Consider the quadhybrid cross (four genes) below: What is the probability of obtaining offspring with the genotype AaBBccDd? O a. Zero (0) O b. 1/32 (or 0.031) O c. 1/16 (or 0.063) O d. 1/4 (or 0.25) O e. 3/8 (or 0.375) AaBbCcDd x AaBBCcDD
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- Which of the following statements illustrates codominance? Select one: a. Red flowers crossed with white flowers produce pink flowers. b. Individuals who are heterozygous for the sickle cell anemia allele are more resistant to malaria. c. Two parents with blood types A and B can have children with blood type O. d. Grey heterozygous chickens have black and white feathers.a. 1 dominant allele will contribute 120/10 = 12 cm to the base height of the plant.b. The height of the parent plant 1 Genotype of the parent plant 1 – D1D1D2D2D3D3d4d4d5d5 The height of the parent plant 2 Genotype of the parent plant 2 – d1d1d2d2d3d3D4D4D5D5Contributing alleles – D4D4D5D5. The height of the plant without any contributing alleles would be 80 cm. The plant with genotype d1d1d2d2d3d3D4D4D5D5 has 4 contributing allele each of which contributes 12 cm to the base. Hence, the height of the plant with genotype d1d1d2d2d3d3D4D4D5D5 would be 80 + 12 + 12 + 12 + 12 = 128 cm. c. Parents – D1D1D2D2D3D3d4d4d5d5 × d1d1d2d2d3d3D4D4D5D5 Gametes – D1D2D3d4d5 × d1d2d3D4D5 F1 generation – D1d1D2d2D3d3D4d4D5d5 The height of the plants of F1 generation = 80 + 12 + 12 + 12 + 12 + 12 = 140 cm Hence, Genotype of the F1 = D1d1D2d2D3d3D4d4D5d5 Phenotype of…What variable in the Hardy-Weinberg equation expresses the heterozygous genotype? A. p^2 + q^2 B. q^2 C. p^2 D. 2pq
- Which are the following are TRUE for test crosses? Select an answer and submit. For keyboard navigation, use the up/down arrow keys to select an answer. a They are useful when the genotype of a dominant phenotype is unknown. b They are always done by crossing the dominant phenotype to a homozygous recessive individual for that trait. c This test is always done in humans. d A + B e A + C f B + CImagine that you are performing a cross involving seed texture in garden pea plants. You cross true-breeding round and wrinkled parents to obtain F1 offspring. Which of the following experimental results in terms of numbers of plants are closest to what you expect in the F2 progeny? a. 8lOroundseeds b. 8lOwrinkledseeds c. 405:395 round seeds:wrinkled seeds d. 610:190 round seeds:wrinkled seedsImagine that you are performing a cross involving seed texture in garden pea plants. You cross true-breeding round and wrinkled parents to obtain F1 offspring. Which of the following experimental results in terms of numbers of plants are closest to what you expect in the F2 progeny?a. 810 round seedsb. 810 wrinkled seedsc. 405:395 round seeds:wrinkled seedsd. 610:190 round seeds:wrinkled seeds
- Assume a biallelic locus in a diploid population. Out of 100 individuals, the following genotypes were observed: 59 A1A1 individuals; 16 A1A2 individuals; and 25 A2A2 individuals. What is the observed (actual; not predicted) allele frequency of the A1 allele? (you may use a calculator here) O A. 0.33 O B. 0.67 O C. 0.16 O D.0.59 O E. 0.25Using the following information perform following crosses using the forked line and predict all possible phenotypes types of the offspring. List the number and description of the phenotypes of the offspring. 1. aa BB Cc Dd Ee X aa bb Cc Dd EE 2. Ff Rr Gg dd Bb X Ff RR Gg Dd Bb 3. Dd gg JJ bb Hh Ee X DD GG Jj Bb Hh EeIf these two individuals with the following characteristics are crossed Parent 1: blood type AB; heterozygous for H gene Parent 2: heterozygous B blood type; hh genotype What is the phenotypic ratio (blood type) of the F, generation? A. 4 AB: 1 A: 1 B: 2 0 B. 2 AB: 2 A: 4 B: no O C. 1 AB: 1 A: 2 B: 4 0 D. 1 AB: 1 A: 1 B: 10
- Red-flowering snapdragons are homozygous for allele R1. White-flowering snapdragons are homozygous for a different allele (R2). Heterozygous plants (R1R2) bear pink flowers. What phenotypes should appear among first-generation offspring of the crosses listed? What are the expected proportions for each phenotype? a. R1R1 × R1R2 c. R1R2 × R1R2 b. R1R1 × R2R2 d. R1R2 × R2RA cross is made between a heterozygote, +++/abc, and a recessive homozygote, abc/abc. Analysis of the progeny gave the following results: 450 10 a + c 70 + b c 210 + b + 65 a + + 200 abc 460 ab+ 15 What is the map distance between the b and c genes? O9.12 map units O 10.81 map units O 12.14 map units O 13.95 map units ONone of the above PrtScn Home End %23 %24 & 7. 3. 4. 7 W R Y H. K C V この #:A cross is carried out between genotypes AaBBCcddEe and AaBbccDdEE. How many phenotypes are possible in the offspring? Assume that the 5 loci assort independently and that, for each gene, the upper-case allele is completely dominant to the lower-case allele. O a. 2 O b. 6 OC. 4 O d. 16 O e. 32 O f. 8