Tut 13 - Genetic Basis for Variation

St Andrew’s Junior College Genetics & Inheritance 2022 H1 Biology 1 Full Name: Civics group: 21S Index no.: Date: Core Idea 2: Genetics and Inheritance Dihybrid cross & Test Cross Tutorial 13 QUESTION 1 A homozygous white -flowered, long -stemmed tobacco plant, Nicotiana affinis , was crossed with homozygous pink fl owered, short-stemmed plant. The alleles for white flower and long stem are dominant to those for pink flower and short stem. The above cross gave rise to the F1 generation. The F1 plants were selfed to produce the F2 generation. Using suitable symbol s, construct a genetic diagram to illustrate the outcome of the above crosses. ………………………………………………………………………………………………..[6] Key: Let W represent the allele for white flowers. Let w reresent the allele for pink flowers. Let L represent the allele for long stems. Let l represent the allele for short stems. The allele for white flowers (W) is dominant to allele for pink flowers (w). The allele for long stems (L) is dominant to allele for short stems (l). Parental phenotypes: white flowers, long stems x pink flowers, short stems Parental genotype: WWLL wwll Parental gametes: WL wl F1 genotypes: WwLl F1 phenotypes: All white flowers, long stems Selfing F1: WwLl x WwLl F1 gametes: WL Wl wL wl WL Wl wL wl F1 genotype, determined by Punnett square: F1 x F1 cross Tip: When 2 homozygous parents of contrasting traits are crossed, traits observed in F1 are dominant traits.

St Andrew’s Junior College Genetics & Inheritance 2022 H1 Biology 2 If no punnett sq: 1 WWLL 1 WWll 1wwLL 1wwll 2 WWLl 2 Wwll 2 wwLl 4 WwLl 2 WwLL F2 phenotypes: White, long stem: White, short stem : pink, long stem: pink, short stem F2 phenotypic ratio: 9 : 3 : 3 : 1 [Total: 6] Mark scheme: 1 Key 2 Parental genotypes 3 Parental gametes – (Gametes must be circled) 4 F1 genotype 5 F1 phenotypic ratio Selfing: 6 F1 genotypes (both heterozygous) 7 F1 gametes – (Gametes must be circled) 8 F2 genotypes / Punnett square 9 Genotypes correspond to phenotypes / legend for Punnett square 10 F2 phenotypic ratio 11 Correct presentation of genetic diagram (correct sequence of presentation) Tip: Points in yellow are likely areas awarded marks. Always provide a full presentation of answers.

St Andrew’s Junior College Genetics & Inheritance 2022 H1 Biology 3 QUESTION 2 In rumbunnies, facial warts (W) is dominant to smooth face (w) and freckled face (F) is dominant to freckle-less face (f). Rumbunnies that are heterozygous for both facial warts and freckles were crossed with smooth-faced and freckle-less rumbunnies. The resultant offspring were as follows: Phenotype No. of rumbunnies Wild phenotype (facial warts, facial freckles) 45 Facial warts, freckle-less 48 Smooth face, freckle-less face 46 Smooth face, Facial freckles 51 (a) Draw a genetic diagram to show the cross described. ………………………………………………………………………………………………..[5] Key: (Given for this question) Let W represent the allele for facial warts. Let w represent the allele for smooth face. Let F represent the allele for facial freckles. Let f represent the allele for freckle-less face. The allele for freckled face (F) is dominant to allele for freckle-less face (f). The allele for facial warts (W) is dominant to allele for smooth face (w). Parental phenotypes: Facial Warts, Freckled face x Smooth face, Freckle-less Parental genotype: WwFf wwff Parental gametes: WF Wf wF wf wf F1 genotype, determined by Punnett square: WF Wf wF wf wf WwFf Facial Warts, Facial Freckles Wwff Facial Warts, Freckleless wwFf Smooth face, Facial Freckles wwff Smooth face, Freckle-less F1 phenotypes: Facial Warts, Facial Freckles ; Facial Warts, Freckleless ; Smooth face, Facial Freckles ; Smooth face, Freckleless F1 phenotypic ratio: 1 : 1 : 1 : 1 If symbols are provided in the question, students do not need to write out the key Question: What kind of cross involves crossing a Rumbunny showing dominant phenotypes with another which is homozygous recessive?

St Andrew’s Junior College Genetics & Inheritance 2022 H1 Biology 4 Mark scheme: 1 Parental phenotype 2 Parental genotypes 3 Parental gametes – (Gametes must be circled) 4 Progeny genotypes / Punnett square Genotypes correspond to phenotypes / legend for Punnett square 5 F1 phenotypic ratio (b) If you are given a Rumbunny with Facial Warts and Facial Freckles, explain how you would derive its actual genotype. ………………………………………………………………………………………………..[5] 1 Perform a test cross between this Rumbunny with unknown genotype (W_F_) with one that is homozygous recessive for both traits (wwff) ; 2 Obtain a large number of progeny (explanation of large number : Use of large sample size for reliability of results); to determine the offspring phenotypic ratio ; 3 Deduce the genotype of the rumbunny based on the phenotypic ratio according to the predicted results in the table below: 4 (2 marks) list all possible genotype and phenotypic ratio Case Possible Genotype Phenotypic ratio of test cross 1 WWFF All Facial warts and freckled face 2 WWFf 1 facial warts and freckled face : 1 facial warts and freckle-less face 3 WwFF 1 facial warts and freckled face : 1 smooth and freckled face 4 WwFf 1 facial warts and freckled face : 1 facial warts and freckle-less face : 1 smooth and freckled face : 1 smooth and freckles face [Total: 10] Tip: Suggest a method, and provide what’s expected to be observed in phenotypic ratios. This allows a way to deduce the genotype clearly.

St Andrew’s Junior College Genetics & Inheritance 2022 H1 Biology 5 QUESTION 3 Fruit flies ( Drosophila melanogaster) have been used extensively in investigation of inheritance. They are particularly suited to study because of their short life cycle and many different mutant forms. In one investigation, red -eyed females with grey bodies were crossed with brown -eyed males with black bodies. All the F1 generation had red eyes and black bodies. When these F1 flies were crossed, the following offspring were produced: Phenotype No. of flies Red-eyed, black-bodied 110 Red-eyed, grey-bodied 39 Brown-eyed, black-bodied 37 Brown-eyed, grey-bodied 13 (a) State the dominant characters for eye color and body color. ………………………………………………………………………………………………..[1] 1 Red eyes and black body ; (b) State the genotypes of the parent flies, using suitable symbols. ………………………………………………………………………………………………..[3] Key : 1 Let R represent the allele for red eyes. Let r represent the allele for brown eyes. Let B represent the allele for black body. Let b represent the allele for grey body. The allele for red eyes (R) is dominant to allele for brown eyes (r). The allele for black body (B) is dominant to allele for grey body (b). 2 Genotype of red-eyed female with grey bodies - RRbb ; 3 Genotype of brown-eyed males with black bodies - rrBB. Tip: When 2 homozygous parents of contrasting traits are crossed, traits observed in F1 are dominant traits Tip: Work backwards, by observing the results of the F2 generation, can you deduce about F1 genotype? To give rise to F1 genotype, what genotype will the parent flies have? Thought process: 1) Red eyed and grey bodied parental flies have minimum 1 copy of allele R and 2 copies of allele b (Genotype: R_bb). 2) Brown eyed and black bodied parental flies are having 2 copies of allele r, minimum 1 copy of allele B (genotype of rrB_ )