F G H A B C D E T 20 20 15 10 20 30 20 20 20 20 From the map given above, the observed frequency of recombinants between alleles of the B and H genes in crosses is most likely to be: a. 115% b. 50% c. 20% d. 75% in tetraploids e. cannot be inferred from the map
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From the map given above, the observed frequency of recombinants between alleles
of the B and H genes in crosses is most likely to be:
a. 115%
b. 50%
c. 20%
d. 75% in tetraploids
e. cannot be inferred from the map](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fec13ab4f-5933-4c78-80c6-824491a34faa%2F79c9d615-b25d-4cc1-a7ef-35e3e14b1a99%2Fu69rzka_processed.jpeg&w=3840&q=75)
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- A cross between yeast strains ab x AB produces the following unordered tetrads: a. Label each group of tetrads as parental ditypes (PD), non-parental ditypes (NPD), or tetratypes (T). AB AB ab ab 43 6 22 b. Are the genes linked or on separate chromosomes? Why? aB aB Ab Ab Ab AB ab aB c. If the genes are linked determine the distance between them.The drug ivacaftor has recently been developed totreat cystic fibrosis in children with the rare G551Dmutant allele of CFTR.a. Do you think that ivacaftor would be effective onlyin patients homozygous for the G551D mutation,or might it work as well in compound heterozygotes in which one copy of chromosome 7 hadG551D and the other copy a different allele ofCFTR, such as the more prevalent allele ΔF508?(The protein encoded by G551D folds up properlyand inserts into the cell membrane, but is inefficient in chloride ion transport. Ivacaftor increasesthe efficiency of G551D’s ion transport. TheΔF508 protein does not fold up properly and therefore does not get inserted into the cell membrane.)b. Why do you think ivacaftor would be more effectivein children than in older cystic fibrosis patients?c. The scientists who developed ivacaftor had a modelfor cystic fibrosis: a line of cells that grow in culture and that are homozygous for G551D. Thesecells accumulate mucus at their surfaces that…. a. A mouse cross A/a ⋅ B/b × a/a ⋅ b/b is made, and inthe progeny there are25% A/a ⋅ B/b, 25% a/a ⋅ b/b,25% A/a ⋅ b/b, 25% a/a ⋅ B/bExplain these proportions with the aid of simplifiedmeiosis diagrams.b. A mouse cross C/c ⋅ D/d × c/c ⋅ d/d is made, and inthe progeny there are45% C/c ⋅ d/d, 45% c/c ⋅ D/d,5% c/c ⋅ d/d, 5% C/c ⋅ D/dExplain these proportions with the aid of simplifiedmeiosis diagrams.
- For questions 22 through 30, please refer to the pictures below: Normal B B B C D D D D' E E 22 23 24 B D G. Y D M H. B C E 25 26 27 28 B D Break through centromere B B A D 29 30 Identify the type of chromosomal aberration in each number above. Please refer to the choices below. Reciprocal translocation Robertsonian translocation Isochromosome Terminal deletion Interstitial deletion Duplication Pericentric inversion Ring chromosome Marker chromosome Paracentric inversionISlate edu/ d2l/le/content/5003190/viewContent/44248878/View Google Tranx 4 My Drive-G X 4. Suppose that a parent Drosophila is e ca* ca The gamete frequency is as follows: e'ca e ca 16% e'ca е са 31% 14% 29% a. Circle the recombinant gametes. b. What is the map distance between the ebony and claret genes?In Section 12.3, ''Laws of Inheritance," an example of epistasis was given for the summer squash. Cross white WAvYy heterozygotes to prove the phenotypic ratio of 12 white:3 yellow:l green that was given in the text.
- Familial retinoblastoma, a rare autosomal dominant defect, arose in a large family that had no prior history of the disease. Consider the following pedigree (the darkly colored symbols represent affected individuals): a. Circle the individual(s) in which the mutation most likely occurred. b. Is the person who is the source of the mutation affected by retinoblastoma? Justify your answer. c. Assuming that the mutant allele is fully penetrant, what is the chance that an affected individual will have an affected child?31 1. In guinea pig, fur color is determined by one gene, and fur length by another independently segregating gene. The black fur allele (B) is dominant to brown fur allele (b) and the short allele (S) is dominant to long fur allele (s). From progeny listed in the following table, provide the probable genotype for the parents of each cross. Phenotype of Progeny Black short black long brown short Bio3110 Genetics Additional problems (Chapter 2): Parents Black short X Black short Black short X Brown long Black long X Black long 90 17 0 32 16 35 29 14 0 brown long 8 15 2. A mouse with the genotype BbccEeFf is mated with another mouse with BBCcEEFF genotype. What is the probability of obtaining the following genotype in the progeny? a. BBCcEeFF b. BbccEEFf C. BbccEeFF 12 3. Determine the type of gametes that are formed from the following individuals with the given genotype? A. AaBbCc B. AABBCc C. AABbCcIn com, male sterility is controlled by maternal cytoplasmic elements. However, the presence of a nuclear fertility restorer gene (F_) restores fertility to male sterile lines. a. What are the crosses male sterile female x FF male? Give the genotypes and phenotypes of the offspring in each cross. Explain.
- In fruit flies, you are mapping three genes in a three point cross. The mutants are hairy body (h), sepia colored eyes (se) and female sterility (g). You cross a heterozygous parent with a homozygous recessive parent and obtain the following results: Type Number h se g. 5 + se + 450 + se g 27 ++g_ h se + + + + h + g. h + + TOTAL is the gene in the middle and the distance in map units between se and g is Oh; 16.4 se; 7.1 Oh; 7.1 70 82 7 327 32 1000 se; 16.41. In a cross between mice the genotypes AB/ab x ab/ab, what is the recombination frequency if the progeny numbers are 72 AB/ab, 68 ab/ab, 17 Ab/ab, and 21 aB/ab? The alleles are shown for each chromosome, separated by a slash (/). . 2. A cross between fruit files with genotypes Aa Bb × aa bb produces the following progeny: 10 Aa Bb 40 Aa bb 40 aa Bb 10 aa bb Were the parental alleles in the coupling or repulsion configuration? Briefly explain. 3. Briefly describe how twin studies can reveal whether or not the appearance of a trait is strongly influenced by genetics.n corn, male sterility is controlled by maternal cytoplasmic elements. This phenotype renders the male part of the corn plants (i.e the tassel) unable to produce fertile pollen; the female parts, however, remain receptive to pollination by pollen from male fertile corn plants. However, the presence of a nuclear fertility restorer gene F restores fertility to male sterile lines Using the cardboard chips, simulate the crosses indicated below. Give the genotypes and phenotypes of the offsprings in each cross, and properly label the nucleus and the cytoplasm each individual in the cross Legend male sterile cytoplasm Male fertile cytoplasm FF nucleus Ff nucleus ff nucleus A. Male sterile female x FF male Explain the phenotype of the offspring B. Male sterile female x Ff male Explain the phenotype of the offspring
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