What is the probability that parents AaBb and aabb will have offspring with genotype AaBb
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The cross of AaBb X AaBb produces offspring with phenotype, the possible number is(a)2(b)4(c)16(d)25(e)50Answer  Hint: The cross of AaBb X AaBb produces offspring with phenotype, the possible number is adequate to the square of the number of gametes formed from the one parent cross of AaBb X AaBb. Complete answer: Note: -On the basis of results of Dihybrid cross, Mendel proposed his second law of inheritance and is called the law of Independent Assortment of factors. This law states that in a hybrid union for two or many traits, the different traits are inherited independently of each other as genes for two characteristics are inherited
independently.
Answer: Cc and Dd Explanation: because this is the near on AaBb hope it's help correct me if Im rwong Answer: The correct answer:The probability of an aabb offspring when AaBb x AaBb parents are crossed is b. 1/16 .
A Punnett Square* shows the genotype*s two individuals can produce when crossed. To draw a square, write all possible allele* combinations one parent can contribute to its gametes across the top of a box and all possible allele combinations from the other parent down the left side. The allele combinations along the top and sides become labels for rows and columns within the square. Complete the genotypes in the square by filling it in with the alleles from each parent. Since all allele combinations are equally likely to occur, a Punnett Square predicts the probability of a cross producing each genotype. Number of traits in cross:
Edit Alleles: Show: Genotype Phenotype* A single trait Punnett Square tracks two alleles for each parent. The square has two rows and two columns. Adding more traits increases the size of the Punnett Square. Assuming that all traits exhibit independent assortment, the number of allele combinations an individual can produce is two raised to the power of the number of traits. For two traits, an individual can produce 4 allele combinations (2^2). Three traits produce 8 combinations (2^3). Independent assortment typically means the genes are on different chromosome*s. If the genes for the two traits are on the same chromosome, alleles for each trait will always appear in the same combinations (ignoring recombination). With one row or column for each allele combination, the total number of boxes in a Punnett Square equals the number of rows times the number of columns. Multi-trait Punnett Squares are large. A three trait square has 64 boxes. A four trait square has 256 boxes. The genotype in each box is equally likely to be produced from a cross. A two-trait Punnett Square has 16 boxes. The probability of a cross producing a genotype in any box is 1 in 16. If the same genotype is present in two boxes, its probability of occurring doubles to 1/8 (1/16 + 1/16). If one of the parents is a homozygote for one or more traits, the Punnett Square still contains the same number of boxes, but the total number of unique allele combinations is 2 raised to the power of the number of traits for which the parent is heterozygous. A commonly discussed Punnett Square is the dihybrid cross. A dihybrid cross tracks two traits. Both parents are heterozygous, and one allele for each trait exhibits complete dominance*. This means that both parents have recessive alleles, but exhibit the dominant phenotype. The phenotype ratio predicted for dihybrid cross is 9:3:3:1. Of the sixteen possible allele combinations:
This pattern only occurs when both traits have a dominant allele. With no dominant alleles, more phenotypes are possible, and the phenotype probabilities match the genotype probabilities. A simpler pattern arises when one of the parents is homozygous for all traits. In this case, the alleles contributed by the heterozygous parent drives all of the variability. A two trait cross between a heterozygous and a homozygous individual generates four phenotypes, each of which are equally likely to occur. More complicated patterns can be examined. In an extreme case when more than two alleles exists for each trait and the parents do not possess same alleles, the total number of genotypes equals the number of boxes in the Punnett Square. It is possible to generate Punnett squares for more that two traits, but they are difficult to draw and interpret. A Punnett Square for a tetrahybrid cross contains 256 boxes with 16 phenotypes and 81 genotypes. A third allele for any one of the traits increases the number of genotypes from 81 to 108. Given this complexity, Punnett Squares are not the best method for calculating genotype and phenotype ratios for crosses involving more than one trait. Test your understanding with the Punnett Square Calculator Problem Set. Video Overview Related Content
What is cross between AaBb and AaBb?The homozygous gametes separate from the dihybrid parents. The heterozygous progeny is formed when these gametes fuse together. As a result, all of the offspring are Heterozygous AaBb. The ratio of F, genotypes between AABB, AaBB, Aabb, and aabb in a hybrid between AABB and aabb would be 1 : 2 : 2 : 1.
How many types of genotypes will be produced in a cross AaBb AaBb?So, the correct answer is, '16.
When AaBb and AaBb are crossed in generation the ratio of AaBb will be?So, the correct answer is 1 AaBB: 1 aaBB.
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