Solution Manual For Genetics: From Genes to Genomes,, 7th Edition

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1-1chapter1Mendel’s Principles of HereditySynopsisChapter 1 covers the basic principles of inheritance that can be summarized as Mendel’s Laws ofSegregation (for one gene) and Independent Assortment (for more than one gene).Key termsgenes and alleles of genes – A gene determines a trait, and different alleles or forms of agene exist. The color gene in peas has two alleles: yellow and green.genotype and phenotype – Genotype is the genetic makeup of an organism (written asalleles of specific genes), while phenotype is how the organism looks.homozygous and heterozygous – When both alleles of a gene are the same, theindividual is homozygous for that gene (or pure breeding). If the two alleles aredifferent, the organism is heterozygous (also called a hybrid).dominant and recessive – The dominant allele is the one that controls phenotype in theheterozygous genotype; the recessive allele controls phenotype only in a homozygote.monohybrid or dihybrid cross – a cross between individuals who are both heterozygotesfor one gene (monohybrid) or for two genes (dihybrid).testcross – performed to determine if an individual with the dominant characteristic ishomozygous or heterozygous: An individual with the dominant phenotype butunknown genotype is crossed with an individual with the recessive phenotype.Key ratios3:1 – Ratio of progeny phenotypes in a cross between monohybrids[Aa×Aa→3A– (dominant phenotype) : 1aa (recessive phenotype)]1:2:1 – Ratio of progeny genotypes in a cross between monohybrids(Aa×Aa→1AA : 2Aa : 1aa)1:1 – Ratio of progeny genotypes in a cross between a heterozygote and a recessive homozygote(Aa×aa→1Aa : 1aa)1:0 – All progeny are the same phenotype. Can result from either of two cases:[AA×– –→A– (all dominant phenotype)][aa×aa→aa (all recessive phenotype)]9:3:3:1 – Ratio of progeny phenotypes in a dihybrid cross(Aa Bb×Aa Bb→9A–B– : 3A–bb : 3aa B – : 1aa bb)

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chapter11-2Problem SolvingThe essential component of solving most genetics problems is to DIAGRAM THE CROSS in aconsistent manner. Usually you will be given information about phenotypes, so the diagram wouldbe:Phenotype of one parent×phenotype of the other parent→phenotype(s) of progenyThe goal is to assign genotypes to the parents and then use these predicted genotypes to generatethe genotypes, phenotypes, and ratios of progeny. If the predicted progeny match the observeddata you were provided, then your genetic explanation is plausible.The points listed below will be particularly helpful in guiding your problem solving:•Remember that two alleles of each gene exist when describing the genotypes ofindividuals. But if you are describing gametes, remember that only one allele of eachgene is in a gamete.•You will need to determine whether a character is dominant or recessive. Two mainclues will help you answer this question.oFirst, if the parents of a cross are true breeding for the alternative characters of thetrait,lookatthephenotypeoftheF1progeny.Theirgenotypemustbeheterozygous, and their phenotype is thus determined by the dominant allele ofthe gene.oSecond, look at the F2progeny (that is, the progeny of the F1hybrids). The 3/4portion of the 3:1 phenotypic ratio indicates the dominant character.•You should recognize the need to set up a testcross (to establish the genotype of anindividual showing the dominant character by crossing this individual to a homozygotefor the recessive allele).•You must keep in mind the basic rules of probability:oProduct rule: If two outcomes must occur together as the result of independentevents, the probability of one outcome AND the other outcome is the product ofthe two individual probabilities.oSum rule: If there is more than one way in which an outcome can be produced,the probability of one OR the other occurring is the sum of the two mutuallyexclusive individual probabilities.•Be aware that sometimes you need to use conditional probability, meaning that anevent’s probability is influenced by its relationship to another event that has alreadyoccurred. You were introduced to conditional probability in Solved Problem III in thischapter, and several of the problems in Section 1.3 require this kind of thinking. Forexample, suppose you are given a pedigree diagram for a disease caused by a recessiveallele. You are asked to determine the chance that an unaffected individual is a carrier(Dd ), when both parents are carriers. As the cross that produced the unaffectedindividual isDd×Dd, you would expect the chance of aDd child to be 1/2. This is true,but it was not the question you were asked! You know something about the individual inquestion—which is that they are unaffected—they cannot bedd. This means that in thiscase, the 1DD : 2Dd : 1dd ratio changes to 1DD : 2Dd, and the chance is 2/3 that theunaffected individual is a carrier. Whensolvingprobability problems in pedigrees,

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chapter11-3always think carefully about what you know (and what you don’t know) about eachindividual.•Remember that Punnett squares are not the only means of analyzing a cross;branched-line diagrams and calculations of probabilities using the product and sumrules are more efficient ways of looking at complicated crosses involving more thanone or two genes.•You should be able to draw and interpret pedigrees. When the trait is rare, look forvertical patterns of inheritance characteristic of dominant traits, and horizontal patternsthat typify recessive traits. Check your work by assigning genotypes to all individuals inthe pedigree and verifying that these make sense.•The vocabulary problem (the first problem in the set) is a useful gauge of how well youknow the terms most crucial for your understanding of the chapter.Vocabulary1.a.phenotype4.observable characteristicb.alleles3.alternate forms of a genec.independent6.alleles of one gene separate into gametes randomlyassortmentwith respect to alleles of other genesd.gametes7.reproductive cells containing only one copy of eachgenee.gene11.the heritable entity that determines a characteristicf.segregation13.the separation of the two alleles of a gene intodifferent gametesg.heterozygote10.an individual with two different alleles of a geneh.dominant2.the allele expressed in the phenotype of theheterozygotei.F114.offspring of the P generationj.testcross9.the cross of an individual of ambiguous genotypewith a homozygous recessive individualk.genotype12.the alleles an individual hasl.recessive8.the allele that does not contribute to the phenotypeof the heterozygotem. dihybrid cross5.a cross between individuals both heterozygous fortwo genesn.homozygote1.having two identical alleles of a given gene

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chapter11-4Section 1.12.Prior to Mendel, people held two basic misconceptions about inheritance. First was thecommon idea of blended inheritance: that the parental characteristics become mixed inthe offspring and forever changed. Second, many people thought that one parentcontributes the most to an offspring’s inherited features. (For example, some peoplethought they saw a fully formed child in a human sperm.)In addition, people who studied inheritance did not approach the problem in anorganized way. They did not always control their crosses. They did not look at traits withclear-cut alternative characteristics. They did not start with pure-breeding lines. They didnot count the progeny types in their crosses. For these reasons, they could not developthe same insights as did Mendel.3.Several advantages exist for using peas for the study of inheritance:•Peas have a fairly rapid generation time (at least two generations per year ifgrown in the field, three or four generations per year if grown in greenhouses).•Peas can either self-fertilize or be crossed artificially by an experimenter.•Peas produce large numbers of offspring (hundreds per parent).•Peas can be maintained as pure-breeding lines, simplifying the ability toperform subsequent crosses.•Because peas have been maintained as inbred stocks, two easily distinguishedand discrete forms of many traits are known.•Peas are easy and inexpensive to grow.In contrast, studying genetics in humans has several disadvantages:•The generation time of humans is very long (roughly 20 years).•No self-fertilization occurs in humans, and it is not ethical to manipulatecrosses.•Humans produce only a small number of offspring per mating (usually onlyone) or per parent (almost always fewer than 20).•Although people who are homozygous for a trait do exist (analogous to pure-breeding stocks), homozygosity cannot be maintained because mating withanother individual is needed to produce the next generation.•Because human populations are not inbred, most human traits show acontinuum of phenotypes; only a few traits have two distinct forms.•People require a lot of expensive care to “grow”.One major advantage exists nonetheless for the study of genetics in humans: Becausemany inherited traits result in disease syndromes, and because the world’s populationnow exceeds 7 billion people, a very large number of people with diverse, variantphenotypes can be recognized. These variations are the raw material of genetic analysis.

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chapter11-5Section 1.24.a.Two phenotypes are seen in the second generation of this cross: normal andalbino. Thus, only one gene with two alleles is needed to control the phenotypesobserved. The 3:1 ratio of these phenotypes in the F2generation will be seen onlyif a single gene is involved.b.Note that the phenotype of the first generation progeny is normal color, and that inthe second generation, there is a ratio of 3 normal : 1 albino. Both observations showthat the allele controlling the normal phenotype (A ) is dominant to the allelecontrolling the albino phenotype (a ).c.In a testcross, an individual showing the dominant phenotype but that has anunknown genotype (the normal-colored female parent in this problem) is mated withan individual that shows the recessive phenotype and is therefore homozygous for therecessive allele. In this case, the individual with the recessive phenotype must be bothmale and an albino, so the male parent’s genotype isaa . The normally coloredoffspring must receive anA allele from the mother, so the genotype of the normaloffspring of the testcross isAa. The albino offspring must receive ana allele fromthe mother, so the genotype of the albino offspring of the testcross isaa . Thus, thefemale parent must be heterozygousAa .5.Because two different phenotypes result from the mating of two cats of the samephenotype, and because the ratio of the short-haired to long-haired progeny is 3:1, only asingle gene is involved, and the short-haired parent cats must have been heterozygous.The phenotype expressed in the heterozygotes (the parent cats) is the dominantphenotype. Therefore, short hair is dominant to long hair.6.a.Two affected individuals have an affected child and a normal child. This outcome isnot possible if the affected individuals were homozygous for a recessive alleleconferring piebald spotting, and if the trait is controlled by a single gene. Therefore,piebald must be the dominant characteristic.b.If the trait is dominant, the piebald parents could be either homozygous (PP ) orheterozygous (Pp). However, because the two affected individuals have an unaffectedchild (pp), they both must be heterozygous (Pp ). A diagram of the cross follows:piebald×piebald→1 piebald : 1 normalPpPpPpppNote that although the apparent ratio is 1:1, this is not a testcross but is instead a crossbetween two monohybrids. The reason for this discrepancy is that only two progenywere obtained, so this number is insufficient to establish what the true ratio would be(it should be 3:1) if many progeny resulted from the mating.7.You would conduct a testcross between your normal-winged fly (W–) and a short-wingedflythatmustbehomozygousrecessive(ww ). The possible results arediagrammed here; the first genotype in each cross is that of the normal-winged fly whosegenotype was originally unknown. Note that if the normal-winged fly is a homozygote,

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chapter11-6all the progeny should have normal wings, but if the normal-winged fly is aheterozygote, half the progeny should have normal wings and the other half shouldhave short wings.WW×ww→allWw (normal wings)Ww×ww→1/2Ww (normal wings) : 1/2ww (short wings)8.First diagram the crosses:closed×open→F1all open→F2145 open : 59 closedF1open×closed→81 open : 77 closedThe results of the crosses fit the pattern of inheritance of a single gene, with openbeing dominant to closed. The first cross is similar to those Mendel did with pure-breeding parents, although you were not provided with the information that the startingplants were true-breeding. The F1plants are open, indicating that open is dominant.The closed parent must be homozygous for the recessive allele. Because only onephenotype is seen among the F1plants, the open parent must be homozygous for thedominantallele.Thus,theparentalcucumberplantswereindeedtrue-breedinghomozygotes.Self-fertilization of the F1plants results in a 3:1 ratio of open : closed among the F2progeny. The 3:1 ratio in the F2shows that a single gene controls the trait and that theF1plants are all monohybrids (that is, they are heterozygotes).The final cross verifies that the F1plants from the first cross are heterozygotes becausethis testcross yields a 1:1 ratio of open: closed progeny. In summary, all the data areconsistent with the trait being determined by one gene with two alleles, and open beingthe dominant characteristic. Rewritten as genotypes for a gene with allelesO ando, thecrosses were:oo (closed)×OO (open)→F1Oo (open)→F2145O– (open): 59oo (closed)F1Oo (open)×oo (closed)→81Oo (open) : 77oo (closed)9.The dominant characteristic (short tail) is easier to eliminate from the population byselective breeding. The reason is you can recognize every animal that has inheritedtheshort tail allele, because only one such dominant allele is needed to see thecharacteristic. If you prevent all the short-tailed animals from mating, then the allelewould become extinct.On the other hand, the recessivedilute allele can be passed unrecognized fromgeneration to generation in heterozygous mice (who arecarriers ). The heterozygous miceare not dilute, so they cannot be distinguished from homozygous dominant mice withnormal coat color. You could prevent the homozygous recessive mice with the dilutecharacteristic from mating, but thedilute allele would remain among the carriers, whichyou could not recognize.10.The problem states that only one gene is involved in this trait, and that the dominantallele is dimpled (D ) while the recessive allele is nondimpled (d ). (We are usingMendelian symbols here instead of human symbols for genotypes to emphasize whichallele is dominant and which is recessive.)

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chapter11-7a.Diagram the cross described in this part of the problem:nondimpled♂×dimpled♀→proportion of dimpled F1?Note that the dimpled woman in this cross had add (nondimpled) mother, so thedimpled woman MUST be heterozygous. We can thus rediagram this cross withgenotypes:dd (nondimpled)♂×Dd (dimpled)♀→1/2Dd (dimpled) : 1/2dd (nondimpled)One half of the children produced by this couple would be dimpled.b.Diagram the cross:dimpled (D? )♂×nondimpled (dd )♀→nondimpled F1(dd )Because they have a nondimpled child (dd ), the man must have ad allele tocontribute to the offspring. The man is thus of genotypeDd .c.Diagram the cross:dimpled (D? )♂×nondimpled (dd )♀→8 F1, all dimpled (D–)TheD allele in the children must come from their father. The father could be eitherDD orDd, but it is most probable that the father’s genotype isDD . We cannot ruleout completely that the father is aDd heterozygote. However, if this was the case, theprobability that all 8 children would inherit theD allele from aDd parent is only(1/2)8= 1/256.11.a.The 3:1 ratio in the progeny of cross 1×4 suggests that a single gene controlsflowering time, and that late (F ) is dominant to early (f ).b.Plants 1 and 4 areFf because their progeny exhibit a 3 late (F–) :1 early (ff ) ratio.Plant 2 isf f because when crossed with plant 1 (Ff×ff ), the progeny are in a 1:1ratio of late (Ff ) to early (ff ). Plant 3 isFF because when crossed with either 1, 3,or 4, all the progeny are late (F−).c.Selfing of Plant 1 or Plant 4 (Ff ) would produce progeny in a phenotypic ratio of3 late : 1 early (3F−: 1f f ). Selfing of Plant 2 (f f ) would produce all early (f f )progeny, and selfing of Plant 3 would result in progeny that are all late (FF ).12.a.You need to realize that the results tabulated are unlike those of the plants in theprevious problem in an important way: The parents are not two individuals who had30–50 progeny. Instead, the entries in the table show the progeny of many pairs ofparents with the same characteristics, but who could have different genotypes. Thismeans that if sticky and dry are controlled by alternate alleles of a single gene, in anyrow of the table involving parents with the dominant phenotype, those parents couldactually be a mixture of homozygous dominant and heterozygous individuals.Thus, in a row where both parents have the dominant phenotype, some of thecrosses could beSs×Ss, while in a row where one parent has the dominantphenotype and the other parent has the recessive phenotype, some of the crossescould beSs×ss. Both cases could produce some progeny who will be homozygousrecessive (ss ). Only crosses between two homozygous recessive individuals will resultin progeny that all have the recessive phenotype. The only crosses that fit this

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chapter11-8criterion are in the row: dry×dry→all dry. Therefore, dry is the recessivephenotype (ss ) and sticky is the dominant phenotype (S – ).b.3:1 or 1:1 ratios are not observed in the table because some sticky individuals ineach of the first two rows areSS while others areSs .In the first row of the table, the sticky×sticky matings in this human populationare not simply crosses between heterozygotes, but instead a mix of three differentkinds of matings: between two heterozygotes (Ss×Ss), between two homozygotes(SS×SS ), and between a homozygote and heterozygote (SS×Ss ). The progeny fromthe two latter crosses obscure the 3:1 ratio that would result from crosses betweenheterozygotes.In the second row of the table, the sticky×dry matings include bothSs×ss andSS×ss. The progeny of the latter crosses obscure the 1:1 ratio that would result fromanySs×ss testcrosses in the same row.13.Diagram the cross:black×red→1 black : 1 redNo, you cannot tell how coat color is inherited from the results of this one mating.The 1:1 ratio indicates that this was a testcross—a cross between a heterozygote and ahomozygous recessive. However, we cannot know from the testcross whether red or blackis the dominant phenotype. To determine which phenotype is dominant, remember thatan animal with a recessive phenotype must be homozygous. Thus, if you mate severalred horses to each other and also mate several black horses to each other, the crossesthat always yield only offspring with the parental phenotype must have been betweenhomozygous recessives. For example, if all the black×black matings result in only blackoffspring, black is recessive. Some of the red×red crosses (that is, crosses betweenheterozygotes) would then result in both red and black offspring in a ratio of 3:1. Toestablish this point, you might have to do several red×red crosses, because some of thesecrosses could be between red horses homozygous for the dominant allele. You could ofcourse ensure that you were sampling heterozygotes by using the progeny of black×redcrosses (such as that described in the problem) for subsequent [black×black] or[red×red] crosses.14.a.1/6 because a die has 6 different sides.b.Three even numbers exist (2, 4, and 6). The probability of obtaining any one of theseis 1/6. Because the 3 events are mutually exclusive, use the sum rule: 1/6 + 1/6 + 1/6= 3/6 = 1/2.c.You must roll either a 3 or a 6, so 1/6 + 1/6 = 2/6 = 1/3.When thinking about probabilities involving 2 dice (in the diagram that follows, awhite one and apink one), it helps to realize that: (1) The probabilities calculatedfor rolling both dice simultaneously will be the same as those calculated for rollingthem in succession (first one, then the other), and (2) as shown in the diagram thatfollows, 36 different outcomes are possible:

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chapter11-9d.Each die is independent of the other, thus the product rule is used: 1/6×1/6 = 1/36.(Two 6s is one of the 36 possible outcomes.)e.The probability of getting an even number on one die is 3/6 = 1/2 [see part (b)]. Thisis also the probability of getting an odd number on the second die. This result couldhappen either of 2 ways—you could get the odd number first and the even numbersecond, or vice versa, and these are mutually exclusive possibilities. Thus, theprobability of both occurring is (1/2×1/2) + (1/2×1/2) = 1/4 + 1/4 = 1/2.Another way of thinking about this problem is that the first die can land any way(probability = 1), but the second die must show an odd number (probability = 1/2) ifthe first die was even, and an even number (probability = 1/2) if the first die was odd.Therefore, no matter how the first die lands, the second die has a 1/2 chance oflanding on a number that satisfies the stated criterion, and so the probability that thecriterion is satisfied is 1×1/2 = 1/2. (You can see in the diagram above that 18 of the36 possible outcomes satisfy the criterion that one die has an odd number and theother die has an even number.)f.The probability of any specific number on a die = 1/6. The probability of the samenumber on the other die = 1/6. The probability of both occurring at the same time is1/6×1/6 = 1/36. The same probability is true for the other 5 possible numbers on thedice. Thus, the probability of any of these mutually exclusive situations occurring is1/36 + 1/36 + 1/36 + 1/36 + 1/36 + 1/36 = 6/36 = 1/6.Another way of thinking of about this problem is that the first die can land anyway at all (probability = 1), but the second die must match the first one (probability= 1/6). The chance of both events happening is 1×1/6 = 1/6. (In other words, 6 ofthe 36 possible outcomes satisfy the criterion that both numbers are the same.)g.The probability of getting two numbers both over four is the probability of getting a 5or 6 on one die (1/6 + 1/6 = 1/3) and a 5 or 6 on the other die (1/3). The results forthe two dice are independent events, so 1/3×1/3 = 1/9. (In other words, 4 of the 36possible outcomes satisfy the criterion both numbers are over 4.)15.a.The probability of drawing a face card = 0.23 (= 12 face cards / 52 cards). Theprobability of drawing a red card = 0.5 (= 26 red cards / 52 cards). The probabilityof drawing a red face card = probability of a red card×probability of a face card= 0.23×0.5 = 0.12. (Alternatively, 6 red face cards / 52 cards = 0.12.)

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chapter11-10b.As seen in part (a), the probability of drawing a face card in a deck of 52 cards is 0.23.If after each draw, the card is returned to the deck, then the deck will have 52 cardsin each draw. Thus, the chance that all four cards picked are face cards is(0.23)4≈0.0028.c.The chance that the first card is a face card is (12/52). If that card was a face card,then the chance that the second card is a face card is (11/51), the chance that the thirdcard is also a face card is (10/50), and the chance that the fourth card is a face card is(9/49). (This is an example of conditional probability, because the probabilitieschange after each draw.) Thus, if the cards are not returned to the deck, the chancethat the first four cards picked are face cards is (12/52)×(11/51)×(10/50)×(9/49)≈0.0018.16.a.TheAa bb CC DD woman can produce 2 genetically different eggs that vary in theirallele of the first gene (A ora). She is homozygous for the other 3 genes and can onlymake eggs with theb C D alleles for these genes. Thus, using the product rule(because the inheritance of each gene is independent), she can make 2×1×1×1 =2 different types of gametes: (A b C D anda b C D).b.Using the same logic, anAA Bb Cc dd woman can produce 1×2×2×1 = 4 differenttypes of gametes:A (B orb) (C orc)d.c.A woman of genotypeAa Bb cc Dd can make 2×2×1×2 = 8 different types ofgametes: (A ora) (B orb)c (D ord ).d.A woman who is a quadruple heterozygote can make 2×2×2×2 = 16 differenttypes of gametes: (A ora) (B orb) (C orc) (D ord ). This problem [like those in parts(a–c) above] can also be visualized with a branched-line diagram.17.a.The probability of any particular progeny phenotype in this cross depends only on thegamete from the heterozygous parent. The reason is that the homozygous recessiveparent always provides a gamete with all recessive alleles (a b c d). The probabilitythatachildwillresemblethequadruplyheterozygousparentisthus:1/2A×1/2B×1/2C×1/2D = 1/16. The probability that a child will resemble thequadruply homozygous recessive parent is: 1/2a×1/2b×1/2c×1/2d = 1/16. Theprobability that a child will resemble either parent is then 1/16 + 1/16 = 1/8. Thiscross will produce 2 different characters for each gene or 2×2×2×2 = 16 potentialphenotypes.aBCDabCdabcdaBcDaBcdabCDabcDaBCdDDDDddddCCccBbaABCDAbCdAbcdABcDABcdAbCDAbcDABCdDDDDddddCCccBbA

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chapter11-11b.The probability of a child resembling the recessive parent is 0; the probability of achild resembling the dominant parent is 1×1×1×1 = 1. The probability that achild will resemble one of the two parents is 0 + 1 = 1. Only one phenotype ispossible in the progeny (dominant for all 4 genes), as (1)4= 1.c.The probability that a child would show the dominant character for any one gene is3/4 in this sort of cross (remember the 3 : 1 phenotypic ratio in the progeny ofmonohybrid crosses), so the probability of resembling the parent for the traitsassociated with all four genes is (3/4)4= 81/256.Two characters are possible foreach gene, so (2)4= 16 different kinds of progeny.d.All progeny will resemble their parents because all the alleles from both parents areidentical, so the probability = 1. Only one phenotype is possible for each gene in thiscross; because (1)4= 1, the child can have only one possible phenotype whenconsidering all four genes.18.a.The combination of alleles in the egg and sperm allows only one genotype for thezygote:aa Bb Cc DD Ee .b.Because the inheritance of each gene is independent, you can use the product rule todetermine the number of different gamete types possible:1×2×2×1×2 = 8 types of gametes. To figure out the genotypes of these gametes,consider the possibilities for each gene separately and then the possible combinationsof genes in a consistent order. For each gene the possibilities are:a, (B :b ), (C :c ),D, and (E :e ). The possibilities can be determined using the productrule. Thus, for the first 2 genes [a]×[B :b] gives [a B :a b]. Next, [a B :a b]×[C :c]gives [a B C :a B c :a b C :a b c]. Next, [a B C :a B c :a b C :a b c]×[D]gives [a B C D :a B c D :a b C D :a b c D]. And finally, [a B C D :a B c D :a b C D :a b c D]×[E :e] gives [(i)a B C D E : (ii)a B C D e : (iii)a B c D E :(iv)a B c D e : (v)a b C D E : (vi)a b C D e : (vii)a b c D E : (viii)a b c D e ].This problem can also be visualized with a branched-line diagram:aBCDEabCDeabcDeaBcDEaBcDeabCDEabcDEaBCDeEEEEeEeeCCccBbaDDDD

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chapter11-1219.Your friend is wrong. As each of these triplets is an independent event, we use theproduct rule to determine the probability of 3 boys: P(BBB) = 1/2×1/2×1/2 = 1/8. Theprobability of 3 girls is the same: P(GGG) = 1/8. However, the probability of 2 boys and1 girl is greater than 1/8 because it includes three different birth orders—BBG, BGB,GBB—each with a probability of 1/8. Thus, we use the sum rule to determine theprobability of 2 boys and 1 girl in any order: 1/8 + 1/8 + 1/8 = 3/8. The same logicapplies to two girls and a boy, the probability of which is also 3/8. Notice that because3 boys, 3 girls, 2 boys + 1 girl, and 2 girls + 1 boy are the only options, the sum of theirindividual probabilities is 1 (that is, 1/8 + 1/8 + 3/8 + 3/8 = 1).20.The first two parts of this problem involve the probability of occurrence of twoindependent traits: the sex of a child and galactosemia. The parents are heterozygous forgalactosemia, so a 1/4 chance exists that a child will be affected (that is, homozygousrecessive). The probability that a child is a girl is 1/2. The probability of an affected girl istherefore 1/2×1/4 = 1/8.a.Fraternal(nonidentical)twinsresultfromtwoindependentfertilizationsandtherefore, the probability that both will be girls with galactosemia is the product oftheir individual probabilities (see above); 1/8×1/8 = 1/64.b.For identical twins, one fertilization gave rise to two individuals. The probability thatboth are girls with galactosemia is 1/8.For parts c–g, remember that each child is an independent fertilization. The sex of thechildren is not an issue in these parts of the problem.c.Both parents are carriers (heterozygous), so the probability of having an unaffectedchild is 3/4. The probability of 4 unaffected children is 3/4×3/4×3/4×3/4 = 81/256.d.The probability that at least one child is affected is the chance of all possible outcomesexcept the one mentioned in part (c). Thus, the probability is 1−81/256 = 175/256.Note that this general strategy for solving problems, where you first calculate theprobability of all outcomes except the one of interest, and then subtract that numberfrom 1, is often useful for problems where direct calculation of the probability ofinterest is difficult.e.The probability of an affected child is 1/4 while the probability of an unaffected childis 3/4. Therefore, 1/4×1/4×3/4×3/4 = 9/256.f.The probability of 2 affected children and 1 unaffected child in any one particularbirth order is 1/4×1/4×3/4 = 3/64. There are 3 mutually exclusive birth orders thatcould produce 2 affected children and 1 unaffected child—unaffected child first born,unaffected child second born, and unaffected child third born. Thus, the chance that2 out of 3 children will be affected is 3/64 + 3/64 + 3/64 = 9/64.g.The genotype of any particular child is independent of all others, so the probabilityof an affected child is 1/4.21.Diagram the cross, whereP is the normal pigmentation allele andp is the albino allele:normal (P? )×normal (P? )→albino (pp)

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chapter11-13An albino must be homozygous recessivepp. The parents are normal in pigmentationand therefore could bePP orPp. Because they have an albino child, both parents mustbe carriers (Pp ). The probability that their next child will bepp is therefore 1/4.22.Diagram the cross:yellow round×yellow round→156 yellow round : 54 yellow wrinkledThe ratio for seed shape is 156 round : 54 wrinkled = 3 round : 1 wrinkled. The parentsmust therefore have been heterozygous (Rr ) for the pea shape gene. All theoffspring areyellow and therefore areYy orYY. The parent plants wereY –Rr×YY Rr . Becauseno green (yy ) progeny exist, you know at least one of the parents must have beenYY .23.Diagram the cross:smooth black♂×rough white♀→F1rough black→F28 smooth white : 25 smooth black : 23 rough white : 69 rough blacka.The 9:3:3:1 phenotypic ratio in the F2gives away the answer to this problem. This F2ratio tells you that the F1must be dihybrids, and that rough and black are dominantto smooth and white. Let’s use the following symbols:R = rough,r = smooth;B = black,b = white. The F1are thusRr Bb, and because all the F1have the samephenotype, the parents must have been homozygotes:rr BB (smooth, black) andRR bb (rough, white). The F2are:R– B– (rough black),R– bb (rough white),rr BB (smooth black), andrr bb (smooth white).b.An F1male is heterozygous for both genes (Rr Bb). The smooth white F2female mustbe homozygous recessive (rr bb). Thus,Rr Bb×rr bb→1/2Rr (rough) :1/2rr (smooth) and 1/2Bb (black) : 1/2bb (white). The inheritance of these genes isindependent, so apply the product rule to find the expected phenotypic ratios amongthe progeny: 1/4 rough black (Rr Bb) : 1/4 rough white (Rr bb) : 1/4 smooth black(rr Bb) : 1/4 smooth white (rr bb).24.Diagram the cross:YY rr×yy RR→allYy Rr→9/16Y–R– (yellow round) : 3/16Y–rr(yellow wrinkled) : 3/16yy R– (green round) : 1/16yy rr (green wrinkled)Each F2pea results from a separate fertilization and so a pod may have peas of differentphenotypes. The probability of 7 yellow round F2peas is (9/16)7= 4,782,969/268,435,456≈0.018.25. a.First diagram the cross, and then figure out the progeny ratios for each gene:Aa Tt×Aa Tt→3/4A– (achoo) : 1/4aa (non-achoo) and3/4T– (trembling) : 1/4tt (non-trembling)The probability that a child will beA– (and have achoo syndrome) is independent ofthe probability that it will lack a trembling chin, so the probability of a child withachoo syndrome but without trembling chin is 3/4A–×1/4tt = 3/16.b.The probability that a child would have neither dominant characteristic is 1/4aa×1/4tt = 1/16.

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Solution Manual For Genetics: From Genes to Genomes,, 7th Edition - Page 15 preview image

chapter11-1426.The F1must be heterozygous for all the genes because the parents were pure breeding(homozygous). The appearance of the F1establishes that the dominant characters forthe four traits are tall, purple flowers, axial flowers and green pods.a.If you cross two heterozygous F1, both dominant and recessive characters for eachtrait will appear among the progeny. Thus, you expect 2×2×2×2 = 16 differentphenotypes when considering the four traits together. The possibilities can bedetermined using the product rule with the pairs of characters for each gene, becausethe traits are inherited independently. Thus: [tall : dwarf]×[green : yellow] gives [tallgreen : tall yellow : dwarf green : dwarf yellow]×[purple : white] gives [tall greenpurple : tall yellow purple : dwarf green purple : dwarf yellow purple : tall green white: tall yellow white : dwarf green white : dwarf yellow white]×[terminal : axial] whichgives (i) tall green purple terminal : (ii) tall yellow purple terminal : (iii) dwarfgreen purple terminal : (iv) dwarf yellow purple terminal : (v) tall green whiteterminal : (vi) tall yellow white terminal : (vii) dwarf green white terminal : (viii)dwarf yellow white terminal : (ix) tall green purple axial : (x) tall yellow purpleaxial : (xi) dwarf green purple axial : (xii) dwarf yellow purple axial : (xiii) tallgreen white axial : (xiv) tall yellow white axial : (xv) dwarf green white axial : (xvi)dwarf yellow white axial. The possibilities can also be determined using a branched-line diagram:b.Designate the alleles:T = tall,t = dwarf;G = green;g = yellow;P = purple,p = white;A = axial,a = terminal. The crossTt Gg Pp Aa (an F1plant)×tt gg pp AA (the dwarfparent) will produce 2 characters for the tall, green and purple genes, but only 1character (axial) for the fourth gene or 2×2×2×1 = 8 different phenotypes. Thefirst 3 genes will give 1 dominant : 1 recessive ratios of the characters, because this isin effect a testcross for each gene. Thus, the proportion of each phenotype in theprogeny will be 1/2×1/2×1/2×1 = 1/8.Using either of the methods described in part (a), the progeny will be 1/8 tallgreen purple axial : 1/8 tall yellow purple axial : 1/8 dwarf green purple axial :talltall, green, purple, axialtall, green, purple, terminaltall, green, white, axialtall, green, white, terminaltall, yellow, purple, axialtall, yellow, purple, terminaltall, yellow, white, axialtall, yellow, white, terminalaxialterminalterminalaxialaxialaxialterminalterminalpurplepurplewhitewhitegreenyellowdwarfdwarf, green, purple, axialdwarf, green, purple, terminaldwarf, green, white, axialdwarf, green, white, terminaldwarf, yellow, purple, axialdwarf, yellow, purple, terminaldwarf, yellow, white, axialdwarf, yellow, white, terminalaxialaxialaxialaxialterminalterminalterminalterminalpurplepurplewhitewhitegreenyellow

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Solution Manual For Genetics: From Genes to Genomes,, 7th Edition - Page 16 preview image

chapter11-151/8 dwarf yellow purple axial : 1/8 tall green white axial : 1/8 tall yellow whiteaxial : 1/8 dwarf green white axial : 1/8 dwarf yellow white axial.27.For each separate cross, determine the number of genes involved. Remember that theexistence of 4 phenotypic classes in the progeny means that 2 genes control thephenotypes. Next, determine the phenotypic ratio for each gene separately. A 3:1phenotypic ratio of progeny tells you which character is dominant and that both parentswere heterozygous. In contrast, a 1:1 ratio results from a testcross where the dominantparent was heterozygous.a.Two genes with alternate alleles are at play in this cross (four phenotypes). One genecontrols flower color (purple or white) with a phenotypic ratio in the offspring of94 + 28 = 122 purple : 32 + 11 = 43 white or ~3 purple : ~1 white. The second genecontrols pod texture (spiny or smooth) with a phenotypic ratio in the offspring of94 + 32 =126 spiny : 28 + 11 = 39 smooth or ~3 spiny : ~1 smooth. Thus, designatethe allelesP = purple,p = white;S = spiny,s = smooth. This is a straightforwarddihybrid cross:Pp Ss×Pp Ss→9P– S– : 3P– ss : 3pp S– : 1pp ss.b.The 1 spiny : 1 smooth ratio indicates a testcross for the pod texture gene. Because allprogeny were purple, at least one parent plant must have been homozygous for theP allele of the flower color gene. The cross was eitherPP Ss×P– ssorP– Ss×PP ss.c.This row is similar to part (b), but here all the progeny were spiny so at least oneparent must have been homozygous for theS allele. The 1 purple : 1 white testcrossratio indicates that the parents were eitherPp S–×pp SS orPp SS×pp S–.d.For color, there are 89 + 31 = 120 purple : 92 + 27 = 119 white. A 1 purple : 1 whiteratio denotes a testcross. For texture, there are 89 + 92 = 181 spiny : 31 + 27 =58 smooth, or a 3 spiny : 1 smooth ratio, indicating that the parents were bothheterozygous for theS gene. The genotypes of the parents werePp Ss×pp Ss.e.A 3 purple : 1 white ratio exists among the progeny, so the parents were bothheterozygous for theP gene. All progeny had smooth pods, so the parents were bothhomozygous recessivess. The genotypes of the parents arePp ss×Pp ss .f.A 3 spiny : 1 smooth ratio exists, indicative of a cross between heterozygotes(Ss×Ss). All progeny were white, so the parents must have been homozygousrecessivepp. The genotypes of the parents arepp Ss×pp Ss.28.Three traits (each controlled by a different gene) are analyzed in this cross. While we canusually tell which alleles are dominant from the phenotype of the heterozygote, we arenot told the phenotype of the heterozygote in this case. Instead, use the phenotypic ratiosfor each trait to determine which allele is dominant and which is recessive for each gene.Consider height first. There are 272 + 92 + 88 + 35 = 487 tall plants and 93 + 31 + 29 +11 = 164 dwarf plants. This is a ratio of ~3 tall : ~1 dwarf, indicating that tall is dominant.Next consider pod shape, where there are 272 + 92 + 93 + 31 = 488 inflated pods and88 + 35 + 29 + 11 = 163 flat pods, or approximately 3 inflated : 1 flat, so inflated isdominant. Finally, consider flower color. There were 272 + 88 + 93 + 29 + 11 = 493

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