Answers
1. Genotypic and phenotypic ratio of the test cross progeny Gg:gg (1:1) Grey guppies : Gold guppies (1:1) 2. i) Genotype of normal eyes in goldfish – DD/Dd Genotype of telescope eyes in goldfish – dd ii) Normal eyes are dominant 3. Genotype for wild grass carp - AA/Aa Genotype for albino grass carp - aa Wild type - Dominant Albino - Recessive Parents Progeny colouration a) wild x wild wild AA A- A- b) wild x Albino wild AA aa Aa c) Albino x wild wild aa AA Aa d) Albino x Albino Albino aa aa aa e) Albino x wild Wild Albino aa Aa Aa aa 4. When a pure normal eye goldfish is crossed to a telescopic one, 50% of the F2 population will be heterozygous. 5. Blue colour carp will be chosen. Test cross. 6. Genotypes of the parents. a) Pattern x No Pattern Dd dd b) Pattern x Pattern Dd Dd c) No Pattern x No Pattern dd dd d) Pattern x No pattern DD dd e) Pattern x Pattern Dd Dd 7. a. Genotype of the fish - Bb Genotype of the female parent - bb b. Genotype of the male parent - BB/Bb c. Genotype of their offspring - Bb : bb (1 : 1) 8. a) Genotypes of the parents – Dd x Dd When two heterozygous parents were crossed for a particular trait, the offsprings will be obtained in the ratio of 3:1 (dominant to recessive) b) When a pure light yellow pattern carp is crossed with a hybrid light yellow pattern the genotypic ratio of the offsprings will be of 1 homozygous light yellow pattern : 1 hybrid yellow pattern. 9. a) Green (wild) is dominant and red is recessive. b) Mendelian pattern of inheritance - (F2 ratio 3:1 ) 10. Since the recessive albino phenotype appears in the F1 in approximately a 1:1 ratio, we know that the female parent must be heterozygous +a. Furthermore, we know that the normal type F1 progeny must also be heterozygous +a. The normal type F1 females are then crossed with their albino brothers: F1 cross : +a ♀ x aa ♂ Normal albino F2 ½ +a : ½ aa Normal albino |