Genetics Activity 3 - Test Crosses For many genetic characters, the phenotype of a homozygous dominant individual and a heterozygous individual is the same. It is not possible to tell the genotype by the physical appearance of the character. One can determine the genotype by conducting a test cross in which the individual with the dominant phenotype, but unknown genotype, is bred with a homozygous recessive individual (which is usually known). The phenotypic ratio of the F1 offspring will typically reveal the genotype of the parent. For example, cross a corn plant that produces cobs with white kernels (homozygous recessive, rr) with a corn plant that produces cobs with purple kernels (homozygous dominant, RR or heterozygous, Rr). Use Punnett squares to predict the phenotypes of the F1 offspring. RR x rr Rr x rr A cross between a homozygous dominant plant (RR) and a homozygous recessive plant (rr) produces offspring with the phenotypic ratio of: A cross between a heterozygous dominant plant (Rr) and a homozygous recessive plant (rr) produces offspring with a phenotypic ratio of:
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Punnett square: ``` R R d Rr Rr d Rr Rr ``` Phenotypic ratio of the F1 offspring: 4 purple : 0 white **Step 2:** Perform a Punnett square for the cross between a heterozygous dominant plant (Rr) and a homozygous recessive plant (dd). Punnett Show more…
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In pea plants, round (R) seeds are dominant to wrinkled (r) seeds. Parental cross purebred round-seed (RR) plants are bred to purebred wrinkled-seed plants. Draw a Punnett square for this cross and determine the genotypic and phenotypic ratios of the offspring. The resulting plants are then crossed with each other (intercross). Draw a Punnett square and determine the genotypic and phenotypic ratios of the offspring. What are the genotypic and phenotypic ratios that result from crossing a heterozygous round (Rr) plant with a homozygous round (RR) plant? Genotype: It could be RR or Rr. To determine the genotype of a round seed plant, you need to cross it with a plant of known genotype. This is called a test cross. By doing a test cross, you can figure out the genotype of the round seed plant. Since wrinkled is recessive, you will always know the genotype of the wrinkled-seed plants and their offspring from the test cross (Rr). Tell the possible phenotypic ratios of the offspring from the test cross (genotype of the unknown).
Anand J.
Predict phenotype and genotype ratios of the offspring from the following crosses using Punnett squares: 1. Homozygous tall x homozygous tall pea plants 2. Homozygous tall x homozygous short pea plants 3. Heterozygous tall x heterozygous tall pea plants Write the genotypes of the parents when crossing heterozygous purple and heterozygous purple pea plants. Show the Punnett square for this cross. What is a test cross? Show the Punnett square of a test cross of a homozygous short pea plant and a type 0 blood individual. For the cross of type AB x type AB and type A x type B, could these parents produce a type 0 child? Describe the pathogenesis of Erythroblastosis fetalis (after Immunity). See Rh discussion, figure 34.9. Do this after the immunity section of the course. Review the heritability of sickle cell anemia and illustrate a cross of two carriers: Hb hb X Hb hb. Review Hemophilia (sex-linked recessive) and illustrate a cross of a carrier female and a normal male: X^hX X^hY. Illustrate a cross of a normal female and a hemophiliac male: XX X^hY. Make up your own di-hybrid cross (both parents heterozygous for both traits).
Adi S.
A dihybrid cross involves two traits. A cross of parental types AaBb and AaBb can be represented with a Punnett square: This representation clearly organizes all of the possible genotypes and reveals the 9:3:3:1 distribution of phenotypes and a 4×4 grid of 16 cells. Expressed as a fraction of the 16 possible genotypes of the offspring, the phenotypic ratio describes the probability of each phenotype among the offspring: 3 (AA, Aa, aA) × 3 (BB, bB, Bb)/16 = 9/16; 3 (AA, Aa, aA) × 1 (bb) /16 = 3/16; 1 (aa) × 3 (BB, bB, Bb) = 3/16; and 1 (aa) × 1 (bb) = 1/16. A. Using the probability method, calculate the likelihood of these phenotypes from each dihybrid cross: recessive in the gene with alleles A and a from the cross AaBb × aabb dominant in both genes from the cross AaBb × aabb recessive in both genes from the cross AaBb × aabb recessive in either gene from the cross AaBb × aabb A Punnett square representation of a trihybrid cross, such as the self-cross of AaBbCc, is more cumbersome because there are eight columns and rows (2×2×2 ways to choose parental genotypes) and 64 cells. A less tedious representation is to calculate the number of each type of genotype in the offspring directly by counting the unique permutations of the letters representing the alleles. For example, the probability of the cross AaBbCc × AaBbCc is 3 (AA, Aa, aA) × 3 (BB, Bb, bB) × 3 (CC, Cc, cC)/64 = 27/64. B. Using the probability method, calculate the likelihood of these phenotypes from each trihybrid cross: recessive in all traits from the cross AaBbCc × aabbcc recessive in the gene with alleles C and c and dominant in the other two traits from the cross AaBbCc × AaBbCc dominant in the gene with alleles A and a and recessive in the other two traits from the cross AaBbcc × AaBbCc C. The probability method is an easy way to calculate the likelihood of each particular phenotype, but it doesn’t simultaneously display the probability of all possible phenotypes. The forked line representation described in the text allows the entire phenotypic distribution to be displayed. Using the forked line method, calculate the probabilities in a cross between AABBCc and Aabbcc parents: all traits are recessive: aabbcc traits are dominant at each loci, A?B?C? traits are dominant at two genes and recessive at the third traits are dominant at one gene and recessive at the other two
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