Mendel, sex determination, evolutionPractice Questions & MCQs
Practice Mendel, sex determination, evolution questions from Heredity and Evolution (Class 10 Science) - important MCQs with answers and step-by-step solutions. Solve the sample questions below, then practice the full set free on the RankByte app.
Mendel, sex determination, evolutionQuiz - Solve & Score
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.
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.
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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