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AP Bio

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Unit 5

5.8 MC Answers and Review

7 min read•december 10, 2021

AP Biology 🧬

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Answers and Review for Multiple Choice Practice on Heredity

⛔STOP ⛔ Before you look at the answers, make sure you gave this practice quiz a try so you can assess your understanding

of the concepts covered in Unit 5. Click here for the practice questions: AP Bio Unit 5 Multiple Choice Questions.

Image courtesy of Pixabay

Facts about the test: The AP Biology exam has 60 multiple choice questions and you will be given 1 hour 30 minutes to complete the section. That means it should take you around 15 minutes to complete 10 questions. The following questions were not written by College Board and although they cover information outlined in the AP Biology Course and Exam Description the formatting on the exam may be different.

1. Meiosis: Mitosis

A. creates 2 sets of diploid cells: creates 1 set of diploid cells

B. creates 2 haploid cells: creates 2 diploid cells

C. Creates 4 haploid cells: creates 2 diploid cells

D. creates 4 diploid cells: creates 4 haploid cells

Answer: Ploidy refers to the sets of chromosomes, haploid meaning one set and diploid two. Meiosis creates 4 genetically unique daughter cells that become the gametes, or sex cells in sexually reproducing organisms. Mitosis creates two identical cells needed for growth or repair.

📄 Study AP Biology, Unit 5.1: Meiosis

2. During Prophase I in meiosis,

A. homologous chromosomes pair up and exchange genetic information

B. sister chromatids pair up and exchange genetic information

C. chromosomes unwind to their chromatin form

D. the nuclear envelope reappears

Answer: Homologous chromosomes can be described as a set of chromosomes, one maternal and one paternal. During Prophase I the nuclear envelope breaks down and the homologs form a tetrad where the event known as crossing-over can occur. Genetic information from maternal and paternal chromosomes can exchange.

📄 Study AP Biology, Unit 5.1: Meiosis

3. Independent assortment occurs during Metaphase I where homologous chromosomes line up independently along the metaphase plate. What does this accomplish?

A. Ensures genetic recombination between homologous chromosomes

B. Ensures daughters cells remain identical to the parent cell

C. Organizes the maternal and paternal chromosomes on separate poles

D. Increases genetic variation in the daughter cells

Answer: Independent assortment allows maternal and paternal homologous chromosomes to align randomly at the metaphase plate. This increases genetic variation because varying alleles from paternal/maternal chromosomes can end up separate or paired with alleles from other chromosomes. The possibility of different combinations is astounding!

📄 Study AP Biology, Unit 5.2: Meiosis and Genetic Diversity

4. Why must Meiosis have two divisions? Choose the BEST answer:

A. After duplicating DNA, it must be split in half so that the original chromosome set remains the same

B. to create gametes (sex cells) that contain only one set of chromosomes

C. The DNA of the parent cell must equal the daughter cell

D. Meiosis must create unique daughter cells.

Answer: Although meiosis does produce unique daughter cells, the purpose of two divisions is to create daughter cells with half the original chromosome count, these daughter cells should only have one copy of each chromosome (haploid). When fertilization occurs, the new cell will have the original diploid number.

📄 Study AP Biology, Unit 5.2: Meiosis and Genetic Diversity

5. Metaphase I: Metaphase II

A. homologous chromosomes align: sister chromatids align

B. homologous chromosomes cross over: sister chromatids cross over

C. sister chromatids align: homologous chromosomes align

D. sister chromatids condense: homologous chromosomes condense

Answer: Meiosis II resembles mitosis, so it isn't until Metaphase II that sister chromatids line up at the metaphase plate.

📄 Study AP Biology, Unit 5.2: Meiosis and Genetic Diversity

6. In a classic Mendelian cross, two heterozygous parents should yield a phenotypic ratio of:

A. 1:2

B. 1:3

C. 3:1

D. 4:0

Answer: Heterozygous parents would have both alleles (1 dominant and 1 recessive). The cross reveals 1 homozygous dominant genotype and 2 heterozygous, this gives us the 3 for dominant PHENOTYPIC ratio, and 1 homozygous recessive, giving us the 1 for the recessive phenotypic ratio.

📄 Study AP Biology, Unit 5.3: Mendelian Genetics

7. In a cross of AaBb x AaBb, calculate probability of aabb offspring using probability rules.

A. 1/16

B. 1/4

C. 1/2

D. 9:3:3:1

Answer: Although 9:3:3:1 reveals the phenotypic ratio found in this cross, the question focuses only on the genotype containing ALL recessive alleles, or homozygous recessive for BOTH traits. Use probability rules to solve: each parent has a 1 in 4 chance of creating a gamete with both recessive alleles. Use the product rule (multiplication) to predict the chance of BOTH happening (1/4) x (1/4)= 1/16

📄 Study AP Biology, Unit 5.3: Mendelian Genetics

8. Genotype: Phenotype

A. inherited: mutated

B. impacted by environment: not impacted by environment

C. physical expression of trait: allele combination

D. allele combination: physical expression of trait

Answer: While easily remembered YET confused, this is always important to ensure your complete understanding! Phenotype=physical expression of a trait, what we see! Genotype: look for alleles!

📄 Study AP Biology, Unit 5.3: Mendelian Genetics

9. Human blood type AB exhibits a pattern of inheritance called codominance. Explain why:

A. A and B alleles become a blend

B. A and B become recessive when present together

C. A and B cannot express with the O allele

D. Both A and B alleles are dominant

Answer: An example of non-mendelian genetics, with codominance both alleles are expressed, neither is recessive to the other. BOTH traits coded by alleles present on the gene will be seen.

📄 Study AP Biology, Unit 5.4: Non-Mendelian Genetics

10. If you need to perform a "testcross" which would you include?

A. one homozygous dominant parent, one homozygous parent

B. an "unknown" parent and recessive parent

C. two heterozygous parents

D. two recessive parents

Answer: Test-crosses are performed to determine the genotype of an unknown individual. If you cross the unknown with a recessive, you can use the test-cross results to determine in if the unknown's genotype is homozygous dominant or heterozygous based on the phenotypic ratios.

📄 Study AP Biology, Unit 5.3: Mendelian Genetics

11. In a trait that displays sex-linked recessive inheritance, a male offspring with the trait received the affected trait from:

A. his mother

B. his father

C. mitochondrial DNA of the mother

D. too little information to answer

Answer: While mitochondrial DNA is passed down from the mother, with sex-linked recessive inheritance, a male receives his X chromosome only from the mother as the father contributes the Y chromosome.

📄 Study AP Biology, Unit 5.4: Non-Mendelian Genetics

12. Select the true statement regarding genetic linkage maps (or recombination maps):

A. direct measure of the exact physical location of genes on a chromosome

B. show cross-over frequency of sex chromosomes

C. based on recombination frequencies

D. cannot show whether or not gene linkage exits

Answer: Gene linkage maps are made using recombination frequencies. Sometimes the simplest answer IS the best. While we use recombination frequencies to roughly map out where genes are located, it is NOT based on true physical location. It is in reference to other genes.

📄 Study AP Biology, Unit 5.3: Mendelian Genetics

13. During Anaphase II of Meiosis, a spindle fiber breaks. Describe the outcome:

A. 1 daughter cell with an extra chromosome, 1 daughter cell missing a chromosome

B. 2 normal daughter cells, 2 daughter cells with missing chromosomes

C. 2 normal daughter cells, 2 daughter cells with extra chromosomes

D. 4 daughter cells with an extra chromosome

📄 Study AP Biology, Unit 5.2: Meiosis and Genetic Diversity

14. Traits W, X and Z are found on the same chromosome. Recombination frequencies: W and Z = 7%, W and X = 22%m, Z and X: 15%. Find the order of genes on the chromosome:

A. W-X-Z

B. W-Z-X

C. recombination frequencies are used to find linkage, not gene order

D. crossover frequencies under 50% do not show gene linkage

Answer: While recombination frequencies can show whether or not genes sit on the same chromosome (gene linkage), they CAN tell us relative locations/orders on the chromosomes. The smaller the recombination frequency, the closer the genes. Frequencies larger than 50% show that genes are NOT linked (found on separate chromosomes)

📄 Study AP Biology, Unit 5.6: Chromosomal Inheritance

15. Female aphids can grow wings if their environment is too crowded (to fly away): this is an example of:

A. phenotypic plasticity

B. genotypic plasticity

C. epigenetics

D. natural selection

Answer: Many organisms exhibit phenotypic plasticity! This refers to the environment influencing the expression of certain genes. The genotype of these organisms will not change, but the phenotype (physical expression of the trait) can!

📄 Study AP Biology, Unit 5.5: Environmental Effects on Phenotype

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