High school · ScienceAbout 150 minutes
ScienceHigh schoolAbout 150 minutes

Model meiosis to show how crossing over, independent assortment and fertilization produce variation, and use Punnett squares and probability to predict and explain the outcomes of monohybrid and dihybrid crosses

Several sessions: instruction with the worked model, a laboratory or data-analysis investigation, independent practice and a written explanation.

Ohio learning standards this draft targets

science-2018:B.H.3science-2018:B.H.2

Read the standard text

science-2018:B.H.3 Genetic mechanisms and inheritance

science-2018:B.H.2 Structure and function of DNA in cells

Provisional targets. Clause-by-clause educational review is still pending.

Section 1 · LearnLearn the idea
A 4 by 4 Punnett square for two heterozygous pea plants, RrYy x RrYy, with each parent making the gametes RY, Ry, rY and ry equally often. Of the 16 equally likely cells, 9 are round yellow (at least one R and one Y, tan), 3 are round green (R with yy, green), 3 are wrinkled yellow (rr with Y, pale yellow) and 1 is wrinkled green (rryy, grey): the 9 : 3 : 3 : 1 ratio.A 4 by 4 Punnett square for two heterozygous pea plants, RrYy x RrYy, with each parent making the gametes RY, Ry, rY and ry equally often. Of the 16 equally likely cells, 9 are round yellow (at least one R and one Y, tan), 3 are round green (R with yy, green), 3 are wrinkled yellow (rr with Y, pale yellow) and 1 is wrinkled green (rryy, grey): the 9 : 3 : 3 : 1 ratio.Dihybrid Punnett square, RrYy x RrYy (peaseed shape and color)RYRyrYryRRYYRRYyRrYYRrYyRRYyRRyyRrYyRryyRrYYRrYyrrYYrrYyRrYyRryyrrYyrryyRYRyrYry
A 4 by 4 Punnett square for two heterozygous pea plants, RrYy x RrYy, with each parent making the gametes RY, Ry, rY and ry equally often. Of the 16 equally likely cells, 9 are round yellow (at least one R and one Y, tan), 3 are round green (R with yy, green), 3 are wrinkled yellow (rr with Y, pale yellow) and 1 is wrinkled green (rryy, grey): the 9 : 3 : 3 : 1 ratio.

Learning goal: Describe the process of meiosis in relation to the function of DNA and chromosomes in coding the instructions for traits passed from parents to offspring; provide evidence that phenotypic variations result from genetic recombination through meiosis (sorting, recombination, crossing over) and sexual reproduction; use a model of meiosis to demonstrate crossing over and independent assortment during gamete formation and explain how this contributes to variation within a population; and explain the outcomes of monohybrid and dihybrid crosses using Punnett squares and probability.

Chromosomes carry the genes. DNA does not float loose in a eukaryotic cell; it is packaged with proteins into chromosomes, and each chromosome carries a linear series of genes. In animals, including humans, body cells are diploid: they hold two sets of chromosomes, one inherited from each parent, arranged in homologous pairs. The two members of a pair carry the same genes at the same positions but may carry different versions, alleles, of those genes. Humans have 23 pairs, 46 chromosomes in all. Gametes, the eggs and sperm, are haploid, holding one set of 23, so that fertilization restores the diploid number. In animals, meiosis is the division that makes haploid gametes from a diploid cell (in plants, meiosis makes haploid spores and the gametes form later by mitosis), and it is the point in the life cycle where DNA does its hereditary job: coding the instructions for traits and delivering half of them to each offspring.

Meiosis, stage by stage. Before meiosis begins, every chromosome is replicated, so each consists of two identical sister chromatids joined at a region called the centromere. In meiosis I, the homologous pairs line up together, and here two things happen that matter for variation. First, crossing over, in the first stage (prophase I): while the homologs are paired, matching segments are exchanged between a maternal chromatid and a paternal chromatid, so a chromatid that was entirely from one parent now carries a stretch from the other. Second, independent assortment, when the pairs line up at metaphase I: each pair lines up on the cell's midline with a random orientation, so which member of a pair faces which pole is independent of every other pair. The homologs then separate to opposite poles, and the cell divides. In meiosis II, which resembles ordinary mitosis, the sister chromatids of each chromosome separate, and each of the two cells divides again. The result is four haploid cells, each with one chromosome from every pair and, because of crossing over and independent assortment, almost always genetically different from one another. Contrast mitosis, the division that makes body cells: one division, no pairing of homologs, no crossing over, and two cells genetically identical to the parent cell (two diploid cells when the parent cell is diploid, as an animal body cell is). The general rule is that mitosis keeps the number of chromosome sets and meiosis halves it.

Counting the variation. Independent assortment alone produces 2 to the power n combinations in the gametes of an organism with n chromosome pairs. For humans that is 2 to the power 23, which is 8,388,608 different chromosome combinations from one parent. Crossing over multiplies that by creating new chromatids that exist in no parent, and random fertilization pairs any of a mother's roughly 8.4 million combinations with any of a father's, giving about 70 trillion possible combinations before crossing over is even counted. This is the evidence for the claim that phenotypic variation among offspring comes from genetic recombination in meiosis and sexual reproduction: two parents' sorting, recombination and crossing over produce offspring that resemble both parents and match neither, and the variation those offspring carry into a population is what natural selection later acts on.

Mendel's rules stated in modern terms. Working with pea plants in the 1860s, Gregor Mendel found patterns that meiosis explains. Segregation: an individual carries two alleles of each gene and each gamete receives exactly one, because the homologs separate in meiosis I. Independent assortment: the alleles of two different genes are sorted into gametes independently of one another, which holds when the genes are on different chromosomes (or far apart on the same one). Dominance: when the two alleles differ, one, the dominant allele, may fully determine the phenotype, and the recessive allele shows only when no dominant allele is present. Genotype is the pair of alleles an individual carries; phenotype is the observable trait.

A worked monohybrid cross. In peas, purple flowers (P) are dominant to white (p). Cross two heterozygotes, Pp × Pp. Each parent makes P gametes and p gametes in equal numbers. A Punnett square with P and p along the top and down the side gives four equally likely offspring: PP, Pp, Pp, pp. Genotype ratio 1 : 2 : 1; phenotype ratio 3 purple : 1 white. The square is a probability model: it says that each offspring has a 3/4 chance of purple, not that every four offspring contain exactly one white plant.

Identifying an unknown genotype. A purple pea plant could be PP or Pp; its phenotype does not say which. Propose the hypothesis "it is heterozygous," design a test: cross it with a white plant (pp), whose gametes are all p. If the unknown is PP, every offspring is Pp and purple. If it is Pp, half the offspring are expected to be white. A single white offspring proves the unknown carried a p allele; a run of all-purple offspring supports PP but never proves it, because a heterozygote could produce several purple offspring by chance, so the analysis needs many offspring. The same logic identifies genotypes in dog breeds, fruit flies, fast-growing plants and virtual simulations: analyze a population, look at the phenotypes of parents and offspring, and infer.

A worked dihybrid cross. Two genes, seed shape (round R dominant to wrinkled r) and seed color (yellow Y dominant to green y), on different chromosomes. Cross RrYy × RrYy. By independent assortment each parent makes four gamete types in equal numbers: RY, Ry, rY and ry. A 4 × 4 Punnett square holds 16 equally likely offspring genotypes. Counting phenotypes: 9 round yellow, 3 round green, 3 wrinkled yellow, 1 wrinkled green. The shortcut is the product rule of probability. The chance of round is 3/4 and of yellow is 3/4, and because the genes assort independently the chance of round and yellow is 3/4 × 3/4 = 9/16; wrinkled green is 1/4 × 1/4 = 1/16. The 16-cell diagram and the multiplication give the same answer, which is a good check on both.

Misconceptions. "Dominant means more common" is false; dominance describes what a heterozygote looks like, not how frequent an allele is. "The offspring of Pp × Pp will be three purple and one white" treats a probability as a guarantee; with four offspring, all purple happens about 32 percent of the time. "Meiosis makes identical cells" describes mitosis; meiosis makes four haploid cells that are almost always genetically different. "Crossing over makes new genes" is wrong; it makes new combinations of existing alleles on one chromosome.

Where the rest of the standards is taught. This lesson teaches the meiosis half of B.H.2 (the DNA-strand half is in the genome lesson) and, from B.H.3, the meiosis model, Punnett squares and monohybrid and dihybrid crosses. Pedigrees, incomplete dominance, sex-linked traits, statistical comparison and linkage are taught in the non-Mendelian inheritance lesson.

Investigation you can do. Model meiosis with pipe cleaners. Take two long pipe cleaners of one color (a maternal pair member) and two of another color (the paternal member) for one chromosome pair, and two short ones of each color for a second pair; twisting two of the same length together makes a replicated chromosome with two sister chromatids. Pair the homologs, swap a segment between a long red and a long blue chromatid to model crossing over, then line up both pairs on a midline. Flip a coin for each pair to decide which color faces the top: this is independent assortment. Separate the homologs (meiosis I), then separate sisters (meiosis II), and record the contents of the four gametes. Repeat ten times and count how many different gamete types appear; with two pairs and no crossing over there are four possible types, and the crossover segments add more. A result in which the four gametes of one run differ from one another and from those of other runs supports the claim that meiosis generates variation. What the model cannot show: it does not show that real chromosomes carry thousands of genes, that crossover positions vary along the chromosome, or how the cell's machinery moves chromosomes.

Section 2 · ModelWork through the examples

Worked test cross to find an unknown genotype: A black guinea pig (black B dominant to brown b) is crossed with a brown one (bb). Of 12 offspring, 7 are black and 5 are brown. Analysis: a brown offspring must be bb and received a b allele from each parent, so the black parent carries b and is Bb. The prediction for Bb × bb is 1/2 black : 1/2 brown, or 6 : 6 in 12 offspring; 7 : 5 is a small deviation of the kind chance produces in a sample this size. Had all 12 been black, the parent would most likely be BB, though a Bb parent produces 12 black offspring in a row with probability (1/2) to the power 12, about 1 in 4,096, so the conclusion would be strong but not certain.

Worked dihybrid probability: A plant RrYy is crossed with a plant rryy (a dihybrid test cross). The first parent makes RY, Ry, rY and ry gametes with probability 1/4 each; the second makes only ry. Offspring: RrYy round yellow 1/4, Rryy round green 1/4, rrYy wrinkled yellow 1/4, rryy wrinkled green 1/4. A 1 : 1 : 1 : 1 result in a large sample is the evidence that the two genes assort independently; a strong excess of the two parental combinations instead would suggest the genes are linked on one chromosome, a case the next lesson treats.

Worked count of gamete types: An organism has three chromosome pairs, labeled 1, 2 and 3, with a maternal and paternal member of each. Independent assortment gives 2 × 2 × 2 = 8 gamete types, which can be listed: MMM, MMP, MPM, MPP, PMM, PMP, PPM, PPP, where each letter says which member of pairs 1, 2 and 3 the gamete received. Random fertilization with a mate that also makes 8 types gives 8 × 8 = 64 combinations of chromosomes in the offspring. For 23 pairs the same rule gives 2 to the power 23 gamete types, and crossing over raises the number further by making chromosomes that are partly maternal and partly paternal.

Worked comparison of mitosis and meiosis for one cell with 4 chromosomes (2 pairs): Mitosis: chromosomes replicate, line up singly, sisters separate; result 2 cells, each with 4 chromosomes identical to the parent. Meiosis: chromosomes replicate, homologs pair and may cross over, pairs line up and assort independently, homologs separate (2 cells with 2 replicated chromosomes each), sisters separate (4 cells with 2 chromosomes each). Only meiosis halves the number and only meiosis reshuffles alleles into new combinations, which is why it, and not mitosis, is the source of new combinations of existing alleles; new alleles themselves come from mutation, which can occur in any dividing cell.

Section 3 · DoPractice, demonstrate, and explain

Supported record: Draw the 16-cell Punnett square for RrYy × RrYy, shade the four phenotype classes in four colors and count each class to confirm 9 : 3 : 3 : 1. Under it, show the product-rule calculation for each class. Then draw a two-pair cell going through meiosis in four labeled sketches, marking where crossing over and independent assortment occur and how many chromosomes each of the four final cells holds.

Independent performance: Run the pipe-cleaner meiosis model ten times with two chromosome pairs and one crossover per run, and write up the results. Success criteria: (a) a table records the contents of all four gametes for each of the ten runs; (b) the number of distinct gamete types observed is counted and compared with the number expected without crossing over; (c) the explanation uses the words homologous, crossing over, independent assortment, haploid and fertilization correctly and states how each contributes to variation within a population; (d) it explains, using the model, why two siblings are not genetically identical; (e) it works one monohybrid and one dihybrid cross with Punnett squares and states the predicted ratios as probabilities rather than guarantees; (f) it lists two things the pipe-cleaner model cannot show.

Section 4 · Check7-question lesson practice test

An adult may read the prompts aloud. Answer independently with words, objects, drawings or a demonstration before opening the answer guide. Award 1 point for a correct result or claim and 1 point for a matching explanation or demonstration; where a question asks only for a result (a corrected sentence, a spelling, a count), a fully correct result earns both points, and the answer guide shows what a complete response contains. Record prompting separately. The total (two points per question) is local practice feedback, not an Ohio proficiency level.

Problem 1

Explain how meiosis relates to the function of DNA and chromosomes in passing the instructions for traits from parents to offspring.

Show worked solution

Genes are segments of DNA carried on chromosomes, and body cells hold two homologous sets, one from each parent. Meiosis halves that number by separating the homologs and then the sister chromatids, producing haploid gametes that each carry one complete set of instructions, one allele of every gene. Fertilization combines a haploid egg and sperm, restoring the diploid number, so each offspring receives one set of chromosome-borne DNA instructions from each parent.

Problem 2

Describe crossing over and independent assortment, say in which stage of meiosis each occurs, and explain how each contributes to variation within a population.

Show worked solution

Crossing over occurs in prophase I, while homologous chromosomes are paired: matching segments are exchanged between a maternal and a paternal chromatid, producing chromosomes with new combinations of alleles that neither parent carried. Independent assortment occurs at metaphase I, when the pairs line up for the first division: each pair orients randomly, so the maternal or paternal member of one pair goes to a pole independently of the others, giving 2 to the power n combinations. Both mean that gametes, and hence offspring, carry different allele combinations, so a population contains many phenotypes rather than copies.

Problem 3

A pea plant has purple flowers. Design a cross to determine whether it is PP or Pp, state the predicted results for each possibility, and explain why one white offspring settles the question but many purple offspring do not.

Show worked solution

Cross the purple plant with a white plant (pp), whose gametes are all p. If the unknown is PP, all offspring are Pp and purple. If it is Pp, half the gametes carry p, so about half the offspring are expected to be white. A single white offspring must be pp and must have received a p allele from the purple parent, which proves it is Pp. All-purple offspring only support PP, because a Pp parent can produce several purple offspring by chance; the more purple offspring there are, the stronger the support, but it never becomes proof.

Problem 4

Two plants heterozygous for both seed shape and seed color are crossed. Use the product rule to find the probability that an offspring is wrinkled and green, and explain what assumption the rule requires.

Show worked solution

Wrinkled requires rr, with probability 1/4 from Rr × Rr, and green requires yy, with probability 1/4 from Yy × Yy. Multiplying gives 1/4 × 1/4 = 1/16. The rule requires that the two genes assort independently, which is the case when they lie on different chromosomes or far apart on the same one; if the genes were closely linked, the alleles would not sort independently and the product rule would give the wrong answer.

Problem 5

A family has four children, all with the dominant phenotype, although both parents are heterozygous. A student says the Punnett square must be wrong. Explain the error.

Show worked solution

The Punnett square gives probabilities, not guaranteed counts. Each child of two heterozygotes has a 3/4 chance of the dominant phenotype, and the chance that four independent children all show it is (3/4) to the power 4, about 32 percent, which is far from rare. The square correctly predicts that over many families of such parents about one quarter of children show the recessive phenotype; it says nothing certain about any one family of four.

Problem 6

For a diploid animal cell, compare mitosis and meiosis in terms of number of divisions, number and chromosome content of the cells produced, and genetic identity, and state which process is the source of hereditary variation.

Show worked solution

Mitosis involves one division and produces two diploid cells that are genetically identical to the parent cell; it is used for growth and repair. Meiosis involves two divisions and produces four haploid cells, each with one chromosome from every pair and almost always genetically different from one another because of crossing over and independent assortment. Meiosis is the source of new combinations of alleles, because it makes the gametes that combine at fertilization; mitosis copies cells faithfully, apart from occasional mutations, which are the source of new alleles.

Problem 7

Humans have 23 chromosome pairs. Calculate how many different chromosome combinations one parent can produce by independent assortment alone, and explain why the true number of different gametes is larger.

Show worked solution

Independent assortment gives 2 to the power 23 combinations, which is 8,388,608. Crossing over exchanges segments between homologs at positions that vary from one meiosis to the next, creating chromosomes that are partly maternal and partly paternal and so are not any of the 8,388,608 pure combinations. Because crossovers occur in nearly every meiosis, the number of genetically distinct gametes a person can make is effectively unlimited.