Biology 120 Lab Exam 2 Review

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1 Biology 120 Lab Exam 2 Review Student Learning Services and Biology 120 Peer Mentors Thursday, November 22, :00 pm Main Rooms: Arts 263, 217, 202, 212 Important note: This review was written by your Biology Peer Mentors (not the lab instructors)! It is designed to help you test yourself on topics and concepts covered in lab and should not in any way be considered a preview of the actual exam. You are encouraged to attempt to complete this review without the lab manual, textbook, your notes, or the internet. The peer mentors are available while you are working to help you with any questions that cause you difficulties. If you find that the room is too noisy for you to concentrate, Room 213 is a designated Quiet Work Space. Structured Study Session Information (Term 1, ) Mondays 2:30-3:50pm Arts 213 (Cameron) Tuesdays 8:30-9:50am Murray G3 (Kelly) Wednesdays 10:00-11:20am Murray G3 (Mars) Wednesdays 5:00-6:20pm Murray G3 (Sergey) Thursdays 2:30-3:50pm Murray 145 (Apurv) Thursdays 4:30-5:50pm Murray 102 (Justin) Friday 1:30-2:50pm Murray G3 (Judy) References Buchanan, F. (2018). Inheritance Patterns. Lecture. ANSCI 313. January 3, Schmutz, S. M. (2004, February 13). General Genetics References. Retrieved March 21, 2018, from Schmutz, S. M. (2016, May 7). Genetics of Coat Color and Type in Dogs. Retrieved March 21, 2018, from Last updated: :30 AM

2 2 Spot Test 1. No 2. GD, Gd, gd, gd 3. Anaphase II 4. Antheridium/sperm 5. Archegonium/egg/ovum 6. Rhizoids 7. Nothing, normal male 8. From this karyotype we can only infer that it is a human male. 9. Klinefelter Syndrome 10. Lots e.g. wide hips, small testes, etc. 11. Arch and Whorl 12. Dermatoglyphics 13. Haploid 14. Father #1, because he is the only one who could be the source of DNA fragments that the child has that the mother does not. 15. Restriction Fragment Analysis

3 3 Short Answer 1. Fill in the blanks. a. In ferns, cells inside the sporangia/sporangium undergo meiosis to produce spores. b. The fern gametophyte can also be called a prothallus. c. In animals two haploid cells called gametes combine together to form a diploid cell called a zygote in the process known as fertilization. d. Crossing over occurs during prophase 1 of meiosis and the region where this occurs is called the chiasma. e. Female mice produce gametes called eggs in their ovaries and male mice produce gametes called sperm in their testes. 2. The Manx cat is a breed of domestic cats that has a shorter than normal tail. The short tail mutation is dominant to normal tail, but the short tail mutation is homozygous lethal and usually results in miscarriages. A male Manx cat is bred with one Manx female and one normal female. What are the expected genotypic and phenotypic ratios of the kittens born for each cat? Show your work. Manx female kittens- 2 Manx : 1 normal, 2 Aa : 1aa Normal female kitten- 1 Manx: 1 normal, 1 Aa: 1 aa

4 4 3. Oculocutaneous albinism and α-mannosidosis are both genetic disorders that can affect Angus cows. Both disorders are autosomal recessive and are found on different chromosomes. Oculocutaneous albinism causes the animals affected to have white eyes. Animals affected with α-mannosidosis, caused by the lack of the enzyme α-d-mannosidase in the lysosome, are characterized by their failure to thrive and they die within the first year (i.e., before they are able to reproduce). A farmer has one bull and 48 cows. All of the animals are heterozygous for both traits. a. Draw the Punnet Square for the cross. Using the letter A for the gene associated with Oculocutaneous albinism and B for the gene associated with α-mannosidosis: AB Ab ab ab AB AABB AABb AaBB AaBb Ab AABb AAbb AaBb Aabb ab AaBB AaBb aabb aabb ab AaBb Aabb aabb aabb b. What is the expected phenotypic ratio if all of the cows have a calf? 9 normal: 3 Albino: 3 Mannosidosis: 1 Albino, Mannosidosis c. How many calves will be born with white eyes? 12 will have white eyes (and we expect 3 of them to die within the first year)

5 5 4. White Heifer Disease (WHD) is a congenital reproductive abnormality in white female offspring (heifers) in certain breeds of cattle, such as Belgian Blue and Shorthorn. WHD is actually the result of two genes- white fur colour is inherited as an autosomal recessive trait (h), which is linked with an autosomal recessive gene that causes defects in the female reproductive tract (r). a. Draw a Punnett square for a cow heterozygous for both traits being bred to a bull that is also heterozygous for both. Assume no crossing over occurs. HR HR HHRR HhRr hr HhRr hhrr hr b. What is the genotypic frequency of this cross? 0.5 HhRr, 0.25 HHRR, 0.25 hhrr c. If the heterozygous bull and the heterozygous cow had only female offspring what proportion would have normal reproductive systems? 75% d. University of Saskatchewan geneticists discovered that the genes for coat colour and reproductive system are 1 map unit apart. If 200 heterozygous cows were bred to a homozygous recessive bull, how many calves would have white coats and normal reproductive systems? Assume that all offspring are female, and that each cow has only one calf. 1 calf would be white and normal e. Draw a chromosome diagram for a cell of the bull in Anaphase I of Meiosis. Assume that no crossing over occurs. h h r H r H h h r h r h OR H H R R H H h h r r h h If you chose to draw the cell of the homozygous recessive bull from part d) If you chose to draw the cell of the heterozygous bull from part a)

6 6 5. Another genetic disease affecting Belgian Blue cattle is double muscling. It is caused by a single gene mutation that reduces or eliminates the activity of the myostatin protein. Use D to represent the allele responsible for the normal phenotype and d to represent the allele leading to the double muscling phenotype. A Belgian Blue breeder kept track of her herd using a pedigree chart. a. Based on the pedigree, is this likely a sex-linked trait? no b. Give the genotypes for each of the following individuals: 1. Dd 2. Dd 3. DD 4. Dd 5. Dd 6. dd 7. dd 8. dd 9. dd 10. DD c. What proportion of the offspring in the third generation are heterozygous? 75% 6. In domestic cats, an orange coat colour is caused by an X-linked gene that is dominant to a black coat colour. a. An orange female cat had a litter of 9 kittens: 5 orange females, 2 orange males, and 2 black males. What is the genotype of the mother cat? X A X a b. You have two male cats that could have sired this litter of kittens. One is orange and one is black. Is it possible to tell which one of these cats is the father based on the phenotypes of the kittens? If so, which was most likely the sire? Yes, the Orange male is most likely the father because no black females were produced.

7 7 7. Domestic cats can have unique phenotypes. One such phenotype is heterochromia; that is when the cat s eyes are two different colours. Another unique phenotype that can occur in cats is called polydactyly; that means that the cat has an extra digit on each of its paws. A female cat that was homozygous normal for both traits had a litter of kittens sired by a male cat that was homozygous for both heterochromia and polydactyly. All of the cats in the litter had eyes that were different colours and extra digits. a. Is the gene for polydactyly X-linked or autosomal? Autosomal Is the gene for heterochromia X-linked or autosomal? Autosomal b. Is the allele for polydactyly dominant or recessive? Dominant Is the allele for heterochromia dominant or recessive? Dominant Mittens was a polydactyl cat with heterochromia. Her mother was a normal cat in regards to both eye colour and number of digits. Mittens had kittens with a cat named Boots, who had the normal number of digits and eyes of the same colour. c. What is Boots genotype? aabb d. What is the genotype of Mittens mother? aabb e. What is Mittens genotype? AaBb f. If the genes for polydactyly and heterochromia are 18 map units apart, what is the expected phenotypic ratio of their offspring? 4.56 Polydactyly and Heterochromia : 1 Polydactyly : 1 Heterochromia : 4.56 normal g. Draw a chromosome diagram for Mittens cell in Telophase II of Meiosis. Assume no crossing over has occurred. A B A B a b a b

8 8 8. Circle the material best suited for DNA extraction and explain why: Urine Hair Red Blood Cells Urine because it contains nucleated cells (hair must have follicle, red blood cells lack nuclei) 9. What role do these solutions have in the DNA extraction you performed? a. Sports drink: keeps cells intact because it is isotonic to the cheek cell b. Detergent: cell lysis, disrupts the lipid and protein bonds that make the nuclear and cell membranes, releasing the DNA c. 70% ethanol: DNA is insoluble in alcohol and will therefore precipitate in the ethanol 10. What does PCR stand for? polymerase chain reaction 11. What are the four ingredients required for PCR? DNA extract the four dntps primers DNA/Taq polymerase 12. Describe the three steps in the PCR cycle: a. DNA denaturation when temperature increases, double stranded DNA separates into single strands b. anneal primers/primer binding when the temperature drops, primers bind/anneal to single strands of DNA c. extend primers/dna synthesis) when temperature increases again, taq polymerase adds nucleotides to extend the primer

9 9 13. What does Taq polymerase do? Is a heat-stable enzyme that adds dntps to the primers, creating a second strand of DNA that is complementary to the single strand 14. What is a thermal cycler? A piece of lab equipment used in PCR; it raises and lowers the temperature of the solution 15. EcoRI is a type of Restriction enzyme and recognizes/cuts the DNA sequence GAATTC. 16. What is gel electrophoresis? a process that uses the negative charge of DNA and a fibrous gel to separate restriction fragments/dna based on size 17. Gels are made of agarose which is a polysaccharide derived from algae/seaweed. 18. DNA is negatively charged, because of its phosphate backbone, which means that DNA will migrate towards the positive electrode. 19. Smaller fragments travel farther than larger fragments. 20. Which stain can be visualized under white light? Fast Blast. Which stain can only be visualized under UV light? ethidium bromide 21. What is the purpose of the ladder in gel electrophoresis? Contains DNA fragments of known size and allows us to estimate the size of our unknown DNA fragments

10 Use the following gel to put the sample numbers in order from smallest fragment to largest. 1,3,2,4 Ladder

11 Application Question: Labrador Retrievers come in three major coat colours: black, chocolate (brown), and yellow. These three coat colours are determined by mutations in the MC1R (melanocortin receptor 1) and the TYRP1 (tyrosinase related protein 1) genes. The TYRP1 gene has two different alleles. B gives the black phenotype, b gives the brown phenotype. MC1R also has two different alleles. E allows either the black or the brown phenotype to be shown, e only allows the yellow phenotype to be shown. In order to determine the coat colour for a dog, two different gels need to be run. One gel will be run with DNA that is cut with a restriction enzyme to determine which MC1R allele is present and the other gel is run with DNA that is cut with a restriction enzyme to determine the TYRP1 allele present. The restriction enzyme used for the MC1R gel will cut the e allele but not the E allele. The restriction enzyme used for the TYRP1 will cut the b allele but not the B allele. If all the dogs are homozygous at both traits determine the coat colour of each of the four dogs below. MC1R TYRP1 Ladder Ladder Black 2-Yellow 3-Yellow 4-Brown

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