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Chapter 10 and 11 Test review

Chapter 10 and 11 Test review . 2012. b. meiosis and fertilization. The numbers in Figure 10-1 represent the chromosome number found in each of the dog cells shown. The processes that are occurring at A and B are ____. .

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Chapter 10 and 11 Test review

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  1. Chapter 10 and 11 Test review 2012

  2. b. meiosis and fertilization • The numbers in Figure 10-1 represent the chromosome number found in each of the dog cells shown. The processes that are occurring at A and B are ____.

  3. A white mouse whose parents are both white produces only brown offspring when mated with a brown mouse. The white mouse is most probably ____. SC.912.L.16.2 a.homozygous recessive b.heterozygous c.homozygous dominant d.haploid

  4. In chickens, rose comb (R) is dominant to single comb (r). A homozygous rose-combed rooster is mated with a single-combed hen. All of the chicks in the F1 generation were kept together as a group for several years. They were allowed to mate only within their own group. What is the expected phenotype of the F2 chicks? SC.912.L.16.2 a.100% rose comb b.75% rose comb and 25% single comb c.100% single comb d.50% rose comb and 50% single comb

  5. 4. In mink, brown fur color is dominant to silver-blue fur color. If a homozygous brown mink is mated with a silver-blue mink and 8 offspring are produced, how many would be expected to be silver-blue? SC.912.L.16.2 • a.0 bb • b.3 • c.6 • d.8

  6. 5. The diagram in Figure 10-2 shows a diploid cell with two homologous pairs of chromosomes. Due to independent assortment, the possible allelic combinations that could be found in gametes produced by the meiotic division of this cell are ____. SC.912.L.16.2 Figure 10-2 a.Bb, Dd, BB, and DD b.BD, bD, Bd, and bd c.BbDd and BDbd d.Bd and bD only

  7. 6. Using Figure 10-3, which process would result in the formation of chromosome C from chromosomes A and B? Figure 10-3 a. asexual reproduction b. independent assortment c. crossing over d. segregation

  8. 7. What fraction of this cross will be recessive for both traits? SC.912.L.16.2 a.1/2 b.1/4 c.1/8 d.1/16 9:3:3:1 dihybrid cross!!!

  9. 8. Which event during meiosis leads to a reduction in chromosome number from 2n to n? SC.912.L.16.16 a. Pairs of homologous chromosomes line up at the equator. b. DNA undergoes replication. c. Homologous chromosomes travel to opposite sides of the cell. d. Sister chromatids are pulled apart at the centromere.

  10. 9. Crossing over would most likely occur during which stage of the cell cycle? SC.912.L.16.16 a. when DNA is being replicated b. when homologous chromomosomes line up in pairs c. when centromeres are separated d. when cytokinesis begins

  11. 10. The typical human body cell contains 46 chromosomes. How many chromosomes are found in a typical human sperm? SC.912.L.14.2 • a.23 • b.45 • c.46 • d.92

  12. 11. Suppose an animal is heterozygous AaBb, and the traits are not linked. When meiosis occurs, what is the total number of possible combinations of gametes that can be made for these traits? SC.912.L.16.2 • a.2 • b.4 • c.6 • d.8 • AB, Ab, aB, ab

  13. 12. Which stage of meiosis is responsible for the law of independent assortment? SC.912.L.16.16 • a.metaphase I • b.prophase I • c.telophase I • d.metaphase II Random way they line up!

  14. 13. In mice, black is dominant to white color and color is determined by a single gene. Two black mice are crossed. They produce 2 black offspring and one white offspring. If the white offspring is crossed with one of its parents, what percent of the offspring are expected to be white? SC.912.L.16.2 • a.0 • b.25 • c.50 • d.75

  15. 14. Mendel crossed a true-breeding plant that produced green seeds with a true-breeding plant that produced yellow seeds to produce an F1 generation. The entire F1 generation produced yellow seeds. Then he crossed the F1 offspring with each other to produce the F2 generation. From the F2 generation, he counted 6022 yellow seeds. Which of these is the most likely estimate of the number of green seeds he collected from the F2 generation? SC.912.L.16.1 • a.0 • c.6000 • b.2000 is 25% of 8000 • d.18000 75% yellow 25% green

  16. 15. A heterozygous organism is best described as which of these? SC.912.L.16.2 • a. dominant • b. genotype • c. hybrid • d. true-breeding

  17. 16. In which situation are the phenotypes of F2 offspring expected to follow the ratio of 9:3:3:1. SC.912.L.16.2 • a. a monohybrid cross for 2 unlinked traits • b. a monohybrid cross for 2 closely linked traits • c. a dihybrid cross for 2 unlinked traits • d. a dihybrid cross for 2 closely linked traits

  18. 17. Of the following species used in agriculture, which is most likely a polyploid? • a.cow • b. goat • c.hen • d. wheat • Mostly in plants!

  19. 18. A geneticist crossed fruit flies to determine whether two traits are linked. The geneticist crossed a fly with blistery wings and spineless bristles (bbss) with a heterozygous fly that had normal wings and normal bristles (BbSs). Which results in the next generation would suggest these traits are linked? SC.912.L.16.2 a. 35 normal wings, normal bristles, 28 normal wings, spineless bristles, 23 blistery wings, normal bristles, 30 blistery wings, spineless bristles b. 105 normal wings, normal bristles, 101 normal wings, spineless bristles, 111 blistery wings, normal bristles, 115 blistery wings, spineless bristles c. 198 normal wings, normal bristles, 200 normal wings, spineless bristles, 185 blistery wings, normal bristles, 189 blistery wings, spineless bristles d. 222 normal wings, normal bristles, 27 normal wings, spineless bristles, 22 blistery wings, normal bristles, 228 blistery wings, spineless bristles • Unusual ratios! You would expect 25% of each type and all of the other answers have that!

  20. 19. Consider the cell labeled X in Figure 10-9 containing 4 chromosomes. Which of the four cells below it represents a healthy gamete that could be produced from this cell? SC.912.L.16.16 • a.A half of what is in the regular cell! • b.B • c.C • d.D

  21. Refer to Figure 11-1. If individual III-2 marries a person with the same genotype as individual I-1, what is the chance that one of their children will be afflicted with hemophilia? Sex linked SC.912.L.16.2 • a.0% • b.25% • c.50% • d.75% I 1Will be heterozygous and III 2 will be recessive but this will cause all children to be heterozygous ( carriers) and not have hemophilia.

  22. 21. What type of inheritance pattern does the trait represented by the shaded symbols in Figure 11-1 illustrate? SC.912.L.16.2 • a. incomplete dominance • c. codominance • b. multiple alleles • d. sex-linked Males have it And females Carry it!

  23. 22. For the trait being followed in the pedigree, individuals II-1 and II-4 in Figure 11-1 can be classified as ____. SC.912.L.16.2 a. homozygous dominant b. mutants c. homozygous recessive d. carriers

  24. 23. What is the relationship between individual I-1 and individual III-2 in Figure 11-1? • grandfather-granddaughter b. grandmother-grandson c. great aunt-nephew d. mother-son

  25. 24. When roan cattle are mated, 25% of the offspring are red, 50% are roan, and 25% are white. Upon examination, it can be seen that the coat of a roan cow consists of both red and white hairs. This trait is one controlled by ____. SC.912.L.16.2 a. multiple alleles b. codominant alleles c. sex-linked genes d. polygenic inheritance • Both alleles can be expressed

  26. 25. Examine the graph in Figure 11-3, which illustrates the frequency in types of skin pigmentation in humans. Another human trait that would show a similar inheritance pattern and frequency of distribution is _______ SC.912.L.16.2 Figure 11-3 a. Height things that have a range of phenotypes b. blood type c. number of fingers and toes d. incidence of cystic fibrosis

  27. 26. A man heterozygous for blood type A marries a woman heterozygous for blood type B. The chance that their first child will have type O blood is ____. SC.912.L.16.2 • a.0% • b.25% • c.50% • d.75%

  28. 27. According to Figure 11-4, what is the chance that individual A will be afflicted with Huntington's? • Figure 11-4 • a.25% • b.50% • c.75% • d.100% • Huntington’s is a dominant disorder. The male parent must be Hh to have normal children. So Hh x hh will have a 50% chance of a Huntington’s child.

  29. 28. Nondisjunction is related to a number of serious human disorders. How does nondisjunction cause these disorders? SC.912.L.16.16 • a. alters the number of gametes produced • b. alters the number of zygotes produced • c. alters the chromosome structure • d. alters the chromosome number

  30. 29. What occurs during the process of meiosis in humans that can lead to a child with the condition of Down Syndrome? SC.912.L.16.2 a. production of a duplicate chromosome set b. production of gametes which are diploid c. production of gametes with one duplicate chromosome d. production of gametes with one duplicate sex chromosome

  31. 30. Which of the following could only be a result of nondisjunction during meiosis of sperm formation and not egg formation? SC.912.L.16.16 a. XYY b. XXX c. XXY d. XO sperm are XY eggs XX so only the one with a problem with the extra Y

  32. 31. A series of plants, produced through vegetative propagation (asexual reproduction), have been planted at different altitudes. The researcher observes that the higher the altitude, the shorter the plant will grow. Please make a hypothesis to explain the results. SC.912.N.1.1 a. Higher altitudes affect the presence of genes that determine height. b. Higher altitudes decrease the expression of genes that promote height. c. The plants at lower altitudes are genetically different from plants at higher altitudes. d. Lower altitudes cause mutations in genes that determine height. the environment can change how the genes are expressed but not the genes themselves!

  33. This pedigree shows the transmission of a rare disease that is debilitating but not lethal. Carriers are not shown. 32. Which type of heredity does the pedigree in Figure 11-6 demonstrate? SC.912.L.16.2 a. autosomal recessive b. autosomal dominant c. X-linked recessive d. X-linked dominant

  34. 33. For the parents III3 and III4, what would be the benefit of fetal testing? SC.912.L.16.2 • a. If the baby is male, there is a 50% chance he will have the disease. • b. There is a 50% chance that a baby of either sex will have the disease. • c. If the baby is male, there is a 25% chance that he will have the disease. • d. There is a 25% that a baby of either sex will have the disease.

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