Suggested Answers to Structure YourKnowledge

1. Mendel's law of segregation occurs in anaphaseI, when alleles segregate as homologous chro-mosomes 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. Mendel's law of independentassortment relates to the lining up of synapsedchromosomes at the equatorial plate in a ran-dom fashion during metaphase I. Genes on dif-ferent chromosomes will assort independentlyinto gametes.

2. One allele does not "dominate" another. All al-leles operate independently within a cell, codingfor their gene products, usually enzymes, asspecified by their particular sequence of DNAnucleotides. If both alleles code for functionalenzymes or products, then the alleles may becodominant with both traits expressed in theheterozygote, as illustrated by MN or ABbloodtypes. If the recessive allele codes for a nonfunc-tional enzyme, then the resulting recessive traitmay be obvious only when homozygous (suchas 0 blood type in which there are no carbohy-drate molecules on the cell membrane). If onecopy of an allele produces sufficient product thatthe dominant trait is expressed, then we havecomplete dominance where the heterozygotephenotype is indistinguishable from thehomozygote dominant (for example, smoothpeas in which the enzyme converts sugar tostarch). With incomplete dominance, the het-erozygote is distinguishable from homozygousdominants and recessives. The functional alleleis insufficient to produce the full trait (as withthe pink color of a heterozygote in carnations).Genotypes translate into phenotypes by way ofbiochemical pathways, which are controlled byenzymes coded for by genes. Thus molecularprocesses produce physical, physiological, andbehavioral results.

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. 1/4 [1/2 (to get AA) x 1/2 (bb)]b. 1/8 [1/4 (aa) x 1/2 (BB)]c. 1/2 [1 (Aa) x 1/2 (Bb) x 1 (Cc)]d. 1/32 [1/4 (aa) x 1/4 (bb) x 1/2 (cc)]

Answer Section 469

3. Since flower color shows incomplete dominance,there will be six phenotypic classes in the F2instead of the normal four classes found in a9:3:3:1 ratio. You could find the answer with aPunnett square, but multiplying the probabili-ties of the monohybrid crosses is more efficient.Tall red LRR = 3/4 x 1/4 3/16Tallpink T_Rr = 3/4 x 1/2 3/8 or 6/16Tallwhite TJr = 3/4 x 1/4 3/16Dwarf red ttRR = 1/4 x 1/4 1/16Dwarf pink ttRr = 1/4 x 1/2 1/8 or 2/16Dwarf white ttrr = 1/4 X1/4 1/16

4. Determine the possible genotypes of the motherand child. Then find the blood groups for thefather that could not have resulted in a childwith the indicated blood group.a. no groups exoneratedb. AorOc. AorOd.ABe. BorO

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 apply-ing the law ofmultiplication.

Lethal (cc--) = 1/4Normal brown (CCBb) = 1/4x 1/2= 1/8= 1Normal white (CCbb) = 1/4x 1/2= 1/8= 1Deformed brown (CcBb) = 1/2x 1/2= 1/4= 2Deformed white (Ccbb) = 1/2x 1/2= 1/4= 2

Ratio of viable offspring: 1:1:2:2 or 1;6 normalbrown: 1/6 normal white: 2/6 deformed brown:2/6deformed white

6. Father's genotype must be Pp since polydactylyis dominant and he has had one normal child.Mother's genotype is pp. The chance of the nextchild having normal digits is 1/2 or 50% becausethe mother can only donate a p allele and thereis a 50% chance that the father will donate a pallele.

7. The genotypes of the puppies were 3/8 B_S-, 3/8B_ss, 1/8 bbS-, and 1/8 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.

Suggested Answers to Structure YourKnowledge

1. Mendel's law of segregation occurs in anaphaseI, when alleles segregate as homologous chro-mosomes 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. Mendel's law of independentassortment relates to the lining up of synapsedchromosomes at the equatorial plate in a ran-dom fashion during metaphase I. Genes on dif-ferent chromosomes will assort independentlyinto gametes.

2. One allele does not "dominate" another. All al-leles operate independently within a cell, codingfor their gene products, usually enzymes, asspecified by their particular sequence of DNAnucleotides. If both alleles code for functionalenzymes or products, then the alleles may becodominant with both traits expressed in theheterozygote, as illustrated by MN or AB bloodtypes. If the recessive allele codes for a nonfunc-tional enzyme, then the resulting recessive traitmay be obvious only when homozygous (suchas 0 blood type in which there are no carbohy-drate molecules on the cell membrane). If onecopy of an allele produces sufficient product thatthe dominant trait is expressed, then we havecomplete dominance where the heterozygotephenotype is indistinguishable from thehomozygote dominant (for example, smoothpeas in which the enzyme converts sugar tostarch). With incomplete dominance, the het-erozygote is distinguishable from homozygousdominants and recessives. The functional alleleis insufficient to produce the full trait (as withthe pink color of a heterozygote in carnations).Genotypes translate into phenotypes by way ofbiochemical pathways, which are controlled byenzymes coded for by genes. Thus molecularprocesses produce physical, physiological, andbehavioral results.

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. 1/4 [1/2 (to get AA) x 1/2 (bb)]b. 1/8 [1/4 (aa) x 1/2 (BB)]c. 1/2 [1 (Aa) x 1/2 (Bb) x 1 (Cc)]d. 1/32 [1/4 (aa) x 1/4 (bb) x 1/2 (cc)]

Answer Section 469

3. Since flower color shows incomplete dominance,there will be six phenotypic classes in the Fzinstead of the normal four classes found in a9:3:3:1ratio. You could find the answer with aPunnett square, but multiplying the probabili-ties of the monohybrid crosses is more efficient.Tall red LRR = 3/4 x 1/4 3/16Tallpink T_Rr = 3/4 x 1/2 3/8 or 6/16Tallwhite TJr = 3/4 x 1/4 3/16Dwarf red ttRR = 1/4 x 1/4 1/16Dwarf pink ttRr = 1/4 x 1/2 1/8 or 2/16Dwarf white ttrr = 1/4 X1/4 1/16

4. Determine the possible genotypes of the motherand child. Then find the blood groups for thefather that could not have resulted in a childwith the indicated blood group.a. no groups exoneratedb. AorOc. AorOd.ABe. BorO

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 apply-ing the law ofmultiplication.

Lethal (cc--) = 1/4Normal brown (CCBb) = 1/4x 1/2= 1/8= 1Normal white (CCbb) = 1/4x 1/2= 1/8= 1Deformed brown (CcBb) = 1/2x 1/2= 1/4= 2Deformed white (Ccbb) = 1/2x 1/2= 1/4= 2

Ratio of viable offspring: 1:1:2:2 or 1/6 normalbrown: 1/6 normal white: 2/6 deformed brown:2/6deformed white

6. Father's genotype must be Pp since polydactylyis dominant and he has had one normal child.Mother's genotype is pp. The chance of the nextchild having normal digits is 1/2 or 50% becausethe mother can only donate a p allele and thereis a 50% chance that the father will donate a pallele.

7. The genotypes of the puppies were 3/8 B_S-, 3/8B_ss, 1/8 bbS-, and 1/8 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.

Answer Section 471

Answers to Test Your Knowledge Multiple Choice:

Matching: 1. c 4. c 7. c 10. c 13. d

1. I 3. F 5. L 7. H 9. C 2. a 5. b 8. d 11. d 14. d

2. A 4. E 6. J 8. M 10. K 3. e 6. d 9. b 12. a

CHAPTER 15THE CHROMOSOMAL BASIS OF INHERITANCE

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• INTERACTIVE QUESTIONS

15.1

I F2 Generation I

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

GenotypewY

15.2 a. tall, purple-flowered and dwarf, white-flow-ered

b. tall, white-flowered and dwarf, purple-flow-ered

15.3 Since linked genes have their loci on the samechromosome, they travel together during meio-sis and you tend to have more parental offspringproduced. Recombinants are the result of cross-ing over between homologous chromosomes, anevent that is less frequent than 50% for geneslocated close to each other.

15.4 15----

.m--9--j-6-1-6-k---12---

15.5 The gene is sex-linked, so a good notation is Xc,Xc, and Y so that you will remember that the Ydoes not carry the gene. Capital C indicates nor-mal sight. Genotypes are:1. XCy 4. XCXc 7. XCXc2. XCXc 5. XCy3. XCXCor XCXc 6. XcY

15.6 a. A trisomic organism (2n + 1) has an extracopy of one chromosome, usually caused by anondisjunction during meiosis. A triploidorganism (3n) has an extra set of chromo-somes, possibly caused by a total nondisjunc-tion in gamete formation.

b. The genetic balance of a trisomic organismwould be more disrupted than that of an

organism with a complete extra set of chro-mosomes.

15.7 translocation, deletion, and inversion

15.8 Aneuploidies of sex chromosomes appear toupset genetic balance less, perhaps because rela-tively few genes are located on the Y chromo-some and extra X chromosomes are inactivatedas Barr bodies.

Suggested Answers to Structure Your Knowledge

1. Genes that are not linked assort independently,and the ratio of offspring from a testcross with adihybrid heterozygote should be 1:1:1:1 (AaBb xaabb gives AaBb, Aabb, aaBb, aabb offspring).Genes that are linked and do not cross overshould produce a 1:1 ratio in this testcross (AaBband aabb because a heterozygote produces onlyAB and ab gametes). If crossovers occur 50% ofthe time, the heterozygote will produce equalquantities ofAB, Ab, aB, and ab gametes, and thegenotype ratio of offspring will be 1:1:1:1, thesame as it is for unlinked genes. Because each1% of crossovers is equal to 1 map unit, thismeasurement ceases to be meaningful at relativedistances of 50 or more map units. However,crosses with intermediate genes on the chromo-some could establish both that the genes A and Bare on the same chromosome and that they are acertain map unit distance apart.

2. If the gene for the mutant is sex-linked, youcould assume it was on the X chromosome. Across between a mutant female fly and a normalmale should produce all normal females (whoget a wild-type allele from their father) andmutant male flies (who get the mutant allele onthe X chromosome from their mother).xmxm x X+Y produces X+xm and xmy offspring(normal females and mutant males).

3. The serious phenotypic effects that are associat-ed with these chromosomal alterations indicatethat normal development and functioning isdependent on genetic balance. Most genes

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 not dis-rupt the balance of genes, can alter phenotypebecause of effects on gene functioning fromneighboring genes.

Answers to Genetics Problems

1. a. The trait is recessive and probably sex-linked.Two sets of unaffected parents in the secondgeneration have offspring with the trait, indi-cating that it must be recessive. More malesthan females display the trait. Females 3 and5 must be carriers since their father has thetrait, and they each pass it on to a son.

b. Using the symbols XT for the dominant alleleand xt for the recessive:1. xty 5. XTXt2. XTXt 6. XTXt or XTXT3. XTXt 7. XTXt4. XTy

c. #6 has a brother who has the trait, so hermother must be a carrier of the trait, meaningthere is a 1/2 probability that #6 is a carrier.Since she is mated to a phenotypically normalmale, none of her daughters will show thetrait (0 probability). Her sons have 1/2 chanceof having the trait if she is a carrier. 1/2 x 1/2 =1/4 probability that her sons will have thetrait. There is a 1/2 chance that a child wouldbe male, so the probability of an affected childis 1/2 x 1/4, or 1/8.

2.eacbd

CHAPTER 16THE MOLECULAR BASIS OF INHERITANCE

3. The genes appear to be linked because theparental types appear most frequently in the off-spring. Recombinant offspring represent 10 outof 40 total offspring for a recombination frequen-cy of 25 percent, indicating that the genes are 25map 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 6 girls and 3 boys.

5. a. For female chicks to be black, they must havereceived a recessive allele from the male par-ent. If all female chicks are black, the maleparent must have been zrz: If the male par-ent was homozygous recessive, then all maleoffspring will receive a recessive allele, andthe female parent would have to be ZBW toproduce all barred males.

b. For female chicks to be both black and barred,the male parent must have been ZBZb. If thefemale parent were ZBW, only barred malechicks would be produced. To get an equalnumber of black and barred male chicks, thefemale parent must have been rv«

Answers to Test Your Knowledge'

Multiple Choice:

1. e 5. a 9. a 13. c2. a 6. e 10. a 14. a3. b 7. b 11. d 15. e4. d 8. c 12. b 16. d

III INTERACTIVE QUESTIONS

16.1 a. They grew T2 with E. coli with radioactivesulfur to tag phage proteins.

b. T2 was 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 notenter the bacterial cells.

d. In the samples with the labeled DNA, most ofthe radioactivity was found in the bacterialcell pellet.

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

16.2 a. sugar-phosphate backboneb. complementary base pairc. adenined. pyrimidine basese. guaninef. thymineg. purine basesh. hydrogen bondsi. cytosinej. nucleotidek. deoxyribose1. phosphatem.3.4nmn.0.34nm0.2nm