Question 2 a (800 words)

i) When yeast metabolises glucose under anaerobic conditions, it produces ethanol and CO2 as end products of glycolysis but under aerobic conditions, pyruvate is the end-product of the glycolytic pathway. What is the physiological rationale for producing ethanol and CO2 under anaerobic conditions? Fully explain your answer.

ii) The standard free energy change for the oxidation of glucose to CO2 is shown below: Glucose + 6 O2 → 6 H2O + 6 CO2 ∆G’o = - 2840 kJ mol-1 If the free energy change (∆G’o) for the hydrolysis of ATP (ATP + H2O → ADP + Pi) is - 30.5 kJ mol-1, what is the overall percentage efficiency of glycolysis in terms of the free energy “captured” in the form of ATP? Make sure you fully explain your calculations and take care to include units and give your answer to three significant figures.

iii) Clearly explain in your own words the differences between substratelevel phosphorylation and oxidative phosphorylation. iv) What is the physiological rationale for the inhibitory effect of ATP on Phosphofructokinase?

Question 2b

In 800 words, Describe the process of receptor activation in Cell Signalling pathways. Illustrate your answer with a Cell Signalling system of your choice making sure to include the pathway(s) of receptor activation, the second messenger(s) involved and the downstream cellular response(s).

Why might the genetic systems observed to exhibit polymorphism associated with non?Mendelian segregation have unusual properties compared to the set of all drive systems that occur over long evolutionary time periods? In other words, do you expect the properties of polymorphic systems to be different from that of all drive systems. Why?

Given the idea that male expressed drive systems like sex-ratio X chromosomes, though favored by gametic selection, may have costs with respect to other fitness components. Alternatively, the fitness costs associated with female?expressed drive may be less obvious. What basic difference between male and female meiosis may explain some of the difference in the fitness costs of drive in males vs. females (all else being equal) ?

A few days after you have completed the streak plate method, you observe not one single isolated colony on the TSA plate. Explain why this may have happened.

How does dental plaque form?

How does tooth decay occur?

Jeffrey Mitton and his colleagues found three genotypes (R2R2, R2R3, and R3R3) at a locus encoding the enzyme peroxidase in ponderosa pine trees growing at Glacier Lake, Colorado. The observed numbers of these genotypes were: Genotypes: Number observed: R2R2 135 R2R3 44 R3R3 11 a. Determine the allele frequencies for the R2 and R3 alleles. b. Use the Chi square test to determine if this population is in Hardy-Weinberg equilibrium for the peroxidase trait. Generate a Chi square value. c. Interpret the Chi square value from part (b). If the population is not in Hardy-Weinberg equilibrium, suggest a reason why.

1. how is gram stain done?

2. What types of bacteria are stained differently in the Gram stain?

3. What is the key step in the Gram stain that distinguishes between two different cell types? Explain why.

The laboratory reported gram-negative bacteria arranged inside the leukocytes of the pus sample. Other tests confirmed the pathogen was Neisseria gonorrhoeae, indicating Tom had the sexually transmitted infection gonorrhea.

4. Describe the shape of the gram-negative bacteria shown in the photo.

Dr. Nelson called Tom to explain the laboratory results and told him that he was required to report cases of gonorrhea to the Public Health Service. He indicated that Tom should expect a follow-up call so that his sexual partners, who may be infected with the pathogen but not yet showing symptoms, can be tested and treated if necessary. Dr. Nelson was confident that Tom would get the prompt and appropriate treatment he needed.

5. Name an antibiotic that could be effective for an uncomplicated case of gonorrhea.

The yellow mutation is an X-linked recessive mutation in Drosophila, causing yellow body color. A yellow-bodied female is crossed to a male with wild type body color to produce F1 progeny. The F1 female progeny and F1 male progeny are then crossed to each other to produce F2 progeny. Among the F2 females

A. all will be yellow body color.

B. 3/4 will be yellow body color.

C. 1/2 will be yellow body color.

D. 1/4 will be yellow body color.

E. none will be yellow body color.

4. Describe the gametophyte(s) and sporophyte of each of the following plants: moss (e.g. Polytrichum), liverwort (e.g. Marchantia), fern (e.g. Ceratopteris), gymnosperm (e.g. pine), and angiosperm (e.g. lily).

13.     Using the Heart Rate Reserve Method, calculate the target heart rate zone for the Tom:

Tom: Age 20; Resting Heart Rate 60; Desired Target Heart Rate Zone: 70% to 85% of Maximum Heart Rate

Show your calculations to earn full credit.

Tom’s Maximum heart rate: (1pts) Show Calculation

Tom’s Heart Rate Reserve (Working Heart Rate): Show calculation (1pts)

Desired Target heart rate range:   Show all calculations for full credit.

              Low range 70% = _____ bpm (beats per minute)(2pts)

                  High range 85% = _____bpm(2pts)

If Tom counted his heart rate for 6 seconds, what would the number be for the above percentages?   For example, a heart rate of 58 beats per minute would be the number 6 when counting for 6 seconds: (2)

70% = ______

85% = ______

14. Using the Heart Rate Reserve Method, calculate the target heart rate zone for the Sally:

Sally: Age 60; Resting Heart Rate 72; Desired Target Heart Rate Zone: 60% to 75% of Maximum Heart Rate.

Show your calculations to earn full credit. Failure to show calculation means a deduction of at least 1 point for each part of the question.

Sally’s Maximum heart rate: (1) Show Calculation

Sally’s Heart Rate Reserve (Working Heart Rate): Show calculation (1pts)

Sally’s Desired Target heart rate range:   Show all calculations for full credit.

Low range 60% = _____ bpm (beats per minute)(2pts)

High range 75% = _____bpm(2pts)

The feature that defines and distinguishes ALL hominins from the apes is:

the fact that we walk on two feet.

Our lack of a tail

Our larger brains than the earlier hominins (like Austrolapithecus).

Our 3-D vision