EBK CAMPBELL BIOLOGY IN FOCUS
2nd Edition
ISBN: 8220101459299
Author: Reece
Publisher: PEARSON
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Chapter 23.3, Problem 1CC
Explain how new body forms can originate by heterochrony.
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5G JA ATTC 3
3 CTIA A1G5
5 GAAT I I3
3 CTIA AA5
Fig. 5-3: The Eco restriction site (left) would be cleaved at the locations indicated by the arrows.
However, a SNP in the position shown in gray (right) would prevent cleavage at this site by EcoRI
One of the SNPs in B. rapa is found within the Park14 locus and can be detected
by RFLP analysis. The CT polymorphism is found in the intron of the Bra013780
gene found on Chromosome 1. The Park14 allele with the "C" in the SNP has two
EcoRI sites and thus is cleaved twice by EcoRI If there is a "T" in that SNP, one
of the EcoRI sites is altered, so the Park14 allele with the T in the SNP has only
one EcoRI site (Fig. 5-3).
Park14 allele
with SNP(C)
Park14 allele
with SNPT)
839 EcoRI
1101 EcoRI
839 EcoRI
Fig. 5.4: Schematic restriction maps of the two different Park14 alleles (1316 bp long) of B. rapa. Where
on these maps is the CT SNP located?
90
The primers used to amplify the DNA at the Park14 locus (see Fig. 5 and Table 3
of Slankster et…
Chapter 23 Solutions
EBK CAMPBELL BIOLOGY IN FOCUS
Ch. 23.1 - Your measurements indicate that a fossilized skull...Ch. 23.1 - Describe an example from the fossil record that...Ch. 23.1 - WHAT IF? What might a fossil record of life today...Ch. 23.1 - Prob. 4CCCh. 23.2 - Explain the consequences of plate tectonics for...Ch. 23.2 - Prob. 2CCCh. 23.2 - What factors promote adaptive radiations?Ch. 23.2 - Prob. 4CCCh. 23.3 - Explain how new body forms can originate by...Ch. 23.3 - Why is it likely that Hox genes have played a...
Ch. 23.3 - MAKE CONNECTIONS Given that changes in morphology...Ch. 23.4 - How can the Darwinian concept of descent with...Ch. 23.4 - Prob. 2CCCh. 23 - Which factor most likely caused animals and plants...Ch. 23 - Adaptive radiations can be a direct consequence of...Ch. 23 - A researcher discovers a fossil of what appears to...Ch. 23 - A genetic change that caused a certain Hox gene to...Ch. 23 - A swim bladder is a gas-filled sac that helps fish...Ch. 23 - SCIENTIFIC INQUIRY Herbivory (plant eating) has...Ch. 23 - FOCUS ON EVOLUTION Describe how gene flow, genetic...Ch. 23 - FOCUS ON ORGANIZATION You have seen many examples...Ch. 23 - SYNTHESIZE YOUR KNOWLEDGE In 2010, the Soufriere...
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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
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