Answer Section 467

c. Homologous pairs line up independently atmetaphase plated. Homologous pairs of chromosomes separateand homologs move toward opposite poles, sis-ter chromatids remain attached at centromeree. Haploid set of chromosomes, each consistingof two sister chromatids, aligns at metaphase

plate; sister chromatids not identical due tocrossing overf. Sister chromatids separate and move to op-posite poles as chromosomes

3.

results in

both involve cells withhaploid set ofchromosomes

results in

in anaphaseof mitosis

produc~g

then inanaphase II

of

L.._....:..... __ -.:=::....---------- ProdUCing-----------. ••IL..._fo_u_r_cells

ANSWERS TO TEST YOUR KNOWLEDGEMultiple Choice:1. b 4. c2 d 5. e3. e 6. b

10. b11. a12. b

16. b17. d18. b

7. a8. c9. b

13. c14. d15. e

19. e20. b

CHAPTER 14: MENDEL AND THE GENE IDEA

INTERACTIVE QUESTIONS You can use a Punnett square to determine the

14.1 a. R g. r expected outcome of a test cross, but a short-

b. r h. Rr G cut is to use only 1 column for the recessiveindividual's gametes, since they produce onlyc. Fl Generation i. Rr G one type of gamete.d. Rr j, rr e

e. F2 Generation k. 3 round:l wrinkled 14.3 a. all tall purple plantsf. R 1. 1RR:2Rr:lrr b. TtPp

14.2 a. all tall (Tt) plants c. TP, Tp, tP, tpb. 1:1 tall (Tt) to dwarf (tt)

468 Answer Section

d. sperm

1/4@ l/ll'» ll/@ 1/l!iJ

1/4@ ITPP ITPp TtPP TtPp

lIlt» ITPp ITpp TtPp Ttpp<JJ0000Q)

l/lii!J TtPP TtPp

l/l!iJ TtPp Ttpp

e. 9 tall purple:3 tall white:3 dwarf purple:1dwarf white£. 12:4 or 3:1 tall to dwarf;12:4 or 3:1 purple to white

14.4 Dihybrids produce four types of gametes. Theprobability of getting both recessive alleles in a~amete is 7'4, and for two such gametes to join isY4 X 7'4, or %6'

14.5 a. Consider the outcome for each gene as amonohybrid cross. The probability that a cross ofAa X Aa will produce an A_ offspring is %. Theprobability that a cross of Bb X bbwill produce aB_ offspring is 7'2' The probability that a cross of ccX CC will produce a C_ offspring is 1. To have allof these events occur simultaneously, multiplytheir probabilities: % X 7'2 X 1 = %.b. Offspring could be A_bbC--, aaB_C--, orA_B_C. The genotype A_B_cc is not possible. Cany.ou see why?

Probability of A_bbC_ = % X 7'2 X 1 = %Probability of aaB_C_ = % X % X 1 = 1/8

Probability of A_B_C_ = % X 7'2 X 1 = %Probability of offspring showing at least twodominant traits is the sum of these independentprobabilities, or 7/8,

c. There is only one type of offspring that canshow only one dominant trait (aabbCc). Can yousee why? Its probability is 7'4 X 7'2 X 1 = lis.

14.6 a. A: [A[A and [Aib. B: [B[B and [Bi

c. AB: [A[B

d. 0: ii

14.7 The ratio of offspring from this MmBb X MmBbcross would be 9:3:4, a common ratio when onegene is epistatic to another. All epistatic ratiosare modified versions of 9:3:3:1.

- --~ --~ - --

14.8 a. The parental cross produced 25-cm tall -=-.plants, all AaBbCc plants with 3 units of 5 c::::added to the base height of 10 cm.b. As a general rule in the polygenic inheriof a quantitative character, the numberphenotypic classes resulting from a croheterozygotes equals the number of allinvolved plus one. In this case, 6 all(AaBbCc) + 1 = 7. So, there will be 7 dillphenotypic classes in the F2 among .64 possible combinations of the 8 types of -=-gametes. (See why you wouldn't want to "'-through a Punnett square to figure thatThese 7 classes will go from 6 dominant(40 cm), 5 dominant (35 cm), 4 dominant (30and so on, to all 6 recessive alleles (10 crn).

14.9 a. This trait is recessive. If it were dominathen albinism would be present in every 00en:~tion, and it would be impossible to havechildren with two nonalbino (homozvgeesrecessive) parents.b. father Aa; mother Aa, because neitheris albino and they have albino offspring (aac. mate 1 AA (probably); mate 2 Aa; gran4Aad. The genotype of son 3 could be AA orhis wife is AA, then he could be Aa (both hisents are carriers) and the recessive allelewould be expressed in his offspring. Even ::and his wife were both carriers (heterozyg .there would be a 243/1024 (% X % X % X % X 1"24% chance that all five children wouldnormally pigmented.

14.10 a. 7'4b. %. Of offspring with a normal pheno . i:

% would be predicted to be heterozygotesthus, carriers of the recessive allele.

14.11 Both sets of prospective grandparents ill

have been carriers. The prospective parennot have the disorder so they are not horngous recessive. Thus, each has a % chance ofing a heterozygote carrier. The probabilityboth parents are carriers is % X % = %;chance that two heterozygotes will haverecessive homozygous child is 7'4' The 0,chance that a child will inherit the disease ..:

Phenotype Genotype Ratio

Black M_B_ %X%=%6

Brown M_bb % X 1/4 = 3/16White mm- % X 1= 14Od'I6

% X 14= %. Should this couple have a baby thathas the disease, this would establish that theyare both carriers, and the chance that a subse-quent child would have the disease is 14,

VGGESTED ANSWERS TO STRUCTUREOUR KNOWLEDGE

1. Mendel's law of segregation occurs inanaphase I, when alleles segregate as ho-mologs move to opposite poles of the cell. Thetwo cells formed from this division have one-half the number of chromosomes and one copyof each gene (although sister chromatids stillhave to separate in anaphase II).Mendel's lawof independent assortment relates to the liningup of attached homologous chromosomes atthe equatorial plate in a random fashion dur-ing metaphase 1. Genes on different chromo-somes will assort independently into gametes.

2. Aa = 2 different gametes AaBb = 4 gametesAaBbCc = 8 gametes AABbCc = 4 gametesA general formula is Z", where n is the numberof gene loci that are heterozygous.

3. Always look to the Fl heterozygote. If allelesshow complete dominance/recessiveness, thenthe heterozygote will have a phenotype identi-cal to one parent. In incomplete dominance, theFl phenotype will be intermediate between thatof the parents. If codominant, both alleles willbe expressed in the heterozygote.

ANSWERS TO GENETICS PROBLEMS

1. White alleles are dominant to yellow alleles. Ifyellow were dominant, then you should be ableto get white squash from a cross of two yellowheterozygotes.

2. a. % [12(to get AA) X 12(bb)]b. % [% (aa) X % (BB)]c. % [1 (Aa) X 12(Bb) X 1 (Cc)]d. %2 [% (aa) X % (bb) X % (cc)]

3. Since flower color shows incomplete domi-nance, use symbols such as CR and CW for thosealleles. There will be six phenotypic classes inthe F2instead of the normal four classes foundin a 9:3:3:1 ratio. You could find the answerwith a Punnett square, but multiplying theprobabilities of the monohybrid crosses is moreefficient.

Tall redTall pink

% X 14= 3/16

% X 12= % or %6

Answer Section 469

Tallwhite LCwCw % X 14= %6Dwarf red tt CRCR % X % = %6Dwarf pink tt CRCW 14X 12= 18or116

Dwarf white tt CwCw 14X 14= 1164. Determine the possible genotypes of the

mother and child. Then find the blood groupsfor the father that could not have resulted in achild with the indicated blood group.a. no groups exonerated d. AB onlyb. A or 0 e. Bor 0c. AorO

5. The parents are CcBb and Ccbb. Right away youknow that all cc offspring will die and no BBblack offspring are possible because one parentis bb. Only four phenotypic classes are possible.Determine the proportion of each type by ap-plying the law of multiplication.

Lethal (cc _) 1/4 of embryos donot develop

Normal brown (CCBb) 14X 12= %Normal white (CCbb) 14X 12= YsDeformed brown (CcBb) 12X 12= 14Deformed white (Ccbb) 12X 12= 14Ratio of viable offspring: 1:1:2:2

6. Father's genotype must be Pp since poly-dactyly is dominant and he has had one normalchild. Mother's genotype is pp. The chance ofthe next child having normal digits is % or 50%because the mother can only donate a p alleleand there is a 50% chance that the father willdonate a p allele.

7. The genotypes of the puppies were % B_S-,% B_ss, % bbS-, and % bbss. Because recessivetraits show up in the offspring, both parents hadto have had at least one recessive allele for bothgenes. Black.chestnut occurs in a 6:2 or 3:1 ratio,indicating a heterozygous cross. Solid:spottedoccurs in a 4:4 or 1:1 ratio, indicating a crossbetween a heterozygote and a homozygousrecessive. Parental genotypes were BbSs X Bbss.

8. The 1:1 ratio of the first cross indicates a crossbetween a heterozygote and a homozygote.The second cross indicates that the hairlesshamsters cannot have a homozygous genotype,because then only hairless hamsters would beproduced. Let us say that hairless hamsters areHh and normal-haired are HH. The secondcross of heterozygotes should yield a 1:2:1genotype ratio. The 1:2 ratio indicates that thehomozygous recessive genotype may be lethal,with embryos that are hh never developing.

470 Answer Section

9. a. Taillength appears to be a quantitative char-acter.A general rule for the number of alleles ina cross involving a quantitative character is oneless than the number of phenotypic classes.Here there are five phenotypic classes, thusfour alleles or two gene pairs. In this cross, thephenotypic ratio is approximately 1:4:6:4:1.Theratio base of 16 also indicates a dihybrid cross.Each dominant gene appears to add 5 em in taillength. The lS-cm pigs are double recessives;the 3S-cmpigs are double dominants.b. A cross of a lS-cm and a 30-cmpig would beequivalent to aabb X AABb. Offspring wouldhave either one or two dominant alleles andwould have tail lengths of 20 or 25 cm.

10. a. The black parent would be CCh. The

Himalayan parent could be either ChCh or Chc.First list the alleles in order of dominance: C,crc: c. To approach this problem you shouldwrite down as much of the genotype as you canbe sure about for each phenotype involved. Theparents were black (C_) and Himalayan (ChJ.The offspring genotypes would be black (C_)and Chinchilla (cch_). Right away you can tellthat the black parent must have been cc= be-cause half of the offspring are Chinchilla; cch isdominant over Ch and cand could not have beenpresent in the Himalayan parent. The ~enotypeof this parent could be either cc: or C c.A test-cross with a white rabbit would be necessary todetermine this genotype.b. You cannot definitely determine the geno-type of the parents in the second cross. Youcan,however, eliminate some genotypes. The blackparent could not be CCor cc=. because both theC and Ch alleles are dominant to c, andHimalayan offspring were produced. The Chin-chilla parent could not be cchCch for the samereason. One or both parents had to have Ch astheir second allele;one, but not both, might havec as their second allele.

11. a. First figure out possible genotypes: __ E_ =

golden (any B combination with at least 1 E),B.ee = black, bbee = brown. All you know at thestart is that both parents are __ E_. Since yousee black and brown offspring, you know thatboth parents had to be heterozygous for Ee, andat least one was heterozygous for Bb (to getblack and brown offspring). Since blackand brown are in a 1:1 ratio (like a testcrossratio), then one dog was Bb and the other wasbb. You need to consider the results from thesecond cross to know which dog was Bb.

b. For the second cross, you now know thatDog 2 is ?bEe and Dog 3 is B?ee. A ratio of ap-proximately 3:1 black to brown looks like across of heterozygotes, so both parents must beBb. So Dog 3 is Bbee, and Dog 2 must be BbEe.That means that Dog 1must be bbEe.

12.(At least one of thesegrandparents musthave been Tt)

13. Since the parents were true-breeding for twocharacters, the Fl would be dihybrids. Since allF, had red, terminal flowers, those two traitsmust be dominant and their genotypes couldbe represented as RrTt. One would predict anF2 phenotypic ratio of 9 red, terminal:3 red,axial:3 white, terminal:1 white, axial. If 100off-spring were counted, one would expectapproximately 19(%6 X 100) plants with red,axial flowers.

14. Because the F2generation is a ratio based on 16,one can conclude that two genes are involvedin the inheritance of this flower color character.The parent plants were probably double ho-mozygotes (AABB for red and aabb for white).The all-red Fl must be AaBb, and the red colormust require at least one dominant allele ofboth genes. The 9:6:1 ratio of the F2 indicatesthat 9 plants had at least one dominant allele ofboth genes to produce the red color; 1plant wasdoubly homozygous recessive and thus white;and the 6 pale purple were caused by having atleast one dominant allele of only one of the twogenes involved in flower color (A_bb or aaBj

ANSWERS TO TEST YOUR KNOWLEDGE

Matching:1. I 3. F 5. L 7. H 9. C2. A 4. E 6. J 8. D 10. K

Multiple Choice:1. c 5. b 9. b 13. d2. a 6. d 10. c 14. d3. c 7. c 11. d 15. c4. d* 8. d 12. a 16. b

* There are three different ways to get this outcome:HHT, HTH, THH. Each outcome has a probability of ~

Answer Section

CHAPTER 15: THE CHROMOSOMAL BASIS OF INHERITANCE

INTERACTIVE QUESTIONS

15.1

I F2 Generation I Xw+ yova spermxw+ !Xw+xw Xw+y

Xw xr x: XWy

Phenotype

red-eyed female red-eyed female red-eyed male white-eyed male

GenotypeXW+Xw+

15.2 The gene is sex-linked, so a good notation is XN,

Xn, and Y so that you will remember that theY does not carry the gene. Capital N indicatesnormal sight. Genotypes are:

1. XNy2. XNXn

,3. XNXNor xNxn (probably XNXN since four sonsare X'Y)

4. XNXn5. XNy6. Xny7. XNXn

15.3 a. tall, purple-flowered and dwarf, white-floweredb. tall, white-flowered and dwarf, purple-flowered

15.4 If linked genes have their loci close together onthe same chromosome, they travel togetherduring meiosis and more parental offspring areproduced. Recombinants are the result of cross-ing over between nonsister chromatids of ho-mologous chromosomes.

15.5 Solving a linkage problem is often a matter oftrial and error. Sometimes it helps to layout theloci with the greatest distance between them andfit the other genes between or on either side byadding or subtracting map unit distances.

--------15--------m---9-j---6-1-6---k

---------12---------15.6 a. An organism with a trisomy (2n + 1) has an

extra copy of one chromosome, usually causedby a nondisjunction during meiosis. A triploid~rganism (3n) has an extra set of chromosomes,

possibly caused by a total nondisjunctiongamete formation.b. The genetic balance caused by the trisonwould be more disrupted than that of <organism with a complete extra set of chromsomes.

15.7 translocation, deletion, and inversion

15.8 Aneuploidies of sex chromosomes appear 1upset genetic balance less, perhaps becauseatively few genes are located on the Y chrosome and extra X chromosomes are inactivaas Barr bodies.

SUGGESTED ANSWERS TO STRUCTUREYOUR KNOWLEDGE

1. Genes that are not linked assort independendand the ratio of offspring from a testcross, .dihybrid heterozygote should be 1:1:1:1 (Aa -aabb gives AaBb, Aabb, aaBb, aabb offspringrecombination frequency of 50%-50% ofspring are recombinants). Genes that are . ,and do not cross over should produce a 1:1in this testcross (AaBb and aabb because aerozygote derived from a parental cross of _'"1..-

X aabb produces only AB and ab gametescrossovers between distant genes almost ahoccur, the heterozygote will producequantities of AB, Ab, aB, and ab gametes, andgenotype ratio of offspring will be 1:1:1:1~same as it is for unlinked genes. Because earecombination frequency is equal to 1mapthis measurement ceases to be meaningful <I

relative distances of 50 or more map ..However, crosses with intermediate genesthe chromosome could establish both thagenes A and B are on the same chromosome Ch

that they are a certain map unit distance a~

2. A cross between a mutant female fly and amal male should produce all normal female(who get a wild-type allele from their faand all mutant male flies (who get the muallele on the X chromosome from their moxrx: X Xm+y produces Xm+Xmand Xspring (normal females and mutant males).

3. The serious phenotypic effects that are associated with these chromosomal alterations in ..cate that normal development and functio ..are dependent on genetic balance. Most gene

472 Answer Section

appear to be vital to an organism's existence,and extra copies of genes upset genetic balance.Inversions and translocations, which do notdisrupt the balance of genes, can alter pheno-type because of the effect of neighboring geneson gene functioning.

ANSWERS TO GENETICS PROBLEMS

1. a. The trait is recessive and probably sex-linked. Two sets of unaffected parents in thesecond generation have offspring with the trait,indicating that it must be recessive. More malesthan females display the trait. Females #3 and#5 must be carriers since their father has thetrait, and they each pass it on to a son.b. Using the symbols XT for the dominantallele and x' for the recessive:1. xty2. XTXt3. XTXt4. XTy5. XTXt6. XTXt or XTXT7. XTXt

c. Individual #6 has a brother who has the traitand her mother's father had the trait, so hermother must be a carrier of the trait, meaningthere is a 12 probability that #6 is a carrier. Sinceshe is mated to a phenotypically normal male,none ofher daughters will show the trait (0prob-ability). Her sons have a 12 chance of having thetrait if she is a carrier. There is a 12 X % = 14probability that her sonswill have the trait. Thereis a % chance that a child would be male, so theprobability of an affected child is 12 X %, or 18.

2. c, a, b, d

3. The genes appear to be linked because theparental types appear most frequently in theoffspring. Recombinant offspring represent 10out of 40 total offspring for a recombination fre-quency of 25%, indicating that the genes are -map units apart.

4. One of the mother's X chromosomes carries therecessive lethal allele. One-half of male fetuseswould be expected to inherit that chromosomeand spontaneously abort. Assuming an equalsex ratio at conception, the ratio of girl to boychildren would be 2:1,or sixgirls and three boy

5. a. For female chicks to be black, they mushave received a recessive allele from the maleparent. If all female chicks are black, the maleparent must have been ZbZb. If the male parentwas homozygous recessive, then all male off-spring will receive a recessive allele, and the fe-male parent would have to be ZBW to produceall barred males.

b. For female chicks to be both black andbarred, the male parent must have been ZBZb. Ifthe female parent were ZBW, only barred malechicks would be produced. To get an equalnumber of black and barred male chicks, the fe-male parent must have been Z~.

ANSWERS TO TEST YOUR KNOWLEDGE

Multiple Choice:1. e 5.· a 9. a· 13. c 17. e2. a 6. e 10. e 14. a 18. d3. b 7. b 11. d 15. e 19. c4. d 8. c 12. b 16. d 20. d

CHAPTER 16: THE MOLECULAR BASIS OF INHERITANCE

INTERACTIVE QUESTIONS

16.1 a. They grew T2 with E. coli with radioactivesulfur to tag phage proteins.b. T2was grown with E. coli in the presence ofradioactive phosphorus to tag phage DNA.c. Radioactivity was found in the supernatant,indicating that the phage protein did' not enterthe bacterial cells.d. In the samples with the labeled DNA, mostof the radioactivity was found in the bacterialcell pellet.

e. They concluded that viral DNA is injectedinto the bacterial cells and serves as the heredi-tary material for viruses.

16.2 a. sugar-phosphate backboneb. 3' end of chainc. hydrogen bondsd. cytosinee. guaninef. adenine (purine)g. thymine (pyrimidine)h. 5' end of chain