Class 10 · Science · Biology · Heredity and Evolution

Mendel, sex determination, evolution MCQs

Practice Mendel, sex determination, evolution multiple-choice questions from Heredity and Evolution (Class 10 Science) - tap an answer for instant feedback and a step-by-step solution. Practice the full set free on the RankByte app.

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  1. Q1. Which of the following is the BEST evidence to estimate the age of a fossil older than 100 million years?

    • A.Radioactive dating using long-half-life isotopes such as K-40/Ar-40 or U-238/Pb-206
    • B.Carbon-14 dating, because all organic matter contains C-14
    • C.Direct comparison with present-day species of similar appearance
    • D.Counting tree-ring like rings on the fossil bones

    Answer: A. Radioactive dating using long-half-life isotopes such as K-40/Ar-40 or U-238/Pb-206

    What does the chapter say about this? Carbon-14 has a short half-life (~5,730 years) and is reliable only for fossils up to ~50,000 years old. For much older fossils, isotopes with long half-lives such as K-40/Ar-40 (~1.25 billion years) or U-238/Pb-206 (~4.5 billion years) are used to date associated igneous rocks. Visual comparison and bone rings are unreliable for absolute ages. Pick: A) Radioactive dating using long-half-life isotopes such as K-40/Ar-40 or U-238/Pb-206.

  2. Q2. Which of the following is the best example of ANALOGOUS organs?

    • A.Wings of a bird and wings of an insect
    • B.Forelimbs of a horse and forelimbs of a dog
    • C.Flippers of a whale and forelimbs of a bat
    • D.Human appendix and goat appendix

    Answer: A. Wings of a bird and wings of an insect

    NCERT fact (biology, chapter 'Heredity and Evolution'): Analogous organs have similar function but different evolutionary origin. Bird wings are modified forelimbs with feathers and a vertebrate bone plan, while insect wings are outgrowths of the exoskeleton - no common ancestor for these structures. The other three options are HOMOLOGOUS (forelimb plan) or unrelated. Final answer - A) Wings of a bird and wings of an insect.

  3. Q3. A pure-breeding tall pea plant (TT) is crossed with a pure-breeding dwarf (tt). What will be the phenotypic and genotypic ratio of the F2 generation?

    • A.Phenotypic ratio 3 tall : 1 dwarf; Genotypic ratio 1 TT : 2 Tt : 1 tt
    • B.Phenotypic ratio 1 tall : 1 dwarf; Genotypic ratio 1 TT : 1 tt
    • C.Phenotypic ratio 9:3:3:1; Genotypic ratio 1:2:1
    • D.All tall plants with genotype Tt only

    Answer: A. Phenotypic ratio 3 tall : 1 dwarf; Genotypic ratio 1 TT : 2 Tt : 1 tt

    What is the textbook's take? F1 of TT × tt is all Tt (tall). Selfing F1 gives F2: 1 TT : 2 Tt : 1 tt - the classic Mendelian 1:2:1 genotypic ratio, which corresponds to a 3 tall : 1 dwarf phenotypic ratio because TT and Tt both show the dominant tall phenotype. 9:3:3:1 is dihybrid, not monohybrid. F1 (not F2) is all Tt. Pick: A) Phenotypic ratio 3 tall : 1 dwarf; Genotypic ratio 1 TT : 2 Tt : 1 tt.

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