Unit 8 & 9 Review

1. Unit 8 & 9 Review Types of Questions on Test: Multiple Choice True/False Matching Punnett Squares

2. Part 1 • Genotype – D • Phenotype – B • Homozygous – F • Heterozygous – E • Monohybrid Cross – A • Dihybrid Cross – C • Part 2 7. Genetics – F 8. Heredity – G 9. Gene – A 10. Diploid – B 11. Haploid – E 12. Law of segregation – D 13. Law of independent assortment – C

3. Part 3 14. Homologous Chromosomes – C 15. Sex Chromosome – F 16. Autosome – D 17. Cross pollination – B 18. Pure strain – E 19. Allele – G 20. Dominant – A 21. Recessive – H • Review the following terms from Unit 9: • Complete Dominance – D • Incomplete Dominance – F • Codominance– I • Single Allele Trait – A • Multiple Allele Trait – G • Polygenic Trait – J • X-linked Gene – B • Sex Influenced Trait – H • Test cross – E • Linkage Group – C

4. 11. How many genes are found on each chromosome? • Hundreds! More than one gene per chromosome 12. How many strands of DNA (chromosomes) do humans have in a diploid? • 46 chromosomes in each body cell (2 sets of 23 chromosomes) • 44 autosomes and 2 sex chromosomes (XX or XY) in every body cell 13. How many strands of DNA (chromosomes) do humans have in a haploid? • 23 chromosomes in each haploid (sex cell, gamete) • 1 set of 23 chromosomes (22 autosomes and 1 sex chromosome) 14. Give two examples of a diploid cell. • Diploid Cell – Somatic (body) cell • Skin cell, Muscle cell, Blood cell, Cheek cell 15. Give two examples of a haploid cell. • Haploid Cell – Reproductive cell (Gamete) • Sperm (male gamete, 22 autosomes & either X or Y) • Egg (female gamete, 22 autosomes & an X chromosome)

5. 16. A person has a mutation in a diploid cell. Will this affect their offspring? • If the mutation is only in a diploid cell (skin cell, muscle cell) it WILL NOT affect the offspring • Skin cells, muscle cells, etc do not get passed down to the offspring 17. A person has a mutation in a haploid cell. Will this affect their offspring? • A mutation in a haploid cell (sperm/egg) WILL affect the offspring • When an egg is fertilized by a sperm, all the genes (good, bad, neutral) become the offspring’s genes 18. Which parent determines the gender of the offspring? Explain your answer • The male (dad) determines the gender of the offspring • The female (mom, XX) can only pass down an X chromosome • The male (XY) can pass down either the X (produces a girl) or the Y (produces a boy) 19. In order to produce a female offspring, an egg must be fertilized by a sperm carrying a(n) X chromosome. 20. In order to produce a male offspring, an egg must be fertilized by a sperm carrying a(n) Y chromosome.

6. 21. In Gregor Mendel’s experiments, what did he call the • original plants? P Generation (Parent Generation) • What about the first generation? F1 (First Filial) • The second generation? F2 (Second Filial) 22. In a monohybrid cross, what phenotypic ratio did Mendel observe when doing a heterozygous X heterozygous cross (F2 generation cross)? • 3:1 ratio 23. When doing a dihybrid cross, what phenotypic ratio did Mendel observe in a heterozygous X heterozygous (F2 generation) dihybrid cross? • 9:3:3:1 ratio 24. Which gender is more likely to have a recessive sex-linked disorder? Why? • MALES! • Since males only have one X chromosome, if they inherit only one X-linked recessive allele, they would have the disorder • If females (XX) inherit only one recessive allele, they would be a carrier but would not have the disorder

7. 25. Polydactyl (Ff) X Five Fingers (ff) • Genotype of offspring: 2 Ff, 2 ff • Phenotype of offspring: 2 Polydactyl, 2 Five finger • Genotypic ratio: 2:2 • Probability of Polydactyl: 2/4 or 50% • Probability of Five Fingers: 2/4 or 50%

8. 26. In impatient flowers, flower color shows incomplete dominance. Red (RR) is dominant to white (rr), but the heterozygous results in a pink phenotype (Rr). Two pink flowers are crossed. • Genotype of Parents: Rr & Rr • Genotype of offspring: 1 RR, 2 Rr, 1 rr • Phenotype of offspring: 1 Red, 2 Pink, 1 white • Probability of Red: ¼ or 25% • Probability of Pink: 2/4 or 50% • Probability of White: ¼ or 25%

9. 27. In rabbits, the allele for black coat color is dominant over the allele for brown coat color. A homozygous brown coat rabbit is crossed with a heterozygous black coat rabbit. • Genotype of Parents: bb & Bb • Genotype of offspring: 2 Bb, 2 bb • Phenotype of offspring: 2 Black, 2 Brown • Probability of Black: 2/4 or 50% • Probability of Brown: 2/4 or 50%

10. 28. In humans, the gene for the genetic disorder Tay-Sachs disease is recessive. Two parents that are carriers (heterozygous) have a child. • Genotype of Parents: Tt & Tt • Genotype of offspring: 1 TT, 2 Tt, 1 tt • Phenotype of offspring: 3 normal, 1 Tay-Sachs • Genotypic Ratio – 1:2:1 • Phenotypic Ratio – 3:1 • Probability of not having T-S: 3/4 or 75% • Probability of Tay-Sachs: 1/4 or 25%

11. 29. Suppose a parent with homozygous Blood Type B (IB IB) has a child with a person heterozygous for Blood Type A (IAi). • Genotype of Parents: IBIB & IAi • Genotype of offspring: 2 IAIB, 2 IB i • Phenotype of offspring: 2 Type AB, 2 Type B • Probability of Type A: 0/4 or 0% chance • Probability of Type B: 2/4 or 50% chance • Probability of Type AB: 2/4 or 50% chance • Probability of Type O: 0/4 or 0% chance

12. 30. A person heterozygous for Blood Type A (IAi) has a child with a person with Blood Type O ( ii ). • Genotype of Parents: IA I & ii • Genotype of offspring: 2 IAi, 2 ii • Phenotype of offspring: 2 Type A, 2 Type O • Probability of Type A: 2/4 or 50% chance • Probability of Type B: 0/4 or 0% chance • Probability of Type AB: 0/4 or 0% chance • Probability of Type O: 2/4 or 50% chance

13. 31. Colorblindness is caused by a recessive X-linked. A normal vision carrier female (XBXb) has a child with a normal vision male (XB Y). • Genotype of Parents: XBXb& XB Y • Genotype of offspring: 1 XB XB, 1 XBXb, 1 XB Y, 1 XbY • Phenotype of offspring: 2 Normal Female (1 normal, 1 carrier), 1 Normal male, 1 Colorblind Male • Probability of Normal Female: 2/4 or 50% chance • Probability of Colorblind Female: 0/4 or 0% chance • Probability of Normal Male: 1/4 or 25% chance • Probability of Colorblind Male: 1/4 or 25% chance

14. 32. Hemophilia is caused by a recessive X-linked gene. A carrier female (XHXh) has a child with a male who has hemophilia (Xh Y). • Genotype of Parents: XHXh& Xh Y • Genotype of offspring: 1 XHXh, 1 XhXh, 1 XH Y, 1 XhY • Phenotype of offspring: • 1 unaffected female (a carrier), 1 female w/ hemophilia, 1 unaffected male, 1 Male w/ hemophilia • Probability of unaffected Female: 1/4 or 25% chance • Probability of Female w hemophilia: 1/4 or 25% chance • Probability of unaffected Male: 1/4 or 25% chance • Probability of Male w hemophilia: 1/4 or 25% chance

15. 12e. In pea plants, purple flower color is dominant to white. In the gene for seed shape, round is dominant to wrinkled. Suppose a plant heterozygous for both traits is crossed with a white flowered, wrinkeld pea plant. • Purple flower, Tall pea plant genotype: FfRr (FR, Fr, fR, fr) • White flower, Short pea plant : ffrr (fr, fr, fr, fr) • Genotypes of offspring: 4 FfRr, 4 Ffrr, 4 ffRr, 4 ffrr • Phenotypes of offspring: • 4 Purple flower & Round, 4 Purple flower & Wrinkled • 4 white flower & Round, 4 white flower & Wrinkled • Genotypic Ratio: 4:4:4:4 • Phenotypic Ratio: 4:4:4:4 • Probability of a Purple flower, Round plant: 4/16 or 25% • Probability of a Purple flower, Wrinkled plant: 4/16 or 25% • Probability of a White flower, Round plant: 4/16 or 25% • Probability of a White flower, Wrinkled plant: 4/16 or 25%