The sarcoplasmic reticulum Ca2 ATPase (SERCA) pumps Ca2 out of the cytoplasm and into the sarcoplasmic reticulum of muscle cells. Muscle contraction occurs in response to the rapid rise of Ca2 in the cytoplasm. Calcium ions must be removed form the cytoplasm for muscle relaxation to occur. SERCA removes the calcium ions, enabling muscle relaxation. SERCA is a monomeric protein with a nucleotide-binding domain (N), a domain that accepts a phosphoryl group (P), and an actuator domain (A). A transmembrane domain binds calcium ions.

From what you know of the mechanism of Ca2 transport, which of the following binding patterns are observed at any point? Select all that apply. Assume that only the species mentioned are bound. For example, if ADP is not mentioned, it is not bound.

Select all that apply. (Four correct answers)

2 Ca2 , ADP, and Pi bound

ADP bound

2 Ca2 and Pi bound

AMP bound

2 Ca2 bound

Pi bound

2 Ca2 , ATP bound

In bacteria, the transport of many nutrients, including sugars and amino acids, is driven by the electrochemical H⁺ gradient across the plasma membrane. In E. coli, for example, an H⁺–lactose symporter mediates the active transport of the sugar lactose into the cell. Given what you know about coupled transport, which is likely true of the H⁺–lactose symporter?

Choose one:

A. If the H⁺ gradient were reversed, the transporter could serve as an H⁺–lactose antiport.

B. The transporter oscillates randomly between states in which it is open to either the extracellular space or the cytosol.

C. To undergo the conformational change that releases lactose into the cell, the transporter hydrolyzes ATP.

D. Lactose and H⁺ ions bind to two different conformations of the transporter.

E. The transporter goes through an intermediate state in which the lactose-binding site is open to both sides of the membrane.

Why have we NOT found examples in the fossil record of every animal that ever lived on Earth? select all that apply

Sodium and potassium ion channels have several negatively charged residues at the entry to the channel. On what basis do K+ channels specifically select for K+ ions? (In other words, why dont K+ channels enable Na+ ions to cross the membrane?)

-ionic radius, Na+ is to small

-energy cost, it is to energetically costly to dehydrate Na+

-charge, Na is slightly more electronegative than K and does not interact with the channel carbonyl groups

-charge, Na is slightly less electronegative than K and does not interact with the channel carbonyl groups

-ionic radius, Na+ is too big

A family is affected by a rare genetic disease with a Mendelian pattern of inheritance. The father is affected, and the mother is unaffected by this disease. They have three children: an affected son and daughter and an unaffected son. If they have a fourth child, what is the probability that this child will be affected by the disease?

100% 50% 33% 25% 0%

Which of the following statements is true regarding olfaction? list all that apply

• Smell is a chemical sense.  
• Odorant molecules dissolve in mucus before stimulating a receptor.  
• Humans can only distinguish up to 200 odors.  
• Olfactory receptors have hairs on the apical surface that respond to stimuli.
  When olfactory receptors are stimulated a receptor potential is created. 

The enzyme glucose oxidase isolated from the mold Penicillium notatum catalyzes the oxidation of β-D-glucose to D-glucono-δ-lactone. This enzyme is highly specific for the β anomer of glucose and does not affect the α anomer. In spite of this specificity, the reaction catalyzed by glucose oxidase is commonly used in a clinical assay for total blood glucose –that is, for solutions consisting of a mixture of β-D-glucose and α-D-glucose. What makes it possible for this test to measure total blood glucose?

• The α anomer of glucose easily converts to the β anomer.
• Both α-D-glucose and B-D-glucose have a free anomeric carbon
• Glucose interconverts between the glucopyranose and glucofuranose forms.
• The β anomer is present in a much higher concentration than the α anomer. 

• Fehling's reagent can only detect the linear form of glucose.
• Fehling's test cannot be used in the lab to estimate the relative amount of glucose.
• Fehling's reagent is not specific to glucose, detecting other reducing sugars.
• Fehling's test cannot detect the presence of the β anomer.

• All four polysaccharides are homopolymers.
• All four polysaccharides are heteropolymers.
• Starch is a heteropolymer, whereas chitin, cellulose, and glycogen are homopolymers.
• Chitin is a heteropolymer, whereas starch, cellulose, and glycogen are homopolymers.

1) The results you predict as a result of a controlled experiment can be described as an hypothesis, such as “selection of Wisconsin Fast Plants with the most trichomes in the first (parent) generation will result in an increase in trichome number in the plants of the second generation.” You are making a prediction based on scientific knowledge of selection, and are able to quantify the number of trichomes. This is your experimental hypothesis. A null hypothesis for your experiment would predict that there will be no difference between the groups as a result of the treatment. Your experimental goal would be to gather data to reject the null hypothesis. The data presented in Part D shows the results of artificial selection for hairy Wisconsin Fast Plants. Identify the null hypothesis for this investigation. The data presented in Part D shows the results of artificial selection for hairy Wisconsin Fast Plants. Identify the null hypothesis for this investigation.

a) There will be no difference between the mean number of trichomes in the second generation compared to the parent population.

b) If the mean number of trichomes is greater in the second generation than in the parent population, then selection has occurred. c) As a result of selection, the mean number of trichomes will be greater in the second generation.

d) If plants with the most trichomes in the first generation are selected as parents, then the second generation will have more trichomes.

2) In the preceding example, we calculated the probability of obtaining certain genotypes in the offspring based on allelic frequencies, but we can also use this method to determine the genetic makeup of a population. The Hardy-Weinberg equation is p 2 + 2pq + q 2 = 1. What do these variables represent, and how can this equation be used to describe an evolving population? In an earlier part of this investigation, we worked with a pair of alleles that were incompletely dominant to each other. Now let’s generalize and use terminology that can be applied to any genetic trait. You may want to print out the following instructions to use as a reference when working on Hardy-Weinberg problems. For a gene locus that exists in two allelic forms in a population, A and a: Let p = the frequency of A, the dominant allele Let q = the frequency of a, the recessive allele All the dominant alleles plus all the recessive alleles will equal 100% of the alleles for this gene, or, expressed mathematically, p + q = 1 for a population in genetic equilibrium. If this simple binomial is expanded we get the Hardy-Weinberg equation: p 2 + 2pq + q 2 = 1 The three terms of this binomial expansion indicate the frequencies of the three genotypes: p 2 = frequency of AA (homozygous dominant) 2pq = frequency of Aa (heterozygous) q 2 = frequency of aa (homozygous recessive) If we know the frequency of one of the alleles, we can calculate the frequency of the other allele: p + q = 1, so p = 1 – q q = 1 – p Let’s use this equation to solve the following problem: In pea plants, the allele for tall plants (T) is dominant to the allele for dwarf plants (t). If a population of 100 plants has 36 dwarf plants, what is the frequency of each allele? Here is a step-by-step guide: Let p = frequency of the dominant allele (R), and q = frequency of the recessive allele (r). q 2 = frequency of the homozygous recessive = 36%, or 0.36. Since q 2 = 0.36, what is q? Take the square root of 0.36, or q = 0.6. Now, p + q = 1, so subtract q from 1 to find the value of p, or 1 – 0.6 = 0.4; therefore, p = 0.4. That’s it! But let’s go a step further--how many of these plants are heterozygous tall (Tt) if the population is in Hardy-Weinberg equilibrium? Calculate 2pq = 2 × 0.4 × 0.6 = 0.48, or 48%. Since there are 100 plants, 48 are heterozygous tall. Suppose that green seeds (G) are dominant to yellow seeds (g) in peas. In a population of 500 individuals, 25% show the recessive phenotype. How many individuals would you expect to be homozygous dominant for this trait if the population is in Hardy-Weinberg equilibrium?

You need to list equations used and provide steps of problem solving. Providing an answer itself is not enough for full grade.

3. You have sampled a population in which you know that the percentage of the homozygous recessive genotype (aa) is 36%. Calculate the frequency of the heterozygous genotype, homozygous dominant genotype and homozygous recessive genotype.

4. You have sampled a population in which you know that the percentage of the homozygous recessive genotype (aa) is 49%. What is the frequency of the recessive allele, and the frequency of the dominant allele? What is the frequency of heterozygous genotypes?