Individual Feedback Report for: St#: Test: GENETICS UNIT TEST 2score Grade: 3 Score: % (20.00 of 35.00)

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Individual Feedback Report for: St#: 703709 Test: GENETICS UNIT TEST 2score Grade: 3 Score: 57.14 % (20.00 of 35.00) 26. T F The cell wall controls the transfer of nutrients into animal cells. 27. Mitosis produces a A. sperm B. zygote C. body cell D. egg E. gamete Mitosis is the name for the process of asexual reproduction for all cells in your body, except for the sex cells. 28. If there are 8 chromosomes in an onion root tip cell, how many are there in an onion skin cell: A. 2 B. 8 C. 4 D. 16 E. 0 All cells in an organism have the same number of chromosomes. 29. If a cell has 26 chromosomes and it undergoes meiosis what new cells will be produced? A. 2 new cells, each having 13 chromosomes. B. 4 new cells, each having 13 chromosomes. C. 2 new cells, each having 26 chromosomes. D. 4 new cells, each having 26 chromosomes. 32. A mutation is a: A. temporary change in a chromosome B. temporary change in a gene C. permanent change in a gene D. deformation of an organism during development 33. If you receive the same kind of gene from each of your parents you are said to be: A. genotyped B. heterozygous C. homozygous D. phenotyped E. dominant Page 1 of 4 December 21, 201810:28:02 AM

Individual Feedback Report for: St#: 703709 37. Breeding a male horse and a female donkey results in an offspring called a mule. What is the 2n chromosome number for the mule? A. 126 B. 62 C. 64 D. 63 38. Which of the following describes the genotype of the female parent if all the offspring were white? A. heterozygous B. recessive C. dominant D. homozygous Page 2 of 4

Individual Feedback Report for: St#: 703709 45. The sex chromosomes carried by a human sperm may be A. only Y B. a pair of Y's C. X and Y together D. either X or Y E. only X 47. A girl has a hemophiliac father and a normal mother. Her chances of becoming a carrier are A. 100% B. 0% C. 25% D. 50% E. 75% The following diagram represents the family tree for Queen Victoria of the British Empire. Queen Victoria carried the gene for hemophilia, but her husband Albert did not. Hemophilia is a sex-linked recessive gene carried on the X chromosome. Use this diagram to answer questions 48 to 48. 48. Alice had three children, 2 girls and 1 boy. Which statement is NOT true: A. Her husband could not have given any hemophilic genes to their children B. Alice had genes for hemophilia on both X chromosomes C. Alice had genes for hemophilia on only 1 X chromosome D. Alice carried the gene for hemophilia 51. Incomplete dominance results in the production of mauve flowers when pink and blue homozygous plants are crossed. The probable phenotypes of sixteen flowers produced from mauve parents would be: A. B. C. D. 8 pink and 8 blue. 12 pink and 4 blue. 16 mauve. 4 pink, 8 mauve and 4 blue. 54. The possible gene combinations responsible for type O blood are A. Oo B. oo C. oo D. Oa E. OO Page 3 of 4

Individual Feedback Report for: St#: 703709 55. If a man with blood type A (whose mother had blood type 0), marries a woman with blood type AB, what is the probability of their first child having type A blood? A. 75% B. 0% C. 25% D. 50% 56. A red bull and a white cow have a roan colored calf. Roan coloration results from a mixture of red and white hairs. What are the genotypes of the parents? A. RR x Rr B. RR x rr C. Rr x Rr D. Rr x rr Page 4 of 4

Individual Feedback Report for: St#: 1277424 Test: GENETICS UNIT TEST 2score Grade: 3 Score: 60.00 % (21.00 of 35.00) 28. If there are 8 chromosomes in an onion root tip cell, how many are there in an onion skin cell: A. 2 B. 8 C. 4 D. 16 E. 0 All cells in an organism have the same number of chromosomes. 32. A mutation is a: A. temporary change in a chromosome B. temporary change in a gene C. permanent change in a gene D. deformation of an organism during development 33. If you receive the same kind of gene from each of your parents you are said to be: A. genotyped B. heterozygous C. homozygous D. phenotyped E. dominant Page 1 of 5 December 21, 201810:28:03 AM

Individual Feedback Report for: St#: 1277424 37. Breeding a male horse and a female donkey results in an offspring called a mule. What is the 2n chromosome number for the mule? A. 126 B. 62 C. 64 D. 63 43. A Manx cat has no tail as shown below. A dominant gene M causes this tailless condition. A homozygous dominant fetus will not develop. Which of the following shows the phenotypic ratios of the living offspring if two Manx cats are crossed? A. 2 Manx: 1 long tail B. 2 Manx: 0 long tail C. 2 manx: 2 long tail D. 1 Manx: 0 long tail Page 2 of 5

Individual Feedback Report for: St#: 1277424 44. Which of the following genotypes would characterize an organism that is heterozygous for two pairs of genes? A. RyWx B. RRYY C. RrYy D. RyRy 45. The sex chromosomes carried by a human sperm may be A. only Y B. a pair of Y's C. X and Y together D. either X or Y E. only X 46. Hemophilia is A. B. C. D. E. a genetic disease which only appears if the affected person's mother does not eat nutritious foods while she is pregnant a contagious disease, easily transmitted from one person to another a contagious disease which is extremely difficult to transmit from one person to another a genetic disease, transmitted from parents to children a genetic disease, transmitted from parents to children under certain environmental conditions 50. A woman who is heterozygous and a man with colour-blindness are considering having children. What is the probability of having a child who is both female and colour-blind? A. B. C. D. 75% 100% 0% 25% Page 3 of 5

Individual Feedback Report for: St#: 1277424 51. Incomplete dominance results in the production of mauve flowers when pink and blue homozygous plants are crossed. The probable phenotypes of sixteen flowers produced from mauve parents would be: A. B. C. D. 8 pink and 8 blue. 12 pink and 4 blue. 16 mauve. 4 pink, 8 mauve and 4 blue. 52. What is the most likely genotype of gray moths based on the results? A. AB B. bb C. Bb D. None of the above allele combinations would yield the results shown. E. BB 54. The possible gene combinations responsible for type O blood are A. Oo B. oo C. oo D. Oa E. OO 55. If a man with blood type A (whose mother had blood type 0), marries a woman with blood type AB, what is the probability of their first child having type A blood? A. 75% B. 0% C. 25% D. 50% Page 4 of 5

Individual Feedback Report for: St#: 1277424 58. Amniocentesis reveals that an embryo has Tay-Sachs disease, which is a recessive trait. What are the genotypes of the parents, if they appear normal? (t=tay-sachs) A. any one of the above is possible B. Both tt C. Both TT D. Both Tt Page 5 of 5

Individual Feedback Report for: St#: 704487 Test: GENETICS UNIT TEST 2score Grade: 4 Score: 71.43 % (25.00 of 35.00) 29. If a cell has 26 chromosomes and it undergoes meiosis what new cells will be produced? A. 4 new cells, each having 13 chromosomes. B. 2 new cells, each having 26 chromosomes. C. 4 new cells, each having 26 chromosomes. D. 2 new cells, each having 13 chromosomes. 32. A mutation is a: A. permanent change in a gene B. temporary change in a chromosome C. temporary change in a gene D. deformation of an organism during development 36. Given the following word association: Phenotype: appearance What would be properly associated with genotype? A. homologous pair B. gene C. combination of alleles D. genome E. combination of chromosomes 40. The dimples allele is dominant and the no dimples allele is recessive. Which of the crosses below have the same probability of producing heterozygous dimpled individuals? D = dimples, d = no dimples I Dd x Dd II Dd x dd III DD x Dd IV DD x dd A. B. C. D. I, II, and III only I, II and IV only I, III, and IV only I, II, III, and IV Page 1 of 3 December 21, 201810:28:04 AM

Individual Feedback Report for: St#: 704487 43. A Manx cat has no tail as shown below. A dominant gene M causes this tailless condition. A homozygous dominant fetus will not develop. Which of the following shows the phenotypic ratios of the living offspring if two Manx cats are crossed? A. 1 Manx: 0 long tail B. 2 Manx: 0 long tail C. 2 Manx: 1 long tail D. 2 manx: 2 long tail 47. A girl has a hemophiliac father and a normal mother. Her chances of becoming a carrier are A. 0% B. 25% C. 50% D. 75% E. 100% The following diagram represents the family tree for Queen Victoria of the British Empire. Queen Victoria carried the gene for hemophilia, but her husband Albert did not. Hemophilia is a sex-linked recessive gene carried on the X chromosome. Use this diagram to answer questions 48 to 48. 48. Alice had three children, 2 girls and 1 boy. Which statement is NOT true: A. Alice carried the gene for hemophilia B. Alice had genes for hemophilia on only 1 X chromosome C. Alice had genes for hemophilia on both X chromosomes D. Her husband could not have given any hemophilic genes to their children Page 2 of 3

Individual Feedback Report for: St#: 704487 54. The possible gene combinations responsible for type O blood are A. OO B. Oo C. oo D. oo E. Oa 55. If a man with blood type A (whose mother had blood type 0), marries a woman with blood type AB, what is the probability of their first child having type A blood? A. 0% B. 25% C. 50% D. 75% 57. A person with Down's Syndrome has: A. 45 chromosomes. B. 47 chromosomes. C. XXY chromosomes. D. XXX chromosomes. Page 3 of 3

Individual Feedback Report for: St#: 704031 Test: GENETICS UNIT TEST 2score Grade: 6 Score: 88.57 % (31.00 of 35.00) 36. Given the following word association: Phenotype: appearance What would be properly associated with genotype? A. homologous pair B. gene C. combination of alleles D. genome E. combination of chromosomes 38. Which of the following describes the genotype of the female parent if all the offspring were white? A. recessive B. dominant C. homozygous D. heterozygous 47. A girl has a hemophiliac father and a normal mother. Her chances of becoming a carrier are A. 0% B. 25% C. 50% D. 75% E. 100% Page 1 of 2 December 21, 201810:28:04 AM

Individual Feedback Report for: St#: 704031 52. What is the most likely genotype of gray moths based on the results? A. BB B. Bb C. bb D. AB E. None of the above allele combinations would yield the results shown. Page 2 of 2