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BIO162Exam4Genetics
| Question | Answer |
|---|---|
| Genotype | The genetic letter combination of an organism (BB/Aa) |
| Phenotype | The physical appearance or visible trait of an organism (tall/purple) |
| Heterozygous | Having two different alleles for a trait (Tt/Bb) |
| Homozygous | Having two of the same alleles for a trait (AA/cc) |
| Gametes | Sex cells (sperm/egg) that carry only one allele for each trait (parent with genotype Bb makes gametes B and b) |
| Incomplete dominance | A pattern where the heterozygous phenotype is a blend or medium mix of the two parents (Red+Yellow= Orange flowers) |
| Co-dominance | A pattern where both traits appear fully and separately in the offspring (AB blood; not a blend like incomplete dominance) |
| Sex-linked trait | A trait found on the X chromosome that affects males more often than females because males only have one X chromosome. |
| Test cross | A method to determine the genotype of a dominant-looking individual by crossing it with a homozygous recessive individual. |
| XX | female |
| XY | male |
| Allele | a specific version of a gene, represented as a single letter within genotypes |
| Carrier | An individual who has the gene for a recessive trait but does not show the trait themselves. They are always heterozygous. |
| A dominant trait is | represented by a captial letter; it masks the recessive trait/allele. |
| A recessive trait is | represented by a lower-case letter; it is masked by the dominant trait/allele. Only applies if the individual is homozygous recessive (aa) and there is no dominant trait to mask it. |
| How to solve: "For the following genotypes, list the gametes (eggs or sperm) they can make: BB, aa, Rr, BBrr, BbRr). | For monohyrids (one trait), simply split the letters. So BB makes B,B / aa makes a,a /Rr makes R, r. For dihybrids (two traits), use the FOIL method (first, outer, inner, last). So BbRr makes BR, Br, bR, br / BBrr makes Br,Br,Br,Br. (INCLUDE COMMAS!!!!!!) |
| Work through this problem: In pea plants, tall is dominant to short. If a short plant is crossed with a tell heterozygous plant, what are the phenotypes and genotypes of the offspring? | 1. Assign labels. Tall is dominant to short. So make tall A and short a. Short plant then must be aa. Tall heterosygous plant must be Aa. 2. Punnet square. Aa x aa. 3. Genotypes = Aa, Aa, aa, aa. 4. Phenotypes = 50%tall, 50% short. |
| Work this problem: In pea plants tall is dominant to short and purple flowers are dominant over white flowers. If a short white is crossed with a tall purple that is heterozygous for both traits, what is the expected phenotypic ratio? | 1. Assign labels. Tall = A. Short = a. Purple is dominant so purple = B. White =b. Short white plant = aabb. tall purple plant = AaBb. 2. Use FOIL to break those apart for the punnet square. aabb=ab, ab, ab, ab. AaBb=AB,Ab,aB,ab. 3. Result is 1:1:1:1. |
| Work this problem:Color blindness is a sex-linked trait on the X chromosome and is recessive. A man who is colorblind marries and has kids with a woman who has normal vision, but her father was colorblind. What is the chance boys and girls will be c-blind | 1. Dad is colorblind (Xb Y). Mom is a carrier (XB Xb). 2. Cross (Xb Y) x (XB Xb). 3. Results = (XB Xb), (XB Y), (Xb Xb), (Xb Y). 4. Chance of colorblindness = 50% boys, 50% girls. |
| Work this problem regarding incomplete dominance: If you cross a red flower with a yellow flower and all the resulting flowers are orange. If you cross two orange flowers, what flower color ratio do you end up with? | 1. Assign labels. AA=red, aa=yellow. Orange = Aa. 2. Cross orange and orange: Aa x Aa 3. Results = AA, Aa, Aa, aa = 1:2:1. (1 red, two orange, 1 yellow). |
| Work this problem: Being polydactyl (extra fingers and toes) is a dominant trait. If I have a polydactyl cat and want to figure out his genogype, what could I do? Explain and show punnet square. | This is a test cross problem. 1. Cat has a dominant trait, so we know its either AA or Aa (can't be aa). 2. Cross those two possiblies with aa. So we need two separate punnet squares. AA x aa and Aa x aa. |
| Work this problem: In an example of co-dominance with blood type, a person with AB blood has children with a person with type O blood. Could a child with AB blood be their child? The answer is it's impossible, but make a punnet square to show why. | 1. Assign genotypes: Parent 1 = IA IB. They have one A allele and one B allele. Parent 2 is type O, so their genotype must be ii. Type O is always recessive, meaning they have two blank alleles (neither A nor B). 2. Punnet square: cross (IA IB) with ii. |
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