5.4 Non-Mendelian Genetics

1. AAlleles on one chromosome stay together unless a rare mutation changes one of them.
   A student who thinks new combinations of alleles come from mutation picks this. Mutation is far too rare to produce recombinants at several percent; the new combinations come from crossing over in meiosis.
2. BAlleles on one chromosome stay together unless crossing over occurs between the genes in meiosis. Correct
   Linked alleles travel together on one chromosome into gametes. Only when a crossover happens between the two genes in prophase I are they separated, and for genes close together this is infrequent, so recombinants are a small percentage.
3. CEach gamete receives both chromosomes of the pair, so the parental pairing of alleles is kept.
   A student who thinks gametes receive both homologous chromosomes picks this. Meiosis gives each gamete one chromosome of each pair; the parental combination is kept because both alleles are on that one chromosome.
4. DParental combinations carry dominant alleles, which are passed on more often than recessive alleles.
   A student who thinks dominant alleles are transmitted or found more often picks this. Each allele of a heterozygote goes to half the gametes, and parental classes can carry recessive alleles; linkage, not dominance, explains the pattern.

CED 5.4.A.1.i · Read this in Fix

1. AIn cross 1 and in cross 2, every son has the same eye color as his father.
   A student who expects sons to inherit X-linked traits from their fathers picks this. In cross 1 the father is white-eyed but the sons are red-eyed; in cross 2 the sons match their white-eyed mother, from whom they receive their X.
2. BCross 1 and cross 2 give the same results in the daughters and in the sons.
   A student who expects sex-linked genes to behave like autosomal ones picks this. The table shows different results: all sons are red-eyed in cross 1 but white-eyed in cross 2.
3. CWhite-eyed sons appear in cross 1 and in cross 2, but white-eyed daughters do not.
   A student who expects sex-linked traits to appear in the sons of every cross picks this. In cross 1 all sons are red-eyed; white-eyed sons appear only in cross 2.
4. DDaughters and sons differ in eye color in cross 2 but not in cross 1. Correct
   In cross 1 all daughters and all sons are red-eyed. In cross 2 all daughters are red-eyed and all sons white-eyed. So the sexes differ only in cross 2, and the two reciprocal crosses give different results, which points to an X-linked gene.

CED 5.4.A.2 · Read this in Fix

1. AAlleles of several genes add together to set the height of a plant.
   A student who confuses pleiotropy with polygenic inheritance picks this. Many genes affecting one trait is polygenic inheritance, the reverse of one gene affecting many traits.
2. BTwo genes close together on a chromosome are usually inherited together.
   A student who confuses pleiotropy with linkage picks this. Genes inherited together because they are on one chromosome are linked genes; pleiotropy involves a single gene.
3. COne gene in a rabbit has four alleles, giving four forms of coat color.
   A student who thinks pleiotropy means having many alleles picks this. Four alleles giving four forms of one trait is a case of multiple alleles; pleiotropy is one gene affecting several different traits.
4. DOne gene in a mouse affects both its coat color and ear development. Correct
   Pleiotropy is the expression of a single gene producing multiple traits or effects; here one gene affects two traits.

CED 5.4.A.3 · Read this in Fix

1. AFemales with the trait × males without it, and females without it × males without it
   A student who thinks a control is always a group in which no parent has the trait picks this. The second cross shows nothing about whether a father with the trait passes it on; the comparison must swap which parent has the trait.
2. BFemales with the trait × males without it, with only the sons of this cross scored
   A student who thinks a trait passed on by mothers must be X-linked picks this, looking for mother-to-son transmission. A single cross scored in sons cannot show whether the result depends on which parent has the trait; that needs the reciprocal cross.
3. CFemales with the trait × males without it, and females without it × males with it Correct
   These reciprocal crosses differ only in which parent shows the trait. If the trait is maternally inherited, offspring of females with the trait will show it and offspring of males with the trait will not.
4. DFemales with the trait × males without it, repeated with a much larger number of pairs
   A student who thinks more repeats of one cross can answer any question picks this. Repeating one cross makes its result more reliable but cannot show whether the result depends on which parent has the trait.

CED 5.4.A.4 · Read this in Fix

1. AFail to reject the null hypothesis, because χ² = 9.04 is less than the critical value of 9.49.
   A student who counts degrees of freedom as the number of classes picks this. With four classes there are 3 degrees of freedom, not 4, so the critical value is 7.81, and 9.04 exceeds it.
2. BFail to reject the null hypothesis, because χ² = 9.04 is greater than the critical value of 7.81.
   A student who thinks a χ² above the critical value shows a good fit picks this. A large χ² means a large difference between observed and expected counts; when it exceeds the critical value the null hypothesis is rejected.
3. CReject the null hypothesis, because none of the four classes has exactly the expected 100 offspring.
   A student who thinks any departure from the expected numbers disproves independent assortment picks this. Chance alone makes counts differ from 100; the null hypothesis is rejected here only because χ² = 9.04 exceeds the critical value of 7.81.
4. DReject the null hypothesis, because χ² = 9.04 is greater than the critical value of 7.81. Correct
   Total = 116 + 114 + 86 + 84 = 400, so each class is expected to have 100 offspring. χ² = (16² + 14² + 14² + 16²)/100 = 9.04. With 4 − 1 = 3 degrees of freedom the critical value at p = 0.05 is 7.81; 9.04 > 7.81, so the difference is larger than chance would be expected to produce and the null hypothesis is rejected. The data are consistent with the genes being linked.

Working Total offspring = 116 + 114 + 86 + 84 = 400. Independent assortment in GgLl × ggll predicts 1:1:1:1, so e = 100 for each class. χ² = (116 − 100)²/100 + (114 − 100)²/100 + (86 − 100)²/100 + (84 − 100)²/100 = 2.56 + 1.96 + 1.96 + 2.56 = 9.04. Degrees of freedom = 4 − 1 = 3; critical value at p = 0.05 = 7.81. 9.04 > 7.81, so reject the null hypothesis. (Using 4 degrees of freedom would give a critical value of 9.49.)

CED 5.4.A.1 · Read this in Fix

1. AThe four classes occur in a 9:3:3:1 ratio, and any deviation from this ratio is due to chance.
   A student who applies the 9:3:3:1 ratio to any cross of two genes picks this. That ratio comes from crossing two dihybrids; a cross with a homozygous recessive parent gives 1:1:1:1 when genes assort independently.
2. BThe two parental classes will contain more of the offspring than the two recombinant classes.
   A student who writes the expected effect as the null hypothesis picks this. Parental classes outnumbering recombinant classes is the prediction of linkage, the alternative hypothesis; the null hypothesis states no effect.
3. CThe four classes occur in a 1:1:1:1 ratio, and any deviation is due to chance. Correct
   The null hypothesis is the model of no effect: genes that assort independently give four equally likely gamete types from the heterozygous parent, so a testcross gives 1:1:1:1. If the data differ significantly from this, the null hypothesis is rejected, which supports linkage.
4. DEach of the four phenotypic classes will contain exactly one-quarter of the offspring.
   A student who expects ratios to give exact counts picks this. Random fertilization makes the counts vary around 1:1:1:1; the null hypothesis allows deviation due to chance, and the test measures it.

Working For a dihybrid testcross (AaBb × aabb), genes that assort independently give AB, Ab, aB and ab gametes in equal proportions from the heterozygote, so the expected phenotypic ratio is 1:1:1:1. The null hypothesis states that the observed counts fit 1:1:1:1, with deviations due to chance.

CED 5.4.A.1 · Read this in Fix

1. A25 map units Correct
   The two largest classes, tall white and dwarf purple, are parental, so the heterozygous plant had T with p on one chromosome and t with P on the other. The recombinants are tall purple and dwarf white: 26 + 24 = 50 of 200 offspring = 25%, which is 25 map units.
2. B75 map units
   A student who assumes that the both-dominant and both-recessive classes are always parental picks this, and so counts tall white and dwarf purple as recombinant: (76 + 74)/200 = 75%. The data show these are the two largest classes, so they are the parental classes.
3. C33 map units
   A student who divides recombinants by parentals instead of by the total picks this: 50/150 = 33%. Map distance is the percentage of all offspring that are recombinant, 50/200 = 25%.
4. D50 map units
   A student who takes the number of recombinant offspring as the map distance picks this. The 50 recombinants must be expressed as a percentage of the 200 offspring: 25%, or 25 map units.

Working Total = 76 + 74 + 26 + 24 = 200. The two largest classes (tall white, dwarf purple) are parental; the two smallest (tall purple, dwarf white) are recombinant. Recombinants = 26 + 24 = 50. Map distance = 50/200 × 100 = 25% = 25 map units.

CED 5.4.A.1.i · Read this in Fix

1. A2 of 4 sperm Correct
   After the exchange the two inner chromatids carry A with b and a with B. The two outer chromatids still carry the parental combinations A with B and a with b. Each chromatid goes to one sperm, so 2 of the 4 sperm are recombinant.
2. B4 of 4 sperm
   A student who thinks crossing over exchanges segments between the two whole chromosomes picks this. Only two chromatids, one from each homolog, take part in a crossover; their sister chromatids are unchanged.
3. C1 of 4 sperm
   A student who thinks crossing over moves a segment in one direction only picks this. A crossover is a reciprocal exchange, so both chromatids involved become recombinant.
4. D0 of 4 sperm
   A student who thinks genes on the same chromosome are always inherited together picks this. The crossover marked will separate A from B on two chromatids, so two sperm will be recombinant.

Working One crossover between non-sister chromatids exchanges segments between two of the four chromatids. After the crossover the chromatids carry A–B (parental), A–b (recombinant), a–B (recombinant) and a–b (parental). Each of the four chromatids goes to one of the four sperm, so 2 of 4 sperm are recombinant.

CED 5.4.A.1.i · Read this in Fix

1. ANeither testcross will give recombinant offspring, as genes on one chromosome are inherited as a unit.
   A student who thinks genes on one chromosome are always inherited together picks this. Crossing over separates linked genes; the map distances given are themselves percentages of recombinant offspring.
2. BThe D and F testcross will give more recombinant offspring, as genes farther apart cross over more often. Correct
   Map distance measures how often crossing over separates two genes: about 3% of offspring are expected to be recombinant for D and E, and about 30% for D and F.
3. CBoth testcrosses will give the four phenotypic classes equally, as each pair of genes assorts independently.
   A student who applies independent assortment to every pair of genes picks this. Genes on the same chromosome are linked, so parental classes outnumber recombinant classes, by far more for D and E than for D and F.
4. DBoth testcrosses will give equally few recombinants, as these arise only when a mutation occurs.
   A student who thinks new combinations of alleles come from mutation picks this. Recombinants come from crossing over, whose frequency rises with the distance between genes: about 3% for D and E but 30% for D and F.

CED 5.4.A.1.i · Read this in Fix

1. AIᴬ and Iᴮ are codominant, as the two alleles are found at equal frequencies in people.
   A student who links dominance to how common an allele is picks this. Codominance describes how the two alleles act in one person's cells; it says nothing about how common either allele is.
2. BIᴬ and Iᴮ show incomplete dominance, as the heterozygote differs from both homozygotes.
   A student who treats codominance and incomplete dominance as the same picks this. The heterozygote does differ from both homozygotes, but it shows both the A and the B carbohydrate rather than an intermediate phenotype, which is codominance.
3. CIᴬ and Iᴮ show incomplete dominance, as their products blend together in the heterozygote.
   A student who thinks heterozygote phenotypes come from blending picks this. The A and B carbohydrates are separate molecules, both present on the cell; nothing blends, and the alleles are passed on unchanged.
4. DIᴬ and Iᴮ are codominant, as the enzymes from both alleles are made and act in one cell. Correct
   Each allele's enzyme is produced in the heterozygote, so both carbohydrates appear on the same cell. The phenotype from both alleles is expressed, which is codominance.

CED 5.4.A.1.ii · Read this in Fix

1. AThe red allele is dominant to the white allele, and roan animals are heterozygous.
   A student who thinks one allele of every gene is dominant picks this. If red were dominant, heterozygotes would be red and a cross of two heterozygotes would give about 3 red : 1 white, not a separate roan class.
2. BThe red and white alleles show incomplete dominance, and roan is a blended phenotype.
   A student who treats codominance and incomplete dominance as the same picks this. Roan coats contain separate red hairs and white hairs, not hairs of one intermediate color, so both alleles' phenotypes are expressed.
3. CThe red and white alleles are codominant, and roan animals carry one of each allele. Correct
   Roan × roan gives red, roan and white in about a 1:2:1 ratio (52:101:47), the result expected from crossing two heterozygotes when the heterozygote has its own phenotype. Roan coats show both red and white hairs, so the phenotype from both alleles is expressed: codominance.
4. DRoan is produced by a third allele, and the roan animals are homozygous for that allele.
   A student who matches each phenotype to its own allele picks this. Animals homozygous for one allele would produce only roan offspring when crossed, but roan × roan produced red and white offspring as well.

CED 5.4.A.1.ii · Read this in Fix

1. AEach flower has separate red patches and white patches on its petals.
   A student who treats incomplete dominance as the same as codominance picks this. Separate red and white areas would show both phenotypes expressed side by side, which is codominance; incomplete dominance gives one intermediate shade.
2. BAll of the flowers are one shade, such as pink, between red and white. Correct
   In incomplete dominance neither allele masks the other, so every heterozygote has the same intermediate phenotype: a single shade between red and white.
3. CSome plants have red flowers and others have white, as dominance is incomplete.
   A student who reads 'incomplete' as 'not every time' picks this. Incomplete dominance does not mean dominance sometimes fails; it means every heterozygote has an intermediate phenotype.
4. DEach flower is the color produced by the more common allele in the population.
   A student who thinks dominance depends on how common an allele is picks this. A heterozygote's phenotype depends on how its two alleles act, not on allele frequencies in the population.

CED 5.4.A.1.iii · Read this in Fix

1. AAbout one red-flowered : two pink-flowered : one white-flowered
   A student who applies the 1:2:1 ratio to any cross involving incomplete dominance picks this. 1:2:1 comes from crossing two heterozygotes; a heterozygote × homozygote cross gives 1:1, and no red offspring are possible here.
2. BOffspring all of one shade, paler than the pink-flowered parents
   A student who thinks alleles blend in the heterozygote picks this. Alleles are not diluted; the pink parent passes on either the red or the white allele unchanged, so offspring are either pink or white.
3. CAbout equal numbers of pink-flowered and white-flowered offspring Correct
   If pink plants are heterozygous and white plants homozygous, the pink parent's gametes carry the red or the white allele equally often and every white-parent gamete carries the white allele. Half the offspring are heterozygous (pink) and half homozygous (white).
4. DPink-flowered offspring only, as each gamete carries both its alleles
   A student who thinks each gamete receives both of the parent's alleles picks this: every gamete of the pink parent would pass on both the red and the white allele, so every offspring would be pink. In meiosis each gamete receives one allele, so half the offspring of pink × white receive two white alleles and are white.

CED 5.4.A.1.iii · Read this in Fix

1. AII-3 must be a carrier, because her son III-1 is affected but she is not. Correct
   A son receives his only X chromosome from his mother. III-1 is affected, so his X carries the recessive allele and came from II-3. II-3 is unaffected, so her other X carries the dominant allele: she is a heterozygous carrier.
2. BII-4 must carry the allele, because his son III-1 is affected by the trait.
   A student who thinks sons inherit their X chromosome from their fathers picks this. II-4 gave III-1 a Y chromosome; III-1's X came from his mother. II-4 is unaffected, so his single X carries the dominant allele.
3. CII-2 cannot be a carrier, because she does not show the trait herself.
   A student who thinks one copy of a recessive X-linked allele shows in females picks this. A heterozygous female is unaffected; II-2's mother, I-2, is a carrier, so II-2 has a 1/2 chance of being one too.
4. DI-2 cannot be a carrier, because only males carry sex-linked alleles.
   A student who thinks sex-linked traits belong only to males picks this. I-2's son II-1 is affected and his father, I-1, is not, so II-1's X with the recessive allele came from I-2, who must be a carrier.

CED 5.4.A.2 · Read this in Fix

1. AFemales, because a female's one Z chromosome carries her only allele of the gene. Correct
   A ZW female has a single copy of the gene, so she shows the recessive trait whenever her one Z chromosome carries the allele. A ZZ male shows it only if both of his Z chromosomes carry it, which is less likely.
2. BMales, because males of every species have just one copy of sex-linked genes.
   A student who assumes the XY pattern of mammals applies to all species picks this. In birds the female is ZW and has one copy of Z-linked genes; the male is ZZ and has two.
3. CMales, because one recessive allele on either of their Z chromosomes is enough.
   A student who thinks one copy of a recessive sex-linked allele is expressed in the sex with two sex chromosomes picks this. A ZZ male with one recessive allele also has a dominant allele, so he does not show the trait.
4. DBoth sexes equally, as sex-chromosome genes are inherited like autosomal ones.
   A student who treats sex-linked genes like autosomal genes picks this. Males have two copies of Z-linked genes and females one, so the recessive trait shows more often in females.

CED 5.4.A.2 · Read this in Fix

1. AThey will have a defect in one organ, as each gene affects a single trait.
   A student who thinks each gene affects only one trait picks this. Gene K's protein is used for cilia in all three organs, so losing it can affect all of them.
2. BThey will have defects in all three organs: the gills, the kidneys and the inner ear. Correct
   All three organs depend on cilia, and kk fish cannot make the functional protein needed to build cilia, so all three are expected to be affected. One gene with several effects is pleiotropy.
3. CThey will have no organ defects, as the k allele is recessive and is not expressed.
   A student who thinks a recessive allele is never expressed picks this. A recessive allele is masked only when a dominant allele is present; kk fish have no K allele, so they cannot make the functional protein and cannot build normal cilia in any of the three organs.
4. DThey will gain new alleles that rebuild the cilia in the organs most in need.
   A student who thinks mutations arise when an organism needs them picks this. Mutations occur by chance, not in response to need, and a fish's cells do not produce useful alleles on demand.

CED 5.4.A.3 · Read this in Fix

1. AA nuclear gene for chlorophyll is switched on in the green cells and switched off in the white cells.
   A student who thinks all of a cell's genes are in the nucleus picks this. The mutation here is in chloroplast DNA, and the diagram shows white tissue arising where cells receive only mutant chloroplasts.
2. BEach daughter cell gets one of the two chloroplast alleles, as in meiosis, so half are white.
   A student who applies Mendelian segregation to chloroplast genes picks this. Chloroplasts are not inherited as a pair of alleles split equally between daughter cells; the diagram shows varying numbers of each kind shared out at random.
3. CChloroplasts in shaded cells mutate because those cells have no use for their chlorophyll.
   A student who thinks mutations arise where they are needed picks this. Mutations occur by chance; the diagram shows existing mutant chloroplasts being passed on, not new mutations arising.
4. DChloroplasts are shared out at random, so some cell lines end up with mutant chloroplasts alone. Correct
   The diagram shows unequal, random shares of the two kinds of chloroplasts in Cells 2 and 3, and Cell 4 has only mutant ones. A cell like Cell 4 and its descendants make no chlorophyll and form white tissue; cell lines that keep normal chloroplasts form green tissue.

CED 5.4.A.4.i · Read this in Fix

1. AX-linked recessive, because the trait passes from affected mothers to their sons.
   A student who thinks maternal transmission means X-linkage picks this. II-4's daughter III-4 is affected although her father II-5 is not; a daughter with an X-linked recessive trait must receive the allele from both parents.
2. BAutosomal dominant, because the trait appears in every generation of the family.
   A student who treats appearing in every generation as proof of dominance picks this. Mitochondrial traits also appear in every generation, and here every child of affected mothers but no child of the affected father is affected.
3. CMitochondrial, because affected mothers have affected children and affected fathers do not. Correct
   All children of the affected mothers I-1 and II-4 are affected, sons and daughters, and neither child of the affected father II-2 is affected. Mitochondria are transmitted by the egg, so this is the pattern expected for a trait determined by mitochondrial DNA.
4. DAutosomal recessive, because the trait is rare and is found in both males and females.
   A student who thinks rare traits are caused by recessive alleles picks this. A rare autosomal recessive trait would seldom appear in all children of an affected parent and an unrelated partner, as it does here.

CED 5.4.A.4.ii · Read this in Fix

1. ASeedling leaf color depends on the ovule's branch and not on the pollen's branch. Correct
   In every row, the seedlings match the branch that supplied the ovule (green ovule → green, white ovule → white, variegated ovule → a mixture), whatever the pollen source. This is the pattern expected because chloroplasts are transmitted in the ovule and not in the pollen.
2. BSeedling leaf color depends on which branch supplied the pollen, not the ovule.
   A student who thinks pollen passes chloroplasts to offspring picks this. Pollen from a green branch gave white seedlings on a white ovule, and white pollen gave green seedlings on a green ovule.
3. CGreen is dominant, as every cross with a green-branch parent gave only green seedlings.
   A student who expects Mendelian dominance picks this. A white ovule pollinated from a green branch gave all white seedlings, and a variegated ovule with green pollen gave a mixture.
4. DCrosses of green and white branches gave green and white seedlings in a 3:1 ratio.
   A student who expects a Mendelian ratio from any cross picks this. Green ovule × white pollen gave all green seedlings and white ovule × green pollen gave all white seedlings; no cross gave a 3:1 ratio.

CED 5.4.A.4.iii · Read this in Fix