Biology 2e
2nd Edition
ISBN: 9781947172517
Author: Matthew Douglas, Jung Choi, Mary Ann Clark
Publisher: OpenStax
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Chapter 18, Problem 20CTQ
Explain why the statement that a monkey is more evolved than a mouse is incorrect.
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You have been hired as a public relations specialist to give invertebrates a good name. After all, they are much more than just creepy crawly bugs!
Your first task though is to convince yourself that is true. The best way to do that is to start close to home. Find something in your house that is a product obtained directly from an invertebrate or only due to an invertebrate’s actions. Describe the product, its function and utility, as well as any human manufactured alternatives. Be sure to highlight the advantages of obtaining this directly from nature.
Keep in mind, a product can be something you use, wear, eat, or enjoy for its visual appeal.
Use the following tree diagram to answer Questions #8-10.
8) Which of the following two animals
are the most closely related based on the
tree to the left?
a) Pig and camel
b) Hippo and pig
c) Deer and cow
9) CIRCLE on the tree diagram where the
common ancestor between a hippo and a
cow is.
10) Put a SQUARE on the tree diagram
where the common ancestor between a pig
and a peccary is.
Explain:
Healthy Cell Function Overview→ Briefly describe how a healthy cell usually works: metabolism (ATP production), pH balance, glycogen storage, ion transport, enzymes, etc.
Gene Mutation and Genetics Part→ Focus on the autosomal recessive mutation and explain:
How gene mutation affects the cell.
How autosomal inheritance works.
Compare the normal and mutated gene sequences simply. → Talk about possible consequences of a faulty hydrolytic enzyme.
Chapter 18 Solutions
Biology 2e
Ch. 18 - Figure 18.14 Which is most likely to survive,...Ch. 18 - Figure 18.22 If two species eat a different diet...Ch. 18 - Figure 18.23 Which of the following statements is...Ch. 18 - Which scientific concept did Charles Darwin and...Ch. 18 - Which of the following situations will lead to...Ch. 18 - Which description is an example of a phenotype? A...Ch. 18 - Which situation is most likely an example of...Ch. 18 - Which situation would most likely lead to...Ch. 18 - What is the main difference between dispersal and...Ch. 18 - Which variable increases the likelihood of...
Ch. 18 - What is the main difference between autopolyploid...Ch. 18 - Which reproductive combination produces hybrids?...Ch. 18 - Which condition is the basis for a species to be...Ch. 18 - Which situation is not an example of a prezygotic...Ch. 18 - Which term is used to describe the continued...Ch. 18 - Which components of speciation would be least...Ch. 18 - If a person scatters a handful of garden pea plant...Ch. 18 - Why do scientists consider vestigial structures...Ch. 18 - How does the scientific meaning of “theory” differ...Ch. 18 - Explain why the statement that a monkey is more...Ch. 18 - Why do island chains provide ideal conditions for...Ch. 18 - Two species of fish had recently undergone...Ch. 18 - Why can polyploidy individuals lead to speciation...Ch. 18 - What do both rate of speciation models have in...Ch. 18 - Describe a situation where hybrid reproduction...
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- Can you fill out those termsarrow_forwardExplain down bellow what happens to the cell: Decreased pH in mitochondria Increased ATP Decreased pH in cytosol Increased hydrolysis Decreasing glycogen and triglycerides Increased MAP kinase activity Poor ion transport → For each one:→ What normally happens?→ What is wrong now?→ How does it mess up the cell?arrow_forwardAn 1100 pound equine patient was given 20 mg/kg sucralfate 3 times a day, 2.8 mg/kg famotidine twice a day, and 10mg/kg doxycycline twice a day. Sucralfate comes as a 1 gm tablet, famotidine as 20 mg tablets, and doxycycline as 100mg tablets. All are in bottles of 100 tablets.How many total mg are needed for the patient and how many tablets of each would be needed to provide each dose?How many bottles of each would be needed to have available if this patient were to be on this drug regimen for 5 days?arrow_forward
- The patient needs a solution of 2.5% dextrose in Lactated Ringer’s solution to run at 75 ml/hr for at least the next 12hours. LRS comes in fluid bags of 500 ml, 1 Liter, 3 Liters and 5 Liters. How can a 2.5% solution be made by adding50% dextrose to the LRS?arrow_forward“Gretchen” was a 68-pound canine who came to the VMTH as small animal surgery patient. She receivedacepromazine, 0.2 mg/kg from a 10 mg/ml solution and oxymorphone, 0.08 mg/kg from a 1 mg/ml solution before surgery.What are the mechanisms of action of acepromazine and oxymorphone? Why would they be given together?How many mg provide each dose and how many ml of each of these solutions were given?arrow_forwardAfter surgery, “Gretchen” was put on carprofen, 1 mg/pound bid (twice a day). The tablets come in 25, 75 and 100 mgsizes. Which size tablet would be appropriate?What is the mechanism of action of carprofen?An outpatient prescription was written for her so she would have enough for 10 days. How many tablets did she need?What information needs to be on her out-patient prescription?arrow_forward
- Joden Koepp olor in chickens is due to incomplete dominance. BB = Black chicken, WW = White BLOOD TYPES Arhite chicken is In humans, Rh positive blood is dominant (R) over Rh negative blood (r). A man with type 0, Rh positive blood (whose mother had Rh negative blood), marries a woman with type AB, Rh negative blood. Several children were born. is? R R Genotypes Phenotypes RRR RR Rr Rr 4/16 RR R RR RK Rr Rr 4/16 rr 3/4 Rh posi 1/4 Rh negu 1/2 Rr rr rr rrrr 88 888 75 e genotype of the man? the genotype of the woman? The mother of the man had type AB blood.arrow_forwardPlease indentify the unknown organismarrow_forwardPlease indentify the unknown organismarrow_forward
- Please indentify the unknown organismarrow_forwardPlease indentify the unknown organismarrow_forward5G 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…arrow_forward
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