1. Fig 4-1 Figure: 04-01b Caption:
Incomplete dominance shown in the flower color of snapdragons.

2. Fig 4-2 Figure: 04-02a Caption:
The biochemical basis of the ABO blood groups. The H allele, present in almost all humans, directs the conversion of a precursor molecule to the H substance by adding a molecule of fucose to it. The IA and IB alleles are then able to direct the addition of terminal sugar residues to the H substance. The IO allele is unable to direct either of these terminal additions. Gal: galactose; AcGluNH: N-acetyl glucosamine; AcGalNH: N-acetylgalactosamine. Failure to produce the H substance results in the Bombay phenotype where individuals are type O, regardless of the presence of an IA or IB allele.

3. Fig 4-2 Figure: 04-02b Caption:
The biochemical basis of the ABO blood groups. The H allele, present in almost all humans, directs the conversion of a precursor molecule to the H substance by adding a molecule of fucose. Failure to do so results in the Bombay phenotype. The IA and IB alleles are then able to direct the addition of terminal sugar residues to the H substance. The IO allele is unable to direct either of these terminal additions. Gal: galactose; AcGluNH: N-acetyl-D-glucosamine; AcGalNH: N-acetylgalactosamine.

4. Fig 4-3 Figure: 04-03 Caption:
A partial pedigree of a woman displaying the Bombay phenotype. Functionally, her ABO blood group behaves as type O. Genetically, she is type B.

5. Fig 4-2 Figure: 04-04 Caption:
Inheritance patterns in three crosses involving the normal wild-type agouti allele (A) and the mutant yellow allele (AY) in the mouse. Note that the mutant allele behaves dominantly to the normal allele in controlling coat color, but it also behaves as a homozygous lethal allele. The genotype AYAY does not survive.

6. Fig 4-5 Figure: 04-05a Caption:
Calculation of the probabilities in a mating involving the ABO blood type and albinism in humans. The calculation is carried out by using the forked-line method.

7. Fig 4-5 Figure: 04-05b Caption:
Calculation of the probabilities in a mating involving the ABO blood type and
albinism in humans. The calculation is carried out by using the forked-line method.

8. Fig 4-6 Figure: 04-06a Caption:
The outcome of a mating between individuals heterozygous at two genes that determine each individuals ABO blood type. Final phenotypes are calculated by considering both genes separately and then combining the results using the forked-line method.

9. Fig 4-6 Figure: 04-06b Caption:
The outcome of a mating between individuals heterozygous at two genes that determine each individuals ABO blood type. Final phenotypes are calculated by considering both genes separately and then combining the results using the forked-line method.

10. Fig 4-7 Figure: 04-07 Caption:
Generation of the various modified dihybrid ratios from the nine unique genotypes produced in a cross between individuals heterozygous at two genes.

11. Fig 4-8 Figure: 04-08 Caption:
The basis of modified dihybrid F2 phenotypic ratios, resulting from crosses between doubly heterozygous F1 individuals. The four groupings of the F2 genotypes shown in Figure 4–7 and across the top of this figure are combined in various ways to produce these ratios.

12. Fig 4-10 Figure: 04-10a Caption:
A theoretical explanation of the biochemical basis of the four eye-color phenotypes produced in a cross between Drosophila with brown eyes and scarlet eyes. In the presence of at least one wild-type  allele, an enzyme is produced that converts substance b to c, and the pigment drosopterin is synthesized. In the presence of at least one wild-type  allele, substance e is converted to f, and the pigment xanthommatin is synthesized. The homozygous presence of the recessive bw and st mutant alleles blocks the synthesis of these respective pigment molecules. Either one, both, or neither of these pathways can be blocked, depending on the genotype.

13. Fig 4-10 Figure: 04-10b Caption:
A theoretical explanation of the biochemical basis of the four eye-color phenotypes produced in a cross between Drosophila with brown eyes and scarlet eyes. In the presence of at least one wild-type bw+ allele, an enzyme is produced that converts substance b to c, and the pigment drosopterin is synthesized. In the presence of at least one wild-type st+ allele, substance e is converted to f, and the pigment xanthommatin is synthesized. The homozygous presence of the recessive bw and st mutant alleles blocks the synthesis of these respective pigment molecules. Either none, one, or both of these pathways can be blocked, depending on the genotype.

14. Fig 4-10 Figure: 04-10c Caption:
A theoretical explanation of the biochemical basis of the four eye-color phenotypes produced in a cross between Drosophila with brown eyes and scarlet eyes. In the presence of at least one wild-type bw+ allele, an enzyme is produced that converts substance b to c, and the pigment drosopterin is synthesized. In the presence of at least one wild-type st+ allele, substance e is converted to f, and the pigment xanthommatin is synthesized. The homozygous presence of the recessive bw and st mutant alleles blocks the synthesis of these respective pigment molecules. Either none, one, or both of these pathways can be blocked, depending on the genotype.

15. Fig 4-10 Figure: 04-10d Caption:
A theoretical explanation of the biochemical basis of the four eye-color phenotypes produced in a cross between Drosophila with brown eyes and scarlet eyes. In the presence of at least one wild-type bw+ allele, an enzyme is produced that converts substance b to c, and the pigment drosopterin is synthesized. In the presence of at least one wild-type st+ allele, substance e is converted to f, and the pigment xanthommatin is synthesized. The homozygous presence of the recessive bw and st mutant alleles blocks the synthesis of these respective pigment molecules. Either none, one, or both of these pathways can be blocked, depending on the genotype.

16. Fig 4-11 Figure: 04-11a Caption:
Complementation analysis of alternative outcomes of two wingless mutations in Drosophila (musa and mcan). In case 1, the mutations are not alleles of the same gene, while in case 2, the mutations are alleles of the same gene.

17. Fig 4-11 Figure: 04-11b Caption:
Complementation analysis of alternative outcomes of two wingless mutations in Drosophila (musa and mcan). In case 1, the mutations are not alleles of the same gene, while in case 2, the mutations are alleles of the same gene.

18. Fig 4-12 Figure: 04-12a Caption:
The F1 and F2 results of T. H. Morgan’s reciprocal crosses involving the X-linked white mutation in Drosophila melanogaster. The actual F2 data are shown in parentheses. The photographs show white eyes and the brick-red wild-type eye color.

19. Fig 4-13 Figure: 04-13a Caption:
The chromosomal explanation of the results of the X-linked crosses shown in Figure 4-12.

20. Fig 4-13 Figure: 04-13b Caption:
The chromosomal explanation of the results of the X-linked crosses shown in Figure 4-12.

21. Fig 4-14 Figure: 04-14ab Caption:
(a) A human pedigree of the X-linked color-blindness trait. (b) The most probable genotypes of each individual in the pedigree. The photograph is of an Ishihara color-blindness chart. Red-green color-blind individuals see a 3, rather than an 8 visualized by those with normal color vision.

22. Fig 4-20 Figure: 04-20 Caption:
The effect of imprinting on the mouse Igf2 gene, which produces dwarf mice in the homozygous condition. Heterozygous offspring that receive the normal allele from their father are normal in size. Heterozygotes that receive the normal allele from their mother, which has been imprinted, are dwarf.

23. Figure: UN
Caption:
Personate and peloric

24. Prob 43, pg 105 Figure: 04-20-02 Caption:
n watermelons (Citrullus lanatus), the flesh may exhibit numerous colors, including canary, red, salmon, and orange. Three crosses are shown below involving four true-breeding strains:
(a) Consider initially the first two crosses:Analyze the results and put forward a proposal that is consistent with them that explains the inheritance of the four flesh colors.(b) Now consider the third cross: How can you extend your proposal in Part a) to accommodate the results in Cross III? (c) Based on your proposals in Part (a) and (b), what are the complete genotypes of the four true-breeding strains? (d) Propose two further crosses and the predicted results that test and confirm your proposals in Part a) and b).

25. Prob 31, pg 103 Figure: 04-20-01 Caption:
Shown in the following table are five human matings (1–5), including both maternal and paternal phenotypes for ABO, MN, and Rh blood-group antigen status. Each mating resulted in one of the five offspring shown to the right (a–e). Match each offspring with one correct set of parents, using each parental set only once. Is there more than one set of correct answers?