The colors cremello, chestnut and palominocan be explained using a codominant modelCrossParentsOffspringof inheritance and are known to becremello x cremelloall cremellodetermined by alleles at a single locus inhorses.IIchestnut x chestnutall chestnutFor this problem, use the genotypes:cremello x chestnutall palominoIIC'C' = cremelloIVpalomino x palomino% chestnutC?C? = chestnut½ palomino% cremelloC'c? = palomino.If cross IV (palomino X palomino) is represented by C'C? x c'c², what are the genotypic proportions ofthe offspring?а.Locus = is the physical position of an alleleon a chromosome (plural loci). As theposition of a gene (or allelic form of one),the concept of locus becomes importantwhen mapping chromosomes, i.e., locatingthe positions of loci.Cross III (cremello X chestnut) can be represented by C'C' x C?c?, what is the genotype of the palominooffspring?b.Although mass sequencing projects mayc. Why do crosses I (cremello X cremello) and II (chesnut X chesnut) produce all cremello and all chestnut,respectively, if cremello X chestnut produces all palomino?have obviated the need for chromosomemapping, the association (linkage) betweenor among alleles led to importantdiscoveries and applications in genetics. BackgroundProblem N° 8Objective: To understand the role ofprobabilities in the transmission of traits in8.1. An individual of genotype AA Bb Cc DD Ee is test-crossed. Assuming that the loci undergo independentassortment, what fraction of the progeny are expected to have the genotype Aa Bb Cc Dd Ee?living organisms.This problem is about probabilities inclassical geneticsSYou may express your results as fractions,decimals, or, for large populations, %.When necessary, always write a zero beforethe decimal point. It increases clarity and8.2. Found on chromosome 4 in humans, two blood group alleles, LM and LN, are equally expressed inheterozygous individuals. Such heterozygous individuals are said to be in the MN blood group. Assume that aman and woman, both in the MN blood group, have a child.precision.a. What is the probability that this child will be a girl and be in the MN blood group?For part 8.3., think of the family as aformula that goes like this:Family = Parents + Sibshipb. What is the probability that the next two children born to this couple will be in the MM blood group?.. Sibship = Family - Parents8.3. Red-green colorblindness in humans is caused by a recessive gene on the X chromosome. A husband andwife both have normal vision, although both of their fathers are red-green colorblind. What is the probabilitythat their first child will be:а.a son with normal vision?b. a daughter with normal vision?с.a colorblind son?d.a colorblind daughter?

Accepted Answer

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Stage	Probability
Interphase	0.5
Prophase	0.1
Metaphase	0.05
Anaphase	0.2
Telophase	0.15