DUMMY
BIO - Heredity Practice genetics problems 1.) A student recently told me that both she and her brother have blue eyes (bb), and that both of her parents are brown-eyed. Using a Punnett square, list the genotypes and phenotypes showing how it’s possible for this to happen. 2.) In fruit flies, L = long wings and l = short wings. When a long-winged fly is crossed with a short-winged fly, the offspring exhibit a 50:50 phenotypic ratio. What is the genotype of the parent flies? Prove how this is possible. Develop a Punnett square, and then list the genotypes and phenotypes. 3.) In tomatoes, red fruit (R) is dominant over yellow fruit (r). Tallness within plants (T) is dominant over shortness (t). A male tomato plant whose genotype is RrTT is crossed with a female plant whose genotype is rrTt. What are the chances of their offspring being heterozygous for both traits (RrTt)? Develop a Punnett square, and then list the genotypes and phenotypes prior to answering the question. 4.) In dogs, there is a hereditary deafness caused by a recessive gene (d). A kennel owner has a male dog that she wants to use for breeding purposes if possible. The dog can hear, so the owner knows his genotype is either DD or Dd. If the dog’s genotype is Dd, the owner does not wish to use him for breeding so that the deafness gene will not be passed on. This can be tested by breeding the dog to a deaf female (dd). Draw the Punnett squares to illustrate these two possible crosses. List all genotypes and phenotypes showing how many of the offspring would be normal and deaf. Background information for problem #5 In humans there is a gene that controls formation of hemoglobin (the protein in red blood cells {RBC’s} that is responsible for carrying oxygen to the tissues of the body). The normal allele of this gene codes for normal hemoglobin. However, there is another allele for this gene that contains a mutation in the DNA and the resulting hemoglobin protein is defective. An RBC with this defective hemoglobin will, under stress, crystallize within the cell and cause a characteristic sickle shape. While the letters S and s are often used to represent these alleles, in reality neither is dominant to the other (codominant). Someone who is SS makes all normal hemoglobin, someone who is ss makes all abnormal hemoglobin (someone who has sickle-cell anemia), and someone who is Ss essentially has two sets of instructions (and so makes some of each kind of hemoglobin; often referred to as sickle-cell trait). Because the RBC’s of a person who is ss contain all abnormal hemoglobin, they will sickle very easily, with very little stress required to provoke that activity. Those abnormally shaped cells will then get lodged in small capillaries as they move through them, leading to strokes, heart attacks, and pulmonary embolisms that can ultimately lead to death. Because only some of the RBC’s of a person who is Ss contain abnormal hemoglobin, that person usually only has trouble with a lot of cells sickling if they are under a lot of stress trying to meet a higher than normal oxygen demand, and so the chances of a person dying from sickle-cell trait are much lower than for full blown sickle-cell anemia. Now, malaria is a parasitic disease that is prevalent in tropical areas. When a mosquito that is carrying the parasites bites someone, the parasites enter the person’s bloodstream, and invade and live in the person’s RBC’s. However, if a person has sickle-cell anemia (ss), the presence of a parasite in an RBC is so stressful, it causes the RBC to crinkle up and when that happens the parasite is killed before it can multiply and spread to other RBC’s. As a result, a person who is ss is immune to malaria. If a person is Ss and a malaria parasite tries to invade an RBC with abnormal hemoglobin, again the RBC will sickle, killing the parasite before it has a chance to reproduce. If a parasite invades am RBC with normal hemoglobin, it will be able to live and multiply, but if its offspring invade other RBC’s with abnormal hemoglobin, they too, will be killed. As a result, a person who is Ss is resistant (although not totally immune) to malaria. If a person is SS and has all normal hemoglobin, the malaria parasites do just fine. Thus, an SS person usually dies, eventually, from causes related to the malaria. 5.) A man and a woman living in a tropical area where malaria is prevalent and health care is not accessible have seven children during their lifetime. The genotypes of their children are: ss, Ss, SS, ss, Ss, Ss, and SS. What must the genotypes of both parents be? Your answer should include a Punnett square to illustrate your work, and list all the genotypes and phenotypes. 6.) In the previous problem, which of their children would you expect to live to adulthood and reproduce? Explain. 7.) In rural northeast Kansas domesticated cats are becoming wild, and biologists are observing that wildcats occur in three colors. Fur color is controlled by a single gene that exhibits incomplete dominance. A homozygous dominant cat is blue (BB), a homozygous recessive cat is red (bb), and a heterozygous wildcat will be purple (Bb). What would be the expected phenotypes of the offspring if a blue wildcat mates with a red wildcat? Develop a Punnett square and list the genotypes as well. 8.) A naturalist is visiting an island in the middle of Lake Superior and observes a new bird species with three distinct types of beaks. Those with short crushing beaks (BB) feed on oak nuts; those with long delicate beaks (bb) pick the seeds from pine cones; and those with intermediate beaks (Bb) feed on both oak nuts and pine cones. (Note: because their beaks are structurally & functionally inadequate, few Bb offspring successfully reach adulthood). Let’s assume that this difference in beak morphology is the result of incomplete dominance in a single gene. In a year in which most of the food available on the island is in the form of hard oak nuts, which of the mated pairs below will produce the most offspring that will be able to survive to adulthood? Use Punnett squares to back up your answer. List all phenotypes and genotypes. a.) Bb x Bb b.) Bb x bb c.) BB x Bb d.) BB x bb 9.) In humans, when a dark-skinned person reproduces with a very light-skinned person, their children may range in color from dark to brown. This occurs because skin color is controlled by more than one gene (the exact # of genes is not known). For your convenience, a table is presented below that shows several possible genotypes & phenotypes: Genotypes Phenotypes AABB Very dark skin AABb or AaBB Dark skin AaBb, AAbb, or aaBB Medium brown skin Aabb or aaBb Light skin aabb Very light skin If a man with dark skin whose genotype is AaBB reproduces with a woman who has light skin (aaBb), what are the possible skin colors that their children will have? Develop a Punnett square and list the possible genotypes & phenotypes. 10.) What is the probability that the following cross will produce an organism homozygous recessive for all 5 traits? AaBbCcDdEe x AaBbCcDdEe Show your work including the rule(s) of probability you use to answer the question. Sample Questions · Reference: Kilo (k)=1000 Centi (c)= 1/100 Milli (m) = 10-3 Micro (μ) = 10-6 Nano (n) = 10-9 pico (p) = 10-12 Molecular weight of double stranded DNA: 1 base pair = 660 g/mol Please write out answers to both sections, take a picture and upload. a. You want to ligate together a 1.2 kB DNA insert and a 7 kB DNA vector. You measure your vector concentration to be 18 ng/µL and your insert to be 11 ng/µL. Calculate what volume of vector you would use to get 0.1 µg of vector. b. You want to perform a 5:1 insert:vector molar ratio ligation. How many nanograms of insert would you add to your ligation? What volume of insert DNA would you add to your ligation? · For your DNA transformation experiment, you want to add 20 pg of your pUC19 miniprep DNA to the cells. The concentration of your pUC19 miniprep DNA is 442 µg/mL. Calculate what dilutions you would make and what volumes you would pipet to get 20 pg of your pUC19 miniprep DNA. Use serial dilutions instead of one large dilution. · You want to plate 100,000 live cells in one well of 6-well plate. You count cells using the TC20 counter. The result is: Total cells = 1.5 x 106 cells/ml Live cells = 9.7 x 105 cells/ml Calculate what volume of your trypsinized and resuspended cell sample you should add. Also calculate how much fresh media to add to the cells in the well of the 6-well plate (total working volume of 1 well= 3 ml). There will be 25 questions all will be related to CRISPR technique