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Download 5. Intro to Dihybrid Crosses
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An Introduction to Dihybrid Crosses Punnett squares can be used to predict the possible genotypes of the offspring from two parents. Punnett square in which only one gene is considered at called monohybrid crosses. However, these can be extended to crosses involving two genes which affect two traits (called dihybrid crosses). For example: Two red-eyed, long-winged flies, heterozygous for both traits, are crossed. Red eyes is dominant to white eyes and long-winged is dominant to short-winged. Determine the possible genotypes of their offspring. Step 1: Determine the genotypes of the parents - Let R = red eyes, r = white eyes; let L = long-winged, l = short winged. - Parents: RrLl x rrll Step 2: Determine the gametes - In a monohybrid cross, only one gene is considered and each gene has two possible alleles, therefore there are only 2 possible combinations that could be passed on to the offspring. - In a dihybrid cross, two genes are considered, and each gene has two alleles, therefore there are 4 possible combinations for each parent (see below). - Parents: RrLl o Combination 1: RL o Combination 2: Rl o Combination 3: rL o Combination 4: rl Step 3: Create a Punnett square - Because there are more possible combinations in the gametes (eggs/sperm), a Punnett square for a dihybrid cross will be much bigger than for a monohybrid cross. - Just like a monohybrid cross, the genotypes of the parents go across the top & down the side of the square RL RRLL RRLl RrLL RrLl RL Rl rL rl Rl RRLl RRll RrLl Rrll rL RrLL RrLl rrLL rrLl rl RrLl Rrll rrLl rrll TRY THE PROBLEMS BELOW 1. Green-podded (G), axial-flowering (A) peas x yellow-podded (g), terminal-flowering (a) peas, in which green pod is dominant to yellow pod and axial flowers is dominant to terminal flowers. Determine the correct genotypic and phenotypic ratios for the F1 generation in the following situations. a. b. c. d. Assume the green, axial plants are homozygous for both traits. Assume the green, axial plants are homozygous for pod colour, but heterozygous for flower position. Assume the green, axial plants are heterozygous for pod colour but homozygous for flower position. Assume the green, axial plants are heterozygous for both traits. 2. A plant has two independently assorting genes for plant height (D) and yellow leaves (Y) with the alleles: D = tall plant Y = green leaves d = dwarf plant y = yellow leaves Two plants, heterozygous for height and yellow leaves, were crossed. a. What is the chance that the offspring will be heterozygous for both loci? b. What is the chance that the offspring will be homozygous recessive for both loci? c. What is the chance that the offspring will be tall with yellow leaves? d. What is the chance that the offspring will either be tall with green leaves or tall with yellow leaves? Answers Question 1 a. Parents: GGAA x ggaa Gametes: GA and ga Genotypic Ratio: 100% GgAa Phenotypic Ratio: 100% green-pod, axial-flowering b. Parents: GGAa x ggaa Gametes: GA, Ga and ga Genotypic Ratio: ½ GgAa: ½ Ggaa Phenotypic Ratio: ½ Green axial : ½ Green terminal c. Parents: GgAA x ggaa Gametes: GA, gA and ga Genotypic Ratio: 50% GgAa, 50% ggAa Phenotypic Ratio: 50% green axial : 50% yellow axial d. Parents: GgAa x ggaa Gametes: GA, Ga, gA, ga and ga Genotypic Ratio: ¼ GgAa : ¼ Ggaa : ¼ ggAa : ¼ ggaa Phenotypic Ratio: ¼ green axial : ¼ green terminal : ¼ yellow axial : ¼ axial terminal Question 2 Parents: DdYy x DdYy Gametes: DY, Dy, dY, dy a. Chance of DdYy = 4/16 = 1/4 b. Chance of ddyy = 1/16 c. Tall with yellow leaves = D_yy = 3/16 d. Tall with yellow leaves = D_yy = 3/16 Tall with green leaves = D_Y_ = 9/16 Total = 12/16 = 3/4 DY Dy dY dy DY DDYY DDYy DdYY DdYy Dy DDYy DDyy DdYy Ddyy dY DdYY DdYy ddYY ddYy dy DdYy Ddyy ddYy ddyy
