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#dihybrid cross

13 public questions tagged with this topic.

Aptitude and Reasoning

The number of phenotypes in the F2 of the dihybrid will be

This outcome reflects quantitative reasoning, where ratio, percentage and series logic determine correct calculation. Option C captures this correctly because it meets mathematical and logical criteria, consistent with textbook descriptions and experimental observations in aptitude and reasoning.

Ref: Agrawal, Quantitative Aptitude for Competitive Examinations, Chapter 5, Logical Reasoning and Probability, covers ratios, percentages, series completion, time-distance and analytical reasoning explaining problem-solving strategies for competitive assessments.

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Gene intraction

The classical Mendelian dihybrid phenotypic ratio is

Classical Mendelian dihybrid experiment crossed peas differing in two traits such as seed color Yy and shape Rr. F1 heterozygotes YyRr selfed produce gametes YR, Yr, yR, yr equally. Random union yields F2 phenotypic distribution 9 Y-R- both dominant, 3 Y-rr one dominant, 3 yyR- other dominant, 1 yyrr double recessive. This 9:3:3:1 ratio equals product of two independent 3:1 monohybrid ratios, signifying independent assortment of unlinked genes with complete dominance. Any significant distortion suggests linkage creating parental excess or epistasis collapsing classes, making ratio reference st

Ref: Griffiths et al., Introduction to Genetic Analysis, 12th ed., Chapter 3: Classical Dihybrid 9:3:3:1 Ratio

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Genetic Crosses

A test cross of a dihybrid heterozygote results in which phenotypic ratio?

Testing double heterozygote AaBb with homozygous recessive aabb directly exposes its gametic output because tester contributes only ab alleles. Under independent assortment, heterozygote meiosis yields four gamete types AB, Ab, aB, ab equally at 25% each, generating progeny genotypes AaBb, Aabb, aaBb, aabb in equal 1:1:1:1 phenotypic ratio. This equal distribution distinguishes Mendelian independent assortment from linkage where parental allele combinations exceed recombinants. Consequently dihybrid test cross serves both as verification of independent assortment and as quantitative assay for

Ref: Hartl & Jones, Genetics: Analysis of Genes and Genomes, 8th ed., Chapter 4: Dihybrid Test Cross 1:1:1:1

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Genetic Crosses

The classical dihybrid phenotypic ratio 9:3:3:1 indicates

Independent assortment principle states alleles at unlinked loci orient randomly on metaphase plate independently of other loci, producing equal parental and recombinant gamete frequencies. In cross AaBb x AaBb this random orientation yields F2 phenotypic distribution 9 A-B- :3 A-bb :3 aaB- :1 aabb, where 9 represents both dominant traits together. Product of two 3:1 monohybrid ratios mathematically. Significant excess of parental phenotypes indicates linkage with reduced recombination, while collapsing of classes into fewer phenotypes signals epistatic interaction. Therefore intact 9:3:3:1 re

Ref: Griffiths et al., Introduction to Genetic Analysis, 12th ed., Chapter 3: Independent Assortment and 9:3:3:1

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Genetic Crosses

A dihybrid cross AaBb × AaBb produces how many phenotypic classes in F2?

A dihybrid individual AaBb with unlinked genes can form four gamete types AB, Ab, aB, ab through independent assortment during meiosis. Selfing AaBb x AaBb produces 4x4=16 genotypic combinations collapsed by complete dominance at each locus. Phenotypically, genotype classes group into A-B- displaying both dominant traits, A-bb showing first dominant alone, aaB- second dominant alone, aabb both recessive. Thus four distinct phenotypic classes appear despite nine genotypic classes. Distinction highlights phenotype versus genotype numbers and forms basis for modified ratios when epistasis operate

Ref: Pierce, Genetics: A Conceptual Approach, 7th ed., Chapter 3: Phenotypic Classes in Dihybrid F2

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Mendelian Inheritance

In a test cross of a dihybrid, the phenotypic ratio expected is

A dihybrid test cross intercrosses double heterozygote AaBb with homozygous recessive tester aabb, directly exposing gamete constitution. When two genes assort independently, heterozygote meiosis generates four gamete types AB, Ab, aB, ab in equal 25% frequency due to random chromosome alignment at metaphase I and independent segregation. Tester parent contributes only ab gametes, so progeny genotypes AaBb, Aabb, aaBb, aabb appear equally, producing phenotypic ratio 1:1:1:1. Any significant deviation toward excess parental types indicates linkage and permits calculation of recombination freque

Ref: Hartl & Ruvolo, Genetics: Analysis of Genes and Genomes, 9th ed., Chapter 4: Dihybrid Test Cross and Independent Assortment

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