1. ¾ A (AA, Aa) A B: ¾ x ¾ = 9/16 A b: ¾ x ¼ = 3/16 a B: ¼ x ¾ = 3/16 a b: ¼ x ¼ = 1/16 GENE INTERACTIONS Consider two independent genes, A and B; two heterozygous individuals for both genes are crossed. Complete the scheme below: AaBb X AaBb Phenotypes and frequencies for gene A Phenotypes and frequencies for gene B Phenotypes and frequencies for genes A and B ¼ a (aa) ¾ B (BB, Bb) ¼ b (bb) ¾ B (BB, Bb) ¼ b (bb)

2. Because of the gene interaction phenomenon, some phenotypic classes are reduced in number, as two or more classes may display the same phenotype. Which are the phenotypic classes of an F2 in which: a) the dominant allele of a gene (A) masks the effect of the other gene (B)? 9/16 AB3/16 Ab3/16 aB1/16 ab 9/16+3/16= 12/16 b) the recessive homozygote (b) masks the effect of the other gene? 9/16 AB3/16 Ab3/16 aB1/16 ab 3/16+1/16= 4/16 the phenotype due to the dominant allele of a gene is indistinguishable from the phenotype due to the dominant allele of the other gene? Distinguish the two situations: when there is addictive effect and when there isn’t. 9/16 AB3/16 Ab3/16 aB1/16 ab 3/16+3/16= 6/16 9/16 AB3/16 Ab3/16 aB1/16 ab 9/16 + 3/16+3/16= 15/16

3. The A gene encodes the enzyme α, the independent gene B encodes the enzyme β; these enzymes are both involved in the following metabolic pathway for cyanide production. The recessive alleles, a and b, encodes for inactive enzymes. a) Identify the genotypes of P and F1 individuals and determine which phenotypic classes are expected in F2 and with which frequencies.  S1 S2 cyanide P F1 Not producing plant X not producing plant Producing plants The parental plants don’t produce cyanide but,after the cross, F1 plants produce cyanide. In parental plants one of two genes is homozygous recessive (inactive gene), the other gene is homozygous dominant (active enzyme) Since the F1 plants are producing plants, the gene homozygous recessive are different between the parents. (P1) AA bb X (P2) aa BB (F1) Aa Bb

4. b) b) what’s the name of genes that interact in the way showed here? c) which phenotypic classes are expected from the backcross of an F1 plant with the recessive homozygote? Complementary Genes, these genes act on the same metabolic chain. (F1) Aa Bb x Aa Bb (F2) 9:3:3:1 9/16 AB3/16 Ab3/16 aB1/16 ab 9/16 Producing plants 3/16 + 3/16 + 1/16= 7/16 Not producing plants Aa Bb x aa bb 1/4 AB1/4 Ab1/4 aB1/4 ab 1/4 Producing Plants ¼ + ¼ + ¼ = ¾ Not Producing plants

5. The two independent genes P and C control two different consecutive steps in the same metabolic pathway. a) PPcc PC S1 S2 Yellow fruit The dominant alleles P and C encode for active enzymes, while the recessive alleles, p and c, encode for inactive enzymes. Which phenotype is expected for individuals with the following genotype (intermediate substrates are colourless): This plant produces the enzyme encoded by gene P, but not produces the other one (gene C). Metabolic chain is blocked at step S2. Phenotype: white melon b) PpCc The plant has both enzymes thus the metabolic chain works until the end. Phenotype: yellow melon. c) ppCC The plant hasn’t the enzyme encoded by gene P. It produces the enzyme encoded by the gene C but it doesn’t convert S1 in S2. Phenotype white melon. d) Ppcc The plant produces the first enzyme but not the second one. Then the metabolic chain is blocked at S1 step. Phenotype white melon.

6. The two independent genes, A e E, that originate through duplication, regulate the same step in the metabolic pathway that leads to the synthesis of a pigment. The colourless phenotype can be seen only in the recessive homozygote state for both genes. Aa Ee x Aa Ee 9/16 AE3/16 Ae3/16 aE1/16 ae 1/16 white fur9/16 + 3/16 + 3/16= 15/16 brown fur Allele A produces enzyme  that converts S1 in pigment; allele a does not produce pigment. Allele E produces enzyme  that converts S1 in pigment; allele e does not produce pigment. Enzima  S1 pigment Enzima  Which phenotypic ratio is expected from the cross between two di-hybrids?

7. Two pure lines of pepper are crossed: first line with Red fruits and the second one with orage fruites. The plants of F1 produce only red peppers, but crossing two individuals of F1 we obtain: 192 RED FRUITS, 47 ORANGE FRUITS, 14 WHITE FRUITS We know that the color fruit is controlled by 2 genes (A and B), which kind of interaction is present? Determine the genotypes and hypotize a possible action on metabolic chain. (P1) RED FRUITS X (P2) ORANGE FRUITS (F1) RED FRUITS F2: 192 red fruits 47 orange fruits 14 white fruits When the plant has an allele of A the phenotype related to gene B is not shown. Phenotypes AB and Ab produce red fruits When A is homozygous recessive, the plant show the B phenotype: if it has dominant allele B the fruits will be orange, if it is homozygous bb the fruits will be white. AA bbaa BB Aa Bb 9/16 AB + 3/16 Ab 3/16 aB 1/16 ab 192+47+14= 253 (:16= 15.8) The dominant allele A is epistatic on gene B

8. PhenotypesXoHXa(Xo – Xa) 2 : Xa 22 Red19212/16189,755,06189,750,03 Orange473/1647,430,1847,430,01 White141/1615,813,2715,810,21 Total25316/162530,25 We have to compare expected individuals (epistasis hypothesis) with observed individuals. Degrees of freedom = 3-1 = 2  2 = 0,25 P =80 - 90% Conclusion: Hypothesis accepted

9. GENE INTERACTIONS AA molecular mechanism for recessive epistasis. Two representative genes encode enzymes catalyzing successive steps in the synthesis of a blue petal pigment. The substrates for these enzymes are colorless and pink, respectively, so null alleles of the genes will result in colorless (white) or pink petals. The epistasis is revealed in the double mutant because it shows the phenotype of the earlier of the two blocks in the pathway (that is, white). Hence the mutation in the earlier gene precludes expression of any alleles of a gene acting at the later step.epistasismutantphenotypemutationgene The results show that homozygosity for the recessive mutant allele of either gene or both genes causes a plant to have white petals. To have the blue phenotype, a plant must have at least one dominant allele of both genes. COMPLEMANTATIONmutant allelegenephenotypedominant allele