V. Two Gene Pairs (Extra Credit) 1... A. In peas, the allele for tall plants (T) is dominant over the allele for short plants (t). The allele for smooth peas (S) is dominant over the allele for wrinkled peas (s). Use this information to cross the following parents. I 1. heterozygous tall and smooth X heterozygous tall and smooth 2. heterozygous tall, wrinkled X short, wrinkled AL.

V. Two Gene Pairs (Extra Credit) 1... A. In peas, the allele for tall plants (T) is dominant over the allele for short plants (t). The allele for smooth peas (S) is dominant over the allele for wrinkled peas (s). Use this information to cross the following parents. I 1. heterozygous tall and smooth X heterozygous tall and smooth 2. heterozygous tall, wrinkled X short, wrinkled AL.

Biology: The Unity and Diversity of Life (MindTap Course List)

14th Edition

ISBN:9781305073951

Author:Cecie Starr, Ralph Taggart, Christine Evers, Lisa Starr

Publisher:Cecie Starr, Ralph Taggart, Christine Evers, Lisa Starr

Chapter13: Observing Patterns In Inherited Traits

Section: Chapter Questions

Problem 7GP: Several alleles affect traits of roses, such as plant form and bud shape. Alleles of one gene govern...

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Several alleles affect traits of roses, such as plant form and bud shape. Alleles of one gene govern whether a plant will be a climber (dominant) or shrubby (recessive). All F1 offspring from a cross between a true-breeding climber and a shrubby plant are climbers. If an F1 plant is crossed with a shrubby plant, about 50 percent of the offspring will be shrubby; 50 percent will be climbers. Using symbols A and a for the dominant and recessive alleles, make a Punnett- square diagram of the expected genotypes and phenotypes the cross between the F1 offspring and the shrubby plant.
а. a chart that shows the combinations (results) for a cross 8. Genetics b. a picture (diagram) that shows genetics pass from generation (pass from 9. Alleles parents to kids to 10. Punnett grandkids) Square genotype of 2 DIFFERENT alleles (Bb, Dd, Aa...) с. 11. Phenotype d. the information on the alleles / Genes (ex. TT, Tt, tt) 12. Genotype e. a person's Physical appearance; their characteristics (ex. Brown eyes) 13. Homozygous f. a type of gene - we get one from our mom and one from our dad. 14. Heterozygous g. genotype of two of the SAME alleles (AA, aa, BB, bb...) 15. Pedigree h. the study of heredity and the variations of characteristics (how people look differently)
2. The following is a pedigree of a family in which a rare form of X-linked colorblindness is found (filled-in symbols). II 2 II 3 4 a. Label the genotypes of as many family members as the available information permits. Remember that this gene is located on the X-chromosome when writing out genotypes. b. Is the colorblindness trait recessive or dominant? Explain how you know.
4. In humans, cystic fibrosis and albinism are both inherited as independently assorting autosomal recessive traits. Use fto represent the recessive allele for cystic fibrosis and Ffor the allele for the unaffected condition. Use A to represent the dominant allele for normal melanin man is albino and heterozygous for cystic fibrosis. His wife has the unaffected phenotype for both traits, although her father had cystic fibrosis and a for the recessive allele for albinism. and her mother is albino. 4a. What is the genotype of the wife? 4b. What is the probability that a child of this couple is albino and has cystic fibrosis?
9. Make a pedigree for each of the following situations. For each individual, write the individual's genotype (when possible) next to the individual's symbol (e.g. O xty, I Gg): a. Two parents do not have cystic fibrosis and they have a daughter with cystic fibrosis and a son who does not have cystic fibrosis. The daughter grows up and she mates with a male who does not have cystic fibrosis. Their only child is a boy and he has cystic fibrosis. b. A man with hemophilia mates with a female without hemophilia. They have one son and one daughter. The daughter has hemophilia and the son does not have hemophilia. The son grows up, and he marries and mates with a female. Their only child is a boy, and he has hemophilia.
8. In guinea pigs, the allele for black fur (B) is dominant over the allele for white fur (b). Similarly, the allele for straight hair (H) is dominant over the allele for having rough hair (h). Pure breeding rough haired guinea pigs with black fur were crossed with pure breeding straight haired guinea pigs with white fur. a. State the genotype and the phenotype of the F1 individuals produced as a result of this cross.b. Two F1 offspring were mated together. Calculate the expected ratio of phenotypes in the F2 generation.c. Show the completed Punnett square below.
4. Phenylketunoria (PKU) is an inborn error of metabolism of the amino acid phenylalanine. The following pedigree is of an affected family. I II III 1 2 3 5 a. What is the mode of inheritance of PKU? b. Which persons are known to be heterozygous for PKU? Identify from I, II, and III generations. c. What is the probability that III-2 9 is heterozygous? d. If III-3 and III-4 marry, what is the probability that their first child will have PKU?
1. What will be the phenotype of the fly that grows from the fertilized cell shown below? Allele key Dominant Recessive C- Curly wing B- Brown body L-Legs on head R- Red eyes C- normal wing b- black body |- antenna r- orange eyes O A. Normal wings, black body, antenna, and red eyes. • B. Curly wings, brown body, legs on head, and red eyes. OC. Curly wings, brown body, legs on head, and orange eyes. D. Normal wings, brown body, legs on head, and red eyes.
3. Short-tailed pup distribution. In mice, the dominant T allele results in a short tail. Homozygous T/T genotype is lethal, which means mouse embryos with this genotype die before they are born. Homozygous t/t is normal, with normal tail. A cross between two short-tailed mice produces a litter of 5 pups. a). What are the genotype(s) of the two short-tailed mice used in the cross? b). What are the possible genotypes and phenotypes of the 5 pups? What are their perspective ratio?
II. Solve the following problems. 1. Mendel crossed a plant that was heterozygous for height and heterozygous for pea shape with a plant that was homozygous dominant for height and heterozygous for shape. What are the possible genotypes and phenotypes of the offspring? Dominant for height= T (tall) Recessive for height =t (short) Dominant for pea shape = R (round) pea %3D Recessive for pea shape =r(wrinkled) P1: Genotypes: Phenotypes: Genotypic Ratio: Phenotypic Ratio
3. Mendel crossed that was homozygous dominant for flower color and heterozygous for height with a plant that was heterozygous for flower color and homozygous recessive for height. Do a punnet square and calculate possible genotypes and phenotypes of this cross. Dominant for flower color = P (purple) Recessive for flower color = p (white) Dominant for height = T (tall) Recessive for height =t (short) %3D P1: PP X Tt Pp X tt PT Pt PT Pt Pt PPTT PPtt PPTT PPtt Pt PPTT PPtt PPTt PPtt pt PpTt Pptt PpTt Pptt pt PpIt Pptt PpTt Genotypes: Phenotypes: Genotypic Ratio: Phenotypic:
4 *The normal color of snapdragons is red. Some pure lines showing variation in flower color have been found. Analyze the following crosses and do the following:² c. Explain the inheritance of these colors (include number of genes, alleles and dominance relationships). d. Determine the genotypes of the parents and the F1 of each cross. e. Propose a biochemical pathway for flower color production in snapdragons. F₂ Parents 1. orange x yellow 2. red x orange 3. red x yellow 4. red x white F₁ orange red red red red 5. yellow x white 9 red: 3 yellow: 4 white 6. orange x white 9 red: 3 orange: 4 white 7. red x white 9 red: 3 yellow: 4 white Hint: Come up with a possible pathway first, it helps to organize your thoughts. red 3 orange: 1 yellow 3 red: 1 orange 3 red: 1 yellow 3 red: 1 white red