5.3 Mendelian Genetics

1. AFour gametes: AB, Ab, aB and ab
   A student who thinks each cell's meiosis produces every combination of alleles picks this. One arrangement at metaphase I gives only two kinds of gametes; the four kinds appear across the many cells of the individual.
2. BFour gametes: Aa, Aa, Bb and Bb
   A student who thinks segregation separates genes rather than alleles picks this. The two alleles of a gene, on homologous chromosomes, go to different gametes, so each gamete receives one allele of each gene, A or a and B or b.
3. CFour gametes: AB, AB, ab and ab Correct
   In meiosis I, homologous chromosomes move to opposite poles, so the A and B chromosomes on the Pole 1 side go to one cell and the a and b chromosomes go to the other. Meiosis II separates sister chromatids, giving two AB and two ab gametes. In other cells the pairs line up the other way and give Ab and aB.
4. DFour gametes, each of them AaBb
   A student who thinks gametes carry the parent's whole genotype, as cells made by mitosis do, picks this. Meiosis halves the chromosome number, so each gamete has one chromosome of each pair and one allele of each gene.

CED 5.3.A.1 · Read this in Fix

1. APurple-flowered and white-flowered offspring will be found in equal numbers.
   A student who thinks a null hypothesis always predicts equal groups picks this. The 'no difference' is between observed and expected counts, and the expected ratio here is 3:1, not 1:1.
2. BOffspring will be purple and white in a 3:1 ratio, apart from chance deviations. Correct
   A chi-square null hypothesis states that observed counts differ from those expected from the genetic model only by chance. For Pp × Pp with complete dominance, the model predicts 3 purple : 1 white.
3. CThe observed numbers will differ significantly from those expected for 3:1.
   A student who writes the predicted effect as the null hypothesis picks this. A prediction of a significant difference is the alternative hypothesis; the null hypothesis states that there is no difference apart from chance.
4. DExactly three-quarters of the offspring will be purple, with no variation at all.
   A student who thinks Mendelian ratios give exact numbers picks this. Random fertilization makes counts vary around 3:1; the null hypothesis allows chance deviation, and the test decides whether a deviation is too large.

Working For Pp × Pp with complete dominance, P(purple) = 3/4 and P(white) = 1/4, so the expected ratio is 3 purple : 1 white. The null hypothesis states that the observed counts differ from the counts expected for 3:1 only by chance.

CED 5.3.A.2.i · Read this in Fix

1. AFail to reject the null hypothesis; the data prove the genes are on different chromosomes.
   A student who thinks failing to reject proves the null hypothesis picks this. χ² = 2.64 is below 7.81, so the null hypothesis is not rejected, but that shows only that the data are consistent with independent assortment; it cannot prove where the genes are.
2. BFail to reject the null hypothesis; the data are consistent with independent assortment. Correct
   The four classes total 400, so each expected count is 100. χ² = 0.64 + 0.36 + 1.00 + 0.64 = 2.64. With 4 − 1 = 3 degrees of freedom the critical value at p = 0.05 is 7.81, and 2.64 < 7.81, so the differences are small enough to be due to chance. The data fit independent assortment but do not prove it.
3. CReject the null hypothesis; the four classes are unequal, so the genes do not assort independently.
   A student who thinks any departure from the expected ratio disproves it picks this. Chance always makes the counts differ a little from 100 each; the chi-square test shows these differences (χ² = 2.64, below 7.81) are within what chance produces.
4. DReject the null hypothesis; χ² is below the critical value, so the difference is significant.
   A student who reads a small χ² like a small p-value picks this. A χ² below the critical value means the observed counts are close to the expected ones; the null hypothesis is rejected only when χ² exceeds 7.81.

Working Total offspring = 108 + 94 + 90 + 108 = 400. Independent assortment in PpRr × pprr predicts a 1:1:1:1 ratio, so e = 400/4 = 100 for each class. χ² = (108 − 100)²/100 + (94 − 100)²/100 + (90 − 100)²/100 + (108 − 100)²/100 = 0.64 + 0.36 + 1.00 + 0.64 = 2.64. Degrees of freedom = 4 classes − 1 = 3; critical value at p = 0.05 = 7.81. Because 2.64 < 7.81, the null hypothesis is not rejected: the data are consistent with independent assortment.

CED 5.3.A.1 · Read this in Fix

1. AEach offspring forms from a different egg and sperm, each with a different combination of alleles. Correct
   Meiosis gives each parent many gametes with different combinations of alleles. Fertilization joins one egg and one sperm, so each zygote receives one haploid set from each parent but its own new combination of the parents' alleles.
2. BFertilization causes mutations in each zygote, giving every offspring alleles its parents lack.
   A student who thinks new genotypes need new mutations picks this. Fertilization does not cause mutations, and new mutations account for only a small part of the genetic differences between siblings; most come from new combinations of alleles the parents already carry.
3. CThe alleles of the two parents blend together in each zygote, forming a different mixture each time.
   A student who thinks inheritance blends the parents' alleles picks this. Alleles keep their identity in the zygote and can be passed on unchanged to the next generation.
4. DEach offspring receives most of its chromosomes from one parent, a different parent each time.
   A student who thinks an offspring can inherit more from one parent picks this. Fertilization fuses two haploid gametes, so every offspring receives one set of chromosomes from each parent.

CED 5.3.A.2 · Read this in Fix

1. AIt would stay the same, because fertilization restores the diploid number.
   A student who treats 'fertilization restores the diploid number' as an automatic rule picks this. Fertilization restores the diploid number only because each gamete is haploid; fusing two diploid gametes gives twice the diploid number.
2. BIt would halve in each generation, because mitosis halves the chromosome number.
   A student who swaps the outcomes of mitosis and meiosis picks this. Mitosis produces cells with the same chromosome number as the parent cell; it is meiosis that halves it.
3. CIt would stay the same, because each offspring would be a clone of one parent.
   A student who links any use of mitosis with cloning picks this. Each offspring would still form from gametes of two parents, so it would not be a clone of either, and it would carry both parents' full chromosome sets.
4. DIt would double in each generation, because two diploid gametes would fuse. Correct
   Mitosis keeps the chromosome number, so each gamete would carry the parent's full set. Two such gametes fusing would give a zygote with twice the parent's number, and the next generation's gametes would carry that larger number, so the number would double each generation.

CED 5.3.A.2 · Read this in Fix

1. AIt is 1/4, because each child's genotype is independent of the others'. Correct
   Each child receives c from each heterozygous parent with probability 1/2, so P(cc) = 1/2 × 1/2 = 1/4. Each fertilization is independent, so the three earlier children do not change this probability.
2. BIt is 1/256, because four affected children in a row would be very unlikely.
   A student who thinks a run of one outcome must soon be broken picks this: the student takes the small probability of a run of four affected children, (1/4)⁴ = 1/256, as the probability for the fourth child. Fertilization has no memory; the probability is 1/4 for every child of these parents.
3. CIt is 1, because all three of the couple's earlier children have been affected.
   A student who sees a pattern in a short run of chance events and expects it to continue picks this. Three affected children do not change the parents' genotypes or the 1/4 probability for each new child.
4. DIt is 1/2, because the child will either have cystic fibrosis or not have it.
   A student who treats two possible outcomes as equally likely picks this. Of the four equally likely combinations of gametes from Cc × Cc, only cc gives cystic fibrosis, so the probability is 1/4.

Working Cc × Cc. A child has cystic fibrosis if it receives c from each parent: P = 1/2 × 1/2 = 1/4. Each fertilization is independent of the earlier ones, so the probability is 1/4 for the fourth child, as for each earlier child.

CED 5.3.A.2.i · Read this in Fix

1. A0.016
   A student who thinks each box of a Punnett square is a different genotype picks this: one box of 8 × 8 = 64 gives 1/64. AaBbDd fills 8 of the 64 boxes, so its probability is 8/64 = 0.125.
2. B0.422
   A student who reads the genotype AaBbDd as the phenotype 'dominant for all three traits' picks this: (3/4)³ = 27/64. That includes AA, BB and DD offspring; the genotype AaBbDd has probability (1/2)³.
3. C0.037
   A student who treats AA, Aa and aa as equally likely picks this: (1/3)³ = 1/27. From Aa × Aa the genotypes occur as 1/4 : 1/2 : 1/4, so P(Aa) = 1/2 for each gene.
4. D0.125 Correct
   Each gene assorts independently. For each, Aa × Aa gives Aa with probability 1/2. By the product rule, P(AaBbDd) = 1/2 × 1/2 × 1/2 = 1/8 = 0.125.

Working Genes on different chromosomes assort independently, so the product rule applies. For each gene, Aa × Aa gives AA : Aa : aa = 1/4 : 1/2 : 1/4, so P(Aa) = 1/2, and likewise P(Bb) = 1/2 and P(Dd) = 1/2. P(AaBbDd) = 1/2 × 1/2 × 1/2 = 1/8 = 0.125. Distractors: one box of a 64-box square, 1/64 = 0.016; probability of showing all three dominant traits, (3/4)³ = 27/64 = 0.422; AA, Aa and aa taken as equally likely, (1/3)³ = 0.037.

CED 5.3.A.2.i · Read this in Fix

1. ABlack is dominant, because black parents in cross 1 produced no red offspring.
   A student who thinks a trait that breeds true is dominant picks this. Homozygous recessive parents also produce only offspring like themselves; black × black giving only black shows only that these black beetles were homozygous.
2. BBlack is dominant, because the table has more black offspring than red ones.
   A student who thinks the more common phenotype is dominant picks this. The totals (162 black, 126 red) depend on which crosses were made; dominance is shown by which phenotype heterozygotes have.
3. CRed is dominant, because two red parents in cross 2 produced black offspring. Correct
   Two red parents could produce black offspring only if both carried a hidden allele for black, so they were heterozygous and red is dominant. The ratio of 74 red to 25 black is close to the 3:1 expected from two heterozygotes.
4. DNeither is dominant, as cross 3 produced similar numbers of both colors.
   A student who thinks equal numbers mean neither allele 'wins' picks this. About 1:1 is the ratio expected when a heterozygote (red) is crossed with a homozygous recessive (black).

Working Cross 2: red × red gave 74 red : 25 black, about 3:1, so both red parents were heterozygous and red is dominant. Cross 3: 52 red : 49 black, about 1:1, as expected for a heterozygous red parent × a homozygous recessive black parent. Cross 1: two homozygous recessive black parents give only black offspring. Totals (162 black, 126 red) say nothing about dominance.

CED 5.3.A.2.ii · Read this in Fix

1. AAbout three straight-winged flies for every curly-winged fly Correct
   If curly is recessive, the F1 are all heterozygous and straight-winged. Crossing F1 × F1 gives 1/4 homozygous recessive, so about 3 straight : 1 curly.
2. BStraight-winged flies alone, as the curly allele was lost in the F1
   A student who thinks a masked recessive allele is lost picks this. The F1 flies carry the curly allele, hidden by the straight allele, and pass it to half their gametes, so curly wings return in the F2.
3. CFlies with wings intermediate between curly and straight
   A student who thinks inheritance blends traits picks this. Under the hypothesis, the alleles keep their identity; heterozygotes are straight-winged and the F2 shows the two parental phenotypes.
4. DAbout equal numbers of straight-winged and curly-winged flies
   A student who treats the two phenotypes as equally likely picks this. Only one of the four equally likely gamete combinations from two heterozygotes gives the recessive phenotype.

CED 5.3.A.2.ii · Read this in Fix

1. AIts white color dilutes the purple, so the shade of the offspring shows the purple plant's genotype.
   A student who thinks inheritance blends colors picks this. Alleles do not blend: offspring are either purple or white, and the test depends on whether any white offspring appear.
2. BIt passes on only p, so each offspring's color shows the allele it received from the purple plant. Correct
   Every gamete of a pp plant carries p, which does not mask anything. An offspring is purple if it received P from the tested plant and white if it received p, so any white offspring show the tested plant is Pp.
3. CIt has no allele for flower color, so the offspring carry the purple plant's alleles alone.
   A student who thinks a recessive allele is the absence of a gene picks this. The white plant has two copies of the p allele and passes one to every offspring.
4. DIt passes on both of its p alleles, so every offspring is white unless it also receives P.
   A student who thinks gametes carry both of a parent's alleles picks this. Each gamete is haploid and carries one allele of each gene, so each offspring receives a single p from the white plant.

CED 5.3.A.2.ii · Read this in Fix

1. AIt is heterozygous, because A and B are different alleles.
   A student who treats alleles of different genes as alleles of one gene picks this. A and B belong to different genes; zygosity is described gene by gene.
2. BIt is homozygous for both genes, since it shows both dominant traits.
   A student who thinks a dominant phenotype means a homozygous dominant genotype picks this. The plant shows both dominant traits, but its genotype for gene A is Aa, which is heterozygous.
3. CIt is heterozygous for gene A and homozygous for gene B. Correct
   The plant received a from its aaBB parent and A from its AABB parent, so it is Aa, heterozygous for gene A. It received B from each parent, so it is BB, homozygous for gene B.
4. DIt is heterozygous for A, and dominant but not homozygous for B.
   A student who thinks 'homozygous' means two recessive alleles picks this. BB has two identical alleles, so the plant is homozygous (homozygous dominant) for gene B.

CED 5.3.A.2.iii · Read this in Fix

1. AII-2 must be heterozygous, because his sister, II-3, is affected.
   A student who thinks every unaffected relative of an affected person is a carrier picks this. II-2's parents are both heterozygous, so II-2, being unaffected, could be homozygous dominant or heterozygous.
2. BI-1 must be homozygous, because he does not show the trait.
   A student who thinks an individual without the recessive trait is homozygous dominant picks this. I-1's daughter II-3 is affected, so she received a recessive allele from him: I-1 is heterozygous.
3. CII-4 must be homozygous, as neither of his children is affected.
   A student who expects a family of two to match Mendelian ratios exactly picks this. If II-4 were heterozygous, each child would have a probability of 1/2 of being unaffected, so two unaffected children are quite likely; his genotype cannot be decided.
4. DIII-1 must be heterozygous, because her mother, II-3, is affected. Correct
   II-3 is affected, so she is homozygous recessive and passes a recessive allele to every child. III-1 is unaffected, so she also has a dominant allele: she must be heterozygous.

Working II-3 is affected, so she is aa. III-1 is unaffected and received a from II-3, so she is Aa (likewise III-2). I-1 and I-2 are both Aa, because they have an affected daughter. II-1 and II-2 are AA (probability 1/3) or Aa (probability 2/3). II-4 is AA or Aa: if he were Aa, each child would be unaffected with probability 1/2, so two unaffected children (probability 1/2 × 1/2 = 1/4) do not decide his genotype.

CED 5.3.A.2.iii · Read this in Fix

1. AX and Y differ in phenotype, since only Y produced white offspring.
   A student who thinks a phenotype includes every allele carried picks this. Phenotype is the observable expression: both parents have purple flowers, so their phenotypes are the same even though Y carries a hidden recessive allele.
2. BX and Y have the same phenotype, but the data show that their genotypes differ. Correct
   Both parents have purple flowers, the same phenotype. Only Y produced white offspring, so Y must carry the recessive allele (Pp), whereas X's 58 purple offspring suggest it is PP. The same phenotype can come from different genotypes.
3. CX and Y share a genotype, and Y's white offspring are new mutants.
   A student who thinks unexpected offspring must be new mutants picks this. Fifteen of 59 offspring is about the 1/4 expected from a heterozygote, far too many for new mutations; Y carries the white allele.
4. DY fails to follow Mendel's laws, since its offspring are not exactly 3:1.
   A student who thinks Mendelian ratios give exact numbers picks this. 44 : 15 is close to 3:1; chance variation around the expected ratio is what Mendel's laws predict for real counts.

CED 5.3.A.2.iv · Read this in Fix

1. AThe more offspring a cross produced, the closer its percentage was to 25%. Correct
   The distance from 25% falls steadily as the number of offspring rises: 25, 10, 5, 3, 1.5 and 0.3 percentage points. Chance has a larger effect on the proportion in small samples.
2. BOnly the largest crosses followed Mendel's laws; the smaller crosses did not.
   A student who thinks Mendelian ratios give exact numbers picks this. Every cross follows the same probability of 1/4; small crosses simply show more chance variation around it.
3. CLow percentages in some crosses were offset by high ones in others, as chance requires.
   A student who thinks chance must balance out picks this. Each cross is independent; nothing makes a low result in one cross call for a high result in another. The pattern in the graph is about sample size.
4. DScoring more offspring caused the parents to produce white ones at 25%.
   A student who reads the relationship as cause and effect picks this. Counting more offspring does not change how the parents' gametes form or combine; it only reduces the effect of chance on the percentage observed.

Working Distance from the expected 25%: 8 offspring, |50 − 25| = 25; 20 offspring, 10; 40 offspring, 5; 100 offspring, 3; 400 offspring, 1.5; 1000 offspring, 0.3 percentage points. The distance falls as the number of offspring rises.

CED 5.3.A.2.i · Read this in Fix

1. ASex-linked, because the trait skips generation II and then reappears.
   A student who thinks a trait that skips a generation must be sex-linked picks this. Autosomal recessive traits also skip generations. Unaffected II-2 and II-3 have an affected daughter, so the allele is recessive, which rules out X-linked dominant inheritance; the daughter's father, II-3, is unaffected, which rules out X-linked recessive inheritance; and a female cannot show a Y-linked trait.
2. BAutosomal dominant, because the trait is found in both males and females.
   A student who mixes up the autosomal-versus-sex-linked test with the dominant-versus-recessive test picks this. Both sexes being affected says nothing about dominance; an affected child with two unaffected parents rules out a dominant allele.
3. CNot inherited, because neither of the parents of III-2 shows the trait.
   A student who thinks a trait neither parent shows cannot be inherited picks this. Unaffected parents can both carry a recessive allele; II-2 received it from her affected father, I-1.
4. DAutosomal recessive, because unaffected II-2 and II-3 have an affected daughter. Correct
   Two unaffected parents having an affected child shows the allele is recessive and both parents are carriers. An X-linked recessive daughter would need an affected father, and II-3 is unaffected, so the gene is autosomal.

Working III-2 is affected, but her parents II-2 and II-3 are unaffected, so the allele is recessive and both parents are carriers. X-linked recessive inheritance would require an affected daughter's father to be affected; II-3 is unaffected, so the gene is autosomal. II-2 received the allele from her affected father, I-1.

CED 5.3.A.2.v · Read this in Fix

1. AThe probability it has the trait is 1, as anyone with a dominant trait is homozygous.
   A student who thinks a dominant phenotype means a homozygous dominant genotype picks this. III-1 has an unaffected mother, so he must be heterozygous and passes the dominant allele to only half his children.
2. BThe probability it has the trait is 3/4, as dominant traits appear in 3 of every 4 offspring.
   A student who treats 3:1 as a property of every dominant trait picks this. 3:1 comes from two heterozygous parents; here the partner is homozygous recessive, so the cross Tt × tt gives 1/2.
3. CThe probability it has the trait is 1/2, as III-1 got a recessive allele from his mother. Correct
   III-1 is affected, so he has at least one dominant allele, and his mother II-3 is unaffected, so she gave him a recessive allele: III-1 is heterozygous. With an unaffected (homozygous recessive) partner, each child has a probability of 1/2 of receiving the dominant allele.
4. DThe probability it has the trait is 0, as both parents must pass on the allele.
   A student who thinks every trait needs the allele from both parents picks this: the unaffected partner has no dominant allele to pass on, so the student concludes that no child can have the trait. The allele is dominant, so one copy, from III-1, is enough; the probability is 1/2.

Working Let T = dominant allele. III-1 is affected, but his mother II-3 is unaffected (tt), so III-1 received t from her and is Tt. The partner is unaffected, so tt. Tt × tt: P(child Tt, affected) = 1/2.

CED 5.3.A.2.v · Read this in Fix