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Biology for AP® Courses

Critical Thinking Questions

Biology for AP® CoursesCritical Thinking Questions
20 .
Describe one reason why the garden pea was an excellent model system for studying inheritance.
  1. The garden pea has flowers that close tightly to promote cross-fertilization.
  2. The garden pea has flowers that close tightly to prevent cross-fertilization.
  3. The garden pea does not mature in one season and is a perennial plant.
  4. Male and female reproductive parts attain maturity at different times, promoting self-fertilization.
21 .
Consider Mendel's studies of the traits of garden pea plant. Provide a reasoned claim, based on his experiments, for how you would perform a reciprocal cross to test stem height in the garden pea.
  1. First cross is performed by transferring the pollen of a heterozygous tall plant to the stigma of a true breeding dwarf plant. Second cross is performed by transferring the pollen of a heterozygous dwarf plant to the stigma of a true breeding tall plant.
  2. First cross is performed by transferring the pollen of a true breeding tall plant to the stigma of a true breeding dwarf plant. Second cross is performed by transferring the pollen of a true breeding dwarf plant to the stigma of a true breeding tall plant.
  3. First cross is performed by transferring the pollen of a true breeding tall plant to the stigma of a heterozygous dwarf plant. Second cross is performed by transferring the pollen of a heterozygous dwarf plant to the stigma of a true breeding tall plant.
  4. First cross is performed by transferring the pollen of a heterozygous tall plant to the stigma of a heterozygous dwarf plant. Second cross is performed by transferring the pollen of a heterozygous tall plant to the stigma of a heterozygous dwarf plant.
22 .
One of the traits that Mendel examined in pea plants was flower position. Flower position in pea plants is determined by a gene with axial and terminal alleles. Given that axial is dominant to terminal, predict all of the possible F1 and F2 genotypes and phenotypes from a cross involving parents that are homozygous for each trait. Express genotypes with conventional genetic abbreviations.
  1. F1: All AA-axial; F2: AA-Axial and aa-terminal.
  2. F1: All aa-terminal; F2: AA-Axial and Aa-terminal.
  3. F1: AA-axial and AAa-terminal; F2: All AA-axial.
  4. F1: All Aa-axial; F2: AA-Axial, Aa-Axial, and aa-terminal.
23 .
Construct a Punnett square to predict the offspring in a cross between a dwarf pea plant (homozygous recessive) and a tall pea plant (heterozygous). Determine and identify here the correct phenotypic ratio of the offspring.
  1. 1 tall : 1 dwarf
  2. 1 tall : 2 dwarf
  3. 3 tall : 1 dwarf
  4. 1 dwarf : 4 tall
24 .
Use your understanding of the type of inheritance associated with color blindness to provide and justify an answer to the following question: Can a human male be a carrier of red-green color blindness?
  1. Yes, males can be the carriers of red-green color blindness, as color blindness is autosomal dominant.
  2. No, males cannot be the carriers of red-green color blindness, as color blindness is X-linked.
  3. No, males cannot be the carriers of red-green color blindness, as color blindness is Y-linked.
  4. Yes, males can be the carriers of red-green color blindness, as color blindness is autosomal recessive.
25 .
The probability method provides proportions (ratios) of phenotypes and genotypes expected to be exhibited by offspring in a given cross, without using visual tools. Use the probability method to calculate the genotypes and genotypic proportions of a cross between AABBCc and Aabbcc parents.
  1. Possible genotypes are AABBcc, AaBbCc, AaBbcc, and the ratio is 1 : 2 : 1.
  2. Possible genotypes are AABbcc, AaBbCc, AaBbcc, and the ratio is 1 : 3 : 1 .
  3. Possible genotypes are AABbCc, AABbcc, AaBbCc, AaBbcc, and the ratio is 1 : 1 : 1 : 1 .
  4. Possible genotypes are AABbcc, AaBbCC, AaBbcc, and the ratio is 1 : 1 : 1 .
26 .
Justify the claim that segregation of traits results in different combinations at the end of meiosis by providing a detailed description of the mechanisms involved.
  1. The chromosomes randomly align during anaphase I at the equator. Separation of bivalent chromosomes occur during metaphase I of meiosis I. Similarly, separation of sister chromatids occurs at metaphase II of meiosis II. At the end of meiosis II, four different gametic combinations are produced, each containing a haploid set of chromosomes.
  2. The chromosomes randomly align during metaphase I at the equator, and separation of homologous chromosomes occurs during anaphase I. Similarly separation of sister chromatids occurs at anaphase II of meiosis II. At the end of meiosis II, four different gametic combinations are produced, each containing a haploid set of chromosomes.
  3. The chromosomes randomly align during prophase I at the equator, and separation of sister chromatids occurs during metaphase I of meiosis I. Similarly separation of bivalent chromosomes occur at metaphase II of meiosis II. At the end of meiosis II, four different gametic combinations are produced, each containing a diploid set of chromosomes.
  4. The chromosomes randomly align during prophase I at the equator, and separation of bivalent chromosomes occur during anaphase I of meiosis I. Similarly, separation of homologous chromosomes occurs at metaphase II of meiosis II. At the end of meiosis II, four different gametic combinations are produced, each containing a diploid set of chromosomes.
27 .

Inheritance of fruit color in summer squash provides an example of the phenomenon of epistasis. Yellow fruit is produced by a cross involving the homozygous recessive expression of the W gene (ww]) and either the homozygous dominant (YY) or heterozygous (Yy) form of the Y) gene. The wwyy outcome is green fruit. White fruit arises whenever a dominant copy of the W gene is involved in the cross, regardless of the Y alleles present.

A cross of white WwYy heterozygotes produces the phenotypic ratio of 12 white : 3 yellow : 1 green fruits. Determine the best explanation for how these outcomes occur.

  1. 12 offspring are white because the W gene is epistatic to the Y gene. Three offspring are yellow, because w is not epistatic. Green offspring are obtained when the recessive form of both genes (wwyy) are present.
  2. 12 offspring are white becasue W gene is hypostatic to Y gene. Three offspring are yellow because Y is epistatic to w. Green offspring are obtained when the dominant form of both the genes (WWYY) is present.
  3. 12 offspring are white because W gene is dominant. Three offspring are yellow because Y is dominant and w is recessive. Green offspring are obtained when the recessive form of both the genes (wwyy) is present, showing codominance.
  4. 12 offspring are white because W is epistatic to Y gene. Three offspring are yellow because Y is hypostatic to w. Green offspring are obtained when the recessive alleles of both genes (wwyy) are present, showing recessive epistasis.
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