Biology 2e
2nd Edition
ISBN: 9781947172517
Author: Matthew Douglas, Jung Choi, Mary Ann Clark
Publisher: OpenStax
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Chapter 31, Problem 17CTQ
Research the life of Jan Babtista van Helmont What did the van Helmont experiment show?
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From your previous experiment, you found that this enhancer activates stripe 2 of eve expression. When you sequence this enhancer you find several binding sites for the gap gene, Giant. To test how Giant interacts with eve, you decide to remove all of the Giant binding sites from the eve enhancer. What results do you expect to see with respect to eve expression?
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Chapter 31 Solutions
Biology 2e
Ch. 31 - Figure 31.6 Soil compaction can result when soil...Ch. 31 - Figure 31.7 Which horizon is considered the...Ch. 31 - Figure 31.10 Farmers often rotate corn (a cereal...Ch. 31 - For an element to be regarded as essential, all of...Ch. 31 - The nutrient that is part of carbohydrates,...Ch. 31 - Most ____ are necessary for enzyme function....Ch. 31 - What is the main water source for land plants?...Ch. 31 - Which factors affect soil quality? chemical...Ch. 31 - Soil particles that are 0.1 to 2 mm in diameter...Ch. 31 - A soil consists of layers called that taken...
Ch. 31 - What is the term used to describe the solid rock...Ch. 31 - Which process produces an inorganic compound that...Ch. 31 - Through mycorrhization, a plant obtains important...Ch. 31 - What term describes a plant that requires...Ch. 31 - What is the term for the symbiotic association...Ch. 31 - What type of plant problems result from nitrogen...Ch. 31 - Research the life of Jan Babtista van Helmont What...Ch. 31 - List two essential macronutrients and two...Ch. 31 - Describe the main differences between a mineral...Ch. 31 - Name and briefly explain the factors that affect...Ch. 31 - Describe how topography influences the...Ch. 31 - Why is biological nitrogen fixation an...Ch. 31 - What is the main difference, from an energy point...Ch. 31 - Why is a root nodule a nutritional adaptation of a...
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- For short answer questions, write your answers on the line provided. To the right is the mRNA codon table to use as needed throughout the exam. First letter U บบบ U CA UUCPhe UUA UCU Phe UCC UUG Leu CUU UAU. G U UAC TV UGCys UAA Stop UGA Stop A UAG Stop UGG Trp Ser UCA UCG CCU] 0 CUC CUA CCC CAC CAU His CGU CGC Leu Pro CCA CAA Gin CGA Arg CUG CCG CAG CGG AUU ACU AAU T AUC lle A 1 ACC Thr AUA ACA AUG Mot ACG AGG Arg GUU GCU GUC GCC G Val Ala GAC Asp GGU GGC GUA GUG GCA GCG GAA GGA Gly Glu GAGJ GGG AACASH AGU Ser AAA1 AAG Lys GAU AGA CAL CALUCAO CAO G Third letter 1. (+7) Use the table below to answer the questions; use the codon table above to assist you. The promoter sequence of DNA is on the LEFT. You do not need to fill in the entire table. Assume we are in the middle of a gene sequence (no need to find a start codon). DNA 1 DNA 2 mRNA tRNA Polypeptide C Val G C. T A C a. On which strand of DNA is the template strand (DNA 1 or 2)?_ b. On which side of the mRNA is the 5' end (left or…arrow_forward3. (6 pts) Fill in the boxes according to the directions on the right. Structure R-C R-COOH OH R-OH i R-CO-R' R R-PO4 R-CH3 C. 0 R' R-O-P-OH 1 OH H R-C-H R-N' I- H H R-NH₂ \H Name Propertiesarrow_forward4. (6 pts) Use the molecule below to answer these questions and identify the side chains and ends. Please use tidy boxes to indicate parts and write the letter labels within that box. a. How many monomer subunits are shown? b. Box a Polar but non-ionizable side chain and label P c. Box a Basic Polar side chain and label BP d. Box the carboxyl group at the end of the polypeptide and label with letter C (C-terminus) H H OHHO H H 0 HHO H-N-CC-N-C-C N-C-C-N-GC-OH I H-C-H CH2 CH2 CH2 H3C-C+H CH2 CH2 OH CH CH₂ C=O OH CH2 NH2arrow_forward
- please draw in what the steps are given. Thank you!arrow_forwardplease draw in and fill out the empty slots from image below. thank you!arrow_forwardThere is a species of eagle, which lives in a tropical forest in Brazil. The alula pattern of its wings is determined by a single autosomal gene with four alleles that exhibit an unknown hierarchy of dominance. Genetic testing shows that individuals 1-1, 11-4, 11-7, III-1, and III-4 are each homozygous. How many possible genotypes among checkered eagles in the population?arrow_forward
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