Solution Key - 7.013 Problem Set 2 - 2018

Solution key-7.013 Problem Set 2- 2018 Question 1 After receiving many bouquets of roses on Valentine’s Day this year, you become very interested in studying the traits of different plants. Given your surplus of roses, you decide to focus on them first. You decide to study petal color (red or white) in a variety of rose plants. An autosomal Gene A regulates petal color. You cross a plant with red roses (P1) to another plant with white roses (P2) and get 100 plants in the F1 generation, all of which have red roses. a) Based on this information, give the genotype(s) of P1, P2 and F1 plants for petal color. Note: Use “A” to represent the allele of Gene A that is associated with the dominant phenotype and “a” to represent the allele of Gene A that is associated with the recessive phenotype. P1: AA P2: aa F1: Aa b) You cross two F1 plants and obtain 100 plants in the F2 generation. List the genotypes and corresponding ratios of the F2 plants for petal color. i Genotype(s) and corresponding ratios: AA (1): Aa(2): aa(1) ii List the phenotypes and corresponding ratios of the F2 plants for petal color. Phenotype(s) and corresponding ratios: Red roses(3) : White roses(1) c) You want to determine whether a red rose plant is homozygous or heterozygous for the alleles of Gene A. You therefore subject it to a test cross. i. Give the genotype and the phenotype of the plant with which you would cross the red rose plant. Genotype: aa Phenotype: White roses If the red rose plant was homozygous for the allele A of Gene A, what would be the… ii. • Genotype(s) and ratio of the resulting plants: Aa (100%) • Phenotype(s) and ratio of resulting plants: Red roses (100%) If the red rose plant was heterozygous for the allele A of Gene A, what would be the… iii. • Genotype(s) and ratio of the resulting plants: Aa (1): aa (1) • Phenotype(s) and ratio of resulting plants: Red roses (1): White roses (1) 1 Question 2 After running out of roses, you decide to study tulips. In tulips, an autosomal Gene D regulates the leaf shape and an autosomal Gene H regulates flower color. a) You cross true-breeding P1 (narrow leaves/ dark blue flowers) and P2 (wide-leaves/ white flowers) plants to get the F1 plants (narrow leaves/ pale blue flowers). Note: You should assume that “blue” is the dominant trait. Using “D” or “H” for the allele for the domiant phenotypes and “d” or “h” for the alleles for the recessive phenotype, give the genotypes of… i. ii. iii. True breeding P1 plant (narrow leaves/ dark blue flowers): DDHH True breeding P2 plant (wide-leaves/ white flowers): ddhh F1 progeny (narrow leaves/ pale blue flowers): DH/dh b) Briefly explain why the flower color in F1 is different from that in P1 and P2 plants. This reflects that Blue tulip color, regulated by “allele H” of Gene H, is not completely dominant to white tulip color regulated by the “allele h” of Gene H. Instead, Blue tulip color is incompletely dominant to white tulip color. As a result, the F1 plants have pale blue tulips instead of dark blue tulips. c) You cross an F1 plant with a tulip plant that has the same genotype as the P2 plant above. You obtain 1600 F2 progeny. Fill in the table below for F2 progeny based on the assumption that the two genes are unlinked and therefore assort independently as per Mendel’s law of Independent assortment. Genotypes? Corresponding phenotype? Corresponding estimated number? DH/dh Narrow leaves/ pale blue tulips 400 dh/dh Wide leaves/white tulips 400 Dh/dh Narrow leaves/ white tulips 400 dH/dh Wide leaves/ pale blue tulips 400 d) You also cross two F1 plants to each other and obtain 1600 F2 plants. How many F2 plants will have… i. Narrow leaves and dark blue tulips? 300 [genotypes: DH/DH (100), DH/dH (200)] ii. Wide leaves and white tulips? 100 (Genotype: ddhh) iii. Your friend says that the phenotype ratios of phenotypes in the F2 generation are different from those that would be expected from a dihybrid cross that followed Mendel’s laws. Is your friend right? If so, why? If the blue tulip color was completely dominant to white tulip color as narrow leaves is to wide-leaves, and the genes D and H were assorting independently, the dihybrid cross would give us four different classes of progeny in the F2 generation: Narrow leaves/ dark blue tulips: Narrow leaves/ white tulips, wide-leaves/ dark blue tulips & wide-leaves/ white tulips in the ratio of 9: 3: 3: 1 per Mendel’s law of Independent assortment. But we are not getting this ratio for the F2 progeny due to the incomplete dominance of dark blue color over white color. So it is not following Mendel’s laws. 2 Question 3 Many plants are often the source of traditional herbal medicines. For example, the roots of the Kava plant, Piper methysticum, are often used to treat insomnia (the inability to fall asleep). Remember, we talked about sleep and wakefulness in problem set 1! You cross a variety of Kava plant that is true breeding for wide-round leaves (P1) with another kava plant that is true breeding for narrow-pointy leaves (P2). You obtain F1 plants all of which have narrow-pointy leaves. a) Assume that Gene A regulates leaf width (wide or narrow) and Gene B regulates shape (round or pointy). Give the genotypes of the following plants for both traits, using “A” and “B” for the alleles regulating the dominant phenotypes and “a” and “b” for the alleles regulating the recessive phenotypes. P1: aabb P2: AABB F1: AB/ab b) Assuming that Gene A and Gene B were absolutely linked, give the phenotypes, genotypes and the corresponding ratios of the F2 plants you would expect by crossing two F1 plants. i. ii. Genotypes and corresponding ratios: AABB(1): AB/ab(2): ab/ab(1) Phenotypes and corresponding ratios: Narrow/pointy leaves(3): wide-round leaves (1) c) You cross an F1 plant with another plant that has the genotype “aabb”. i. If Gene A and Gene B are 4cM apart, complete the table below for each class of F2 plants. Assume there are 100 F2 plants in total. Genotypes? Corresponding phenotype? Corresponding number? AB/ab Narrow, pointy leaves 48 ab/ab Wide, round leaves 48 Ab/ab Narrow, round leaves 2 aB/ab Wide-pointy leaves 2 ii. In the table above, circle the recombinant (non-parental) F2 phenotypes The last two rows or Narrow, round leaves and wide-pointy leaves d) You also study Gene D, which regulates root color: brown (dominant, associated with allele D) or green (recessive, associated with allele d). Gene D and Gene A are 10cM apart. i. Draw the two possible chromosomal maps between Genes A, B and D and specify the distance (in cM) between them. Map 1: A(4cM)B(6cM)D Map 2: D(10cM)A(4cM)B Design a cross that can help you identify which of the above chromosomal maps is correct. Note: Specify the genotypes of the Parental (P1 and P2), F1 and F2 plants that you will use in your crosses. Also specify the corresponding % of F2 progenies. You want to know the location of Gene D relative to gene B. So you can design many possible crosses. P1 (BBDD) x P2 (bbdd) -> F1 (BD/bd) x (bd/bd) -> See if the % recombinants in F2 is 6cM or 14cM OR P1 (BBdd) x P2 (bbDD) -> F1 (Bd/bD) x (bd/bd)-> See if the % recombinants in F2 is 6cM or 14cM ii. 3 Question 4 The following schematic shows the orientation of Genes A and B located on chromosome 18 in humans. Note: Trisomy of Chromosome 18 can result in Edwards’s syndrome, an inherited genetic disorder. a) On the schematic below, draw the alignment of replicated chromosomes during metaphase of MITOSIS and in the resulting daughter cells and show the arrangement of the alleles of Genes A and B on all the chromatids. A B a b A B a b A B a b A B a b A B a b b) If the germ cell (reproductive cell) drawn below undergoes MEIOSIS, draw the alignment of replicated chromosomes during each phase of meiosis on the schematic below. Also show the arrangement of the alleles of Genes A and B on all the chromatids in each phase. Note: Assume NO recombination between the alleles of Genes A and B. 4 Question 4 continued c) If the “daughter cell 1” from Meiosis –I above undergoes nondisjunction of chromosome 18, how would the arrangement of the alleles of Genes A and B during different phases change compared to what you showed in Part (b)? Show the change(s) on the schematic below. Note: Assume NO recombination occurs between the alleles of Genes A and B. d) Could the nondisjunction events such as the one you drew in part (c) account for genetic disorders such as Trisomy 18 or Edward’s syndrome? Why or why not? Yes since it produces a gamete that has two copies of Chromosome 18. When this gamete fuses with another gamete of the opposite sex, it results in a zygote that shows trisomy 18. This zygote can develop into a newborn that has Edwards syndrome. e) If the germ cell drawn below undergoes MEIOSIS, give the genotypes of the gametes. Note: Assume Recombination between the alleles of Genes A and B. Gametes would have the genotypes: AB (parental), Ab (recombinants), aB (recombinants) and ab (parental) 5 Question 5 (This question will not be graded) Color blindness in humans has an X- linked recessive mode of inheritance. Prof. Amon draws the following three human pedigrees for you. Note: You may assume that the individuals marrying into the family do not have the disease-associated allele and that no other mutation arises within the pedigree. a) Which of the above pedigrees shares the same mode of inherietnce as color blindness? Explain why you selected this pedigree. Color blindness shows an X-linked recessive mode of inheritence. Only Pedigree B follows this pattern. Here the female in G-II receives a wild-type associated allele of the gene from the mom but a disease associated allele of the same gene from her dad. If the G-II female passes on the disease associated allele to her son , he will be colorblind. If the g-II female passes this disease associated allele to her daughter the daughter will be a carrier but will have a normnal phenotype. The trait in Pedigree A shows an autosomal recessive mode of inheritence and that in Pedigree C shows an autosomal dominant mode of inheritence. b) On the pedigree below, shade in one individual so that the pedigree will show the same mode of inheritance as color blindness. Explain why you selected this individual specifically. 9 The pedigree originally shows an autosomal recessive mode of inheritance. If however you shade Individual 8 then the inheritance pattern changes to X-linked recessive. This is because since #4 is affected (XaXa) and #3 has a wild-type phenotype (XaY) their son (#8) will be affected (XaY) but their daughter (#7) maybe a carrier. 6