5.2 Meiosis and Genetic Diversity

1. AAll 23 came either from his mother or from his father, chosen at random for each sperm.
   A student who thinks the maternal and paternal sets stay together picks this. Each pair orients independently, so sperm carry mixtures of maternal and paternal chromosomes.
2. BFor each of the 23 pairs, a sperm got his mother's or his father's chromosome, chosen independently. Correct
   Each pair orients independently at metaphase I, so for every pair a sperm is equally likely to receive the maternal or the paternal chromosome. With 23 homologous pairs, each sperm receives one chromosome of every pair, almost always some inherited from the man's mother and some from his father. The combination differs among sperm; there are 2²³ possible combinations.
3. CThe same 23 chromosomes, some from each parent, are found in every one of his sperm.
   A student who thinks meiosis makes identical copies, as mitosis does, picks this. Different sperm receive different combinations of the man's maternal and paternal chromosomes.
4. DEach of the 23 is a blend of the versions he inherited from his mother and father.
   A student who pictures inheritance as blending picks this. With no crossing over, each sperm chromosome is one whole homolog, either maternal or paternal; homologs do not merge.

CED 5.2.A.1 · Read this in Fix

1. AThe two sister chromatids of one chromosome exchange their segments in prophase I.
   A student who thinks crossing over involves sister chromatids picks this. Sister chromatids are identical, so exchanges between them would make no new combinations; crossing over involves non-sister chromatids.
2. BAt a chiasma, all four chromatids of a homologous pair exchange segments in prophase I.
   A student who thinks crossing over swaps segments between whole homologous chromosomes, so that both chromatids of each homolog change, picks this. At any one chiasma, only two non-sister chromatids, one from each homolog, exchange segments; the other two chromatids are not involved in that exchange.
3. CHomologous pairs swap the sides of the plate they face at metaphase I.
   A student who confuses crossing over with the random orientation of homologous pairs picks this. That orientation is a separate source of variation; crossing over exchanges chromatid segments.
4. DNon-sister chromatids of a homologous pair exchange segments in prophase I. Correct
   During prophase I, homologous chromosomes are paired, and non-sister chromatids (one from each homolog) exchange corresponding segments at chiasmata, producing chromatids with new combinations of alleles.

CED 5.2.A.2 · Read this in Fix

1. AEvery gamete would be genetically identical, since crossing over is the only source of genetic variation.
   A student who thinks crossing over is the only source of variation picks this. Random assortment of homologous pairs still makes gametes differ.
2. BThey would vary less but still differ, because homologous pairs still orient randomly at metaphase I. Correct
   Without crossing over, gametes lose the new allele combinations made within chromosomes, so variation falls. Random orientation of homologous pairs at metaphase I still gives gametes different combinations of maternal and paternal chromosomes.
3. CThey would vary just as much as before, since crossing over just swaps sister chromatid segments.
   A student who thinks crossing over is between sister chromatids, which are identical, expects it to add no variation. It occurs between non-sister chromatids and does add variation.
4. DEach gamete would carry either the full maternal or the full paternal set of chromosomes.
   A student who thinks maternal and paternal chromosomes stay together as sets picks this. Each pair orients independently, so gametes still get mixtures.

CED 5.2.A.3 · Read this in Fix

1. A6
   A student who adds the options for independent choices instead of multiplying them picks this: 2 + 2 + 2 = 6. Independent orientations multiply: 2 × 2 × 2 = 8.
2. B2
   A student who thinks all maternal chromosomes go to one pole together picks this: all-maternal or all-paternal. The pairs orient independently, and the diagram itself shows a mixed arrangement.
3. C1
   A student who thinks meiosis makes identical gametes picks this. Different cells line up the pairs in different ways, giving 8 combinations.
4. D8 Correct
   The cell has three homologous pairs. Each pair can orient with either homolog facing Pole 1, independently of the others, so the number of combinations is 2 × 2 × 2 = 8.

Working Count homologous pairs in the diagram: 3. Each pair has 2 possible orientations, independent of the others. Combinations = 2 × 2 × 2 = 2³ = 8.

CED 5.2.A.1 · Read this in Fix

1. ASister chromatids of one long chromosome did not separate in meiosis I.
   A student who thinks sister chromatids separate in meiosis I picks this. The two long chromosomes in gametes 1 and 2 are a solid and a dashed one, which are homologs, not sister chromatids.
2. BThe homologous chromosomes of the long pair did not separate in meiosis II.
   A student who thinks homologs separate in meiosis II picks this. Homologs are separated in meiosis I; meiosis II separates sister chromatids.
3. CHomologous chromosomes of the long pair did not separate in meiosis I. Correct
   Gametes 1 and 2 contain both long homologs (one solid, one dashed) and gametes 3 and 4 contain none. Homologs normally go to different cells in meiosis I; if the long pair fails to separate, one cell receives both and the other neither, and meiosis II passes this on to all four gametes.
4. DThe two sister chromatids of a long chromosome failed to separate in meiosis II.
   A student who thinks a solid and a dashed copy are sister chromatids picks this. Sister chromatids come from one parent, and a meiosis II error would leave two gametes with normal chromosome numbers.

CED 5.2.A.1 · Read this in Fix

1. AThe percentage stayed low in younger females and rose more and more steeply with age. Correct
   From 2 to 5 years the percentage rose only from 2% to 4%, but it then rose from 4% to 7%, 7% to 13% and 13% to 24%, so each year's increase was larger than the last.
2. BThe data show that aging causes chromosomes in eggs to fail to separate.
   A student who reads observational data as showing a cause picks this. The data show a relationship with age; other differences between older and younger females are not ruled out.
3. CThe percentage rose by about the same amount with each additional year.
   A student who assumes increasing trends are linear picks this. The yearly increases grew from 0 to 1 to 3 to 6 to 11 percentage points.
4. DFemales older than 8 years would produce no eggs with a normal chromosome number.
   A student who extends a trend beyond the data picks this. No females older than 8 years were measured, and at 8 years 76% of eggs were normal.

Working Yearly increases: 2→3 y: 0; 3→4: 1; 4→5: 1; 5→6: 3; 6→7: 6; 7→8: 11 percentage points. Increases grow, so the curve gets steeper. At 8 years, 100 − 24 = 76% of eggs were normal.

CED 5.2.A.1 · Read this in Fix

1. AOnly its reproductive cells would have 11 chromosomes; the rest would have 10.
   A student who thinks inherited chromosome changes are found only in reproductive cells picks this. Every body cell descends from the zygote by mitosis.
2. BEvery body cell would have 11 chromosomes, copied from the zygote by mitosis. Correct
   The zygote has 6 + 5 = 11 chromosomes. All body cells develop from the zygote by mitosis, which gives each daughter cell the same chromosomes as the parent cell, so every body cell has 11.
3. CIts cells would have 10 or 11, since mitosis gives daughter cells different chromosomes.
   A student who thinks mitosis shares out chromosomes unevenly picks this. Mitosis gives both daughter cells copies of every chromosome.
4. DEvery body cell would have 10 chromosomes, because fertilization restores the diploid number.
   A student who treats 'fertilization restores the diploid number' as a rule picks this. Fertilization adds the gametes' chromosomes: 6 + 5 = 11.

Working Egg 6 chromosomes + normal sperm n = 5 → zygote 11. Mitosis copies all 11 into every body cell.

CED 5.2.A.1 · Read this in Fix

1. AAB and AB in one cell; ab and ab in the other
   A student who thinks crossing over is an exchange between sister chromatids picks this. Sister chromatids carry the same alleles, so such an exchange would change nothing. The chiasma joins non-sister chromatids 2 and 3, which swap the segments carrying gene B, so chromatid 2 becomes Ab and chromatid 3 becomes aB.
2. BAB and ab in one cell; Ab and aB in the other
   A student who thinks sister chromatids separate in meiosis I and homologous chromosomes in meiosis II picks this, placing one chromatid from each homolog in each cell. In meiosis I the homologs separate, so chromatids 1 and 2 stay together in one cell and chromatids 3 and 4 in the other.
3. CAB and Ab in one cell; aB and ab in the other Correct
   Crossing over at the chiasma exchanges the segments carrying gene B between non-sister chromatids 2 and 3, so chromatid 2 becomes Ab and chromatid 3 becomes aB; chromatids 1 (AB) and 4 (ab) are unchanged. In meiosis I the homologous chromosomes separate, and each keeps its two sister chromatids joined at the centromere. One cell therefore receives chromatids 1 and 2 (AB and Ab) and the other receives chromatids 3 and 4 (aB and ab).
4. DAb and Ab in one cell; aB and aB in the other
   A student who thinks both chromatids of each homolog exchange segments picks this. A single crossover involves only the two non-sister chromatids that meet at the chiasma, so chromatids 1 and 4 keep AB and ab.

CED 5.2.A.2 · Read this in Fix

1. ADoes the temperature at which males are raised affect their mean number of chiasmata per cell? Correct
   This question names an independent variable that can be set (rearing temperature, one of the ways the two altitudes differ) and a dependent variable that can be measured (mean number of chiasmata per cell), so it can be tested by raising genetically similar males at different temperatures.
2. BWhy do grasshoppers living at high altitude need less genetic variation than others?
   A student who thinks 'why do they need' questions are testable picks this. It asks about purpose, assumes its own answer and names no variable to change or measure.
3. CIs it better for a grasshopper population to have more crossing over or less of it?
   A student who thinks 'better' questions are testable as stated picks this. 'Better' is a value judgement until it is replaced by a measurable outcome.
4. DWhat is the name of the structure where non-sister chromatids join in prophase I?
   A student who equates looking up a fact with experimenting picks this. The answer (chiasma) is a definition, not the result of manipulating a variable.

CED 5.2.A.2 · Read this in Fix

1. AMutant cells and normal cells of the parent strain, both undergoing meiosis in the same conditions Correct
   The mutant and its parent strain differ only in whether crossing over can occur, so any difference in variation among the products of meiosis can be attributed to crossing over.
2. BMutant cells compared with normal cells of a related species, in the same conditions
   A student who thinks any comparison group will do picks this. A different species differs in many genes, so a difference in variation could not be attributed to crossing over.
3. CMutant cells undergoing meiosis at two different temperatures, with no normal cells included
   A student who thinks changing any condition tests the question picks this. Temperature is not the variable in the question; the comparison must vary crossing over.
4. DMutant cells alone, since without crossing over all the cells formed will be identical
   A student who thinks crossing over is the only source of variation assumes the outcome and sees no need for a comparison. Random assortment still makes the products vary, so a comparison with normal cells is needed.

CED 5.2.A.3 · Read this in Fix

1. ACrossing over in prophase I makes each gamete different, and it is the only process that adds any variation.
   A student who thinks crossing over is the only source of variation picks this. Random assortment of pairs and random fertilization also contribute.
2. BFrom each of his parents, each brother got all of that parent's maternal chromosomes or all of the paternal ones.
   A student who thinks maternal and paternal chromosome sets are passed on whole picks this. Each pair assorts independently, so gametes carry mixtures.
3. CEach brother received a different blend of their parents' traits, mixed in different amounts.
   A student who pictures inheritance as blending picks this. Alleles are passed on as separate units and recombined, not mixed.
4. DCrossing over and random orientation of pairs make varied gametes, and fertilization joins two at random. Correct
   Crossing over in prophase I and random assortment of homologous pairs at metaphase I make each parent's gametes genetically varied, and random fertilization combines one egg and one sperm from these many kinds, so siblings receive different combinations.

CED 5.2.A.3 · Read this in Fix