1 SEM ACC W/RAVEN CARDED
12th Edition
ISBN: 9781265486297
Author: Raven
Publisher: MCGRAW-HILL HIGHER EDUCATION
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Chapter 22, Problem 3S
Refer to figure 22 6 In Texas. Drummond s phlox is a dark shade of red where it occurs with the pale blue pointed phlox. Elsewhere, where it occurs by itself, it is a pale pinkish-blue In this case, there is no competition for ecological
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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 22 Solutions
1 SEM ACC W/RAVEN CARDED
Ch. 22.1 - Understand the biological species concept and why...Ch. 22.1 - Define the two kinds of reproductive isolating...Ch. 22.1 - Describe the relationship of reproductive...Ch. 22.2 - Prob. 1LOCh. 22.2 - Prob. 2LOCh. 22.3 - Prob. 1LOCh. 22.3 - Prob. 2LOCh. 22.4 - Compare and contrast sympatric and allopatric...Ch. 22.4 - Prob. 2LOCh. 22.5 - Prob. 1LO
Ch. 22.5 - Prob. 2LOCh. 22.6 - Prob. 1LOCh. 22.6 - Prob. 2LOCh. 22.7 - Prob. 1LOCh. 22.7 - Prob. 2LOCh. 22 - Prob. 1IQCh. 22 - Prob. 2IQCh. 22 - Prezygotic isolating mechanisms include all of the...Ch. 22 - Reproductive isolation is a. a result of...Ch. 22 - Problems with the biological species concept...Ch. 22 - Prob. 4UCh. 22 - Prob. 5UCh. 22 - Prezygotic isolation a. always involves mechanisms...Ch. 22 - Speciation by allopolyploidy a. takes a long time....Ch. 22 - Adaptive radiation a. is the result of enriched...Ch. 22 - Prob. 9UCh. 22 - Character displacement a. arises through...Ch. 22 - Prob. 11UCh. 22 - If reinforcement is weak and hybrids are not...Ch. 22 - Natural selection can a. enhance the probability...Ch. 22 - Hybridization between incompletely isolated...Ch. 22 - Natural selection can lead to speciation a. by...Ch. 22 - Natural selection can lead to the evolution of...Ch. 22 - Prob. 2SCh. 22 - Refer to figure 22 6 In Texas. Drummond s phlox is...Ch. 22 - Refer to figure 22.16. Geospiza fuliginosa and...
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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?arrow_forwardWhat experiment could you do to see if the maternal gene, bicoid, is sufficient to form anterior fates?arrow_forwardYou’re curious about the effect that gap genes have on the pair-rule gene, evenskipped (eve), so you isolate and sequence each of the eve enhancers. You’re particularly interested in one of the enhancers, which is just upstream of the eve gene. Describe an experimental technique you would use to find out where this particular eve enhancer is active.arrow_forward
- 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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