BIOL110 Exam 2 Practice Test Solutions
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1 BIOL110 Exam 2 Practice Test Solutions Problem #1: D The Golgi apparatus receives the products from the ER and glycosylates them. The Endomembrane system is a network of membrane-bound chambers where Protein targeting and processing takes place. Problem #2: D Chloroplasts are only prevalent throughout the Plant kingdom (and some protists), NOT the Animal kingdom. Problem #3: D Smooth ER has no attached ribosomes, where as rough ER has attached ribosomes. The major function of rough ER is to make polypeptides to be secreted from the cell, where as the smooth ER makes carbs and lipids. Problem #4: C Microfilaments are the smallest cytoskeletal fibers, Intermediate Filaments are the medium sized cytoseletal fibers, and Microtubules are the largest cytoskeletal fibers. Problem #5: A. Mitosis gives rise to 2 daughter cells with the same ploidy number. A 2N cell would produce 2 identical 2 N cells. Problem #6: A. When a cell undergoes replication, sister chromatids form. During meiosis I, non sister chromatids separate, and during meiosis II, sister chromatids separate.
2 Problem #7: C. During prophase I, the physical crossing over of genetic material occurs. In metaphase I, homologous chromosomes line up on the metaphase plate independently of one another. Problem #8: C. The number of different combinations for independent assortment is 2 n where n = # of homologous pairs. Problem #9: A. The # different combinations of chromosomes = 2 n. Dragon has a diploid (2n) = 8. So n = 4. For the dragon, # = 2 4 = 16. Unicorn has a haploid (n) = 4. So n = 4. For the unicorn, # = 2 4 = 16. Problem #10: D Trypanosomes are a member of the Kinetoplastids, of the kingdom Euglenozoa. Problem #11: C Transmission of malaria cia mosquitos is caused by Plasmodium, an apicomplexan. Phytoplankton Problem #12: C The small cavities enclosed in membranes that hug the internal cell surface are Alveoli, characteristic of the kingdom Alveolata. The cilia covering indicates a ciliate, a type of alveolate. Problem #13: A Oomycetes display Convergent Evolution (The process by which unrelated organisms that occupy similar environments evolve similar functional traits) with fungi. Problem #14: D The sporophyte produces haploid spores via meiosis and the gametophyte produces haploid gametes via mitosis. Remember, sporophyte is 2N, so it must undergo meiosis to produce N spores. The gametophyte is N, so it can only undergo mitosis to produce N gametes. The only time diploid products are formed is during fertilization.
3 Problem #15: B The nucleus stores DNA, while lysosomes break down unwanted organelles. Problem #16: C After leaving the nucleus via the nuclear pore, mrna binds to the ribosomes on the rough ER to form a polypeptide, which, when complete, is transported to the cis face of Golgi apparatus via a transport vescicle where it undergoes glycosylation, after which it exits the trans face of the Golgi apparatus via a secretory vescicle. Problem #17: A Rough ER uses attached ribosomes for protein systhesis, while smooth ER is for synthesizing carbs and lipids. Problem #18: D Giardia is caused by a diplomonad intestinal parasite. Problem #19: D African Sleeping Sickness and Chagas disease are caused by a type of kinetoplast, not an apicomplexan. Problem #20: A Mosquitoes transmit Plasmodium, an apicomplexan, when they latch on and begin sucking your blood. Problem #21: C It is green algae that are most similar to plants, not red. Problem #22: C Diplomonads are not photosynthetic Problem #23: C Stable conditions imply no need for genetic variation, and so it is must more efficient and effective to reproduce asexually, and thus allow the population to grew much more rapidly.
4 Problem #24: B Remember mitosis makes an exact copy of what you start with, thereby doubling the initial input. Meiosis, on the other hand, makes four distinct cells, each of them haploid. Problem #25: D At the start of meiosis each chromosome replicates itself, making an identical copy known an a homologous chromosome that is genetically identical at first. This later changes due to crossing over, but that happens later. Problem #26: D The 28 chromosomes double up to 56 at the start of meiosis I, but later get split apart during anaphase I, leaving each daughter cell with 28 chromosomes (genetically distinct at this point). At the end of meiosis II, these daughter cells have been split once more, making a total of four genetically distinct cells, each now with only 14 chromosomes. Problem #27: B Easy question chromosomes align on the equator during metaphase. Problem #28: D When homologous chromosomes pair up they offer the opportunity for some genetic information to be switched between the maternal and paternal chromosomes. This recombination increases genetic variation among species. Problem #29: A (NOT ON TEST) Problem #30: C To calculate the number of different combinations you use the equation P = 2 n, where n is the haloid number. If the dragon s diploid number is 2N = 28, then it s haploid number is N = 14, and the unicorn s haploid number is N = 9, so for each species the calculation is: Dragon: P = 2 14 = 16,384 Unicorn: P = 2 9 = 512 Problem #31: B (NOT ON TEST) Problem #32: C (NOT ON TEST)
5 Problem #33: B You must do a quick punnett square for each gene. There is a ½ chance that AA can be made from AA and Aa, a ½ chance that Bb can be made from Bb and bb, and ½ chance that Cc can be made from cc and Cc, and a ½ chance that DD can be made from Dd and DD. By the rule of multiplication, (1/2)(1/2)(1/2)(1/2) = 1/16. Problem #34: D A test cross is used to determine whether an individual is dominant homozygous or heterozygous. Problem #35: B Using 2 punnett squares, you calculate that from PP x Pp, there is a 100% chance of getting a purple flower, and from Yy x yy, there is a ½ chance of getting yellow seeds. By the rule of multiplication, (1)(1/2) = ½. Problem #36: A The only combinations of genotypes you can come up with from AA and Bb is AB or Ab. Problem #37: B Bb x bb gives you ½ Bb and ½ bb. bb is the only genotype that gives you brown hair. Problem #38: D To get a boy, girl, girl in that order: there is a ½ chance of getting a boy, ½ chance of getting a girl, and another ½ chance of getting a girl. By the rule of multiplication, (1/2)(1/2)(1/2) = 1/8. Problem #39: A There are 3 different ways to get a boy and 2 girls: B,G,G or G,B,G or G,G,B. The probability of getting one of these combinations is [(1/2)(1/2)(1/2)] = 1/8. By the rule of addition, (1/8) + (1/8) + (1/8) = 3/8. Problem #40: C In codominance, both alleles are displayed.
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