A housewife fixed lunch for herself and two others. The lunch consisted of home-canned gefilte fish (served cold with horseradish on toast) soft drinks and milk. She ate two portions of gefilte fish, her employee ate one portion and her daughter-in-law ate half a portion. Four hours later, she complained of headache, epigastric distress, hoarseness and slight dyspnea. The epigastric distress continued. She vomited repeatedly and experienced dryness of the mouth, weakness, constipation, and urinary retention. Examination by a physician revealed an anxious woman with labored respirations (26 per minute); a blood pressure of 80/58 mm Hg, and a pulse rate of 110 beats per minute. The pupils were equal in size but somewhat dilated; they were reactive to light. Extraocular eye movements were normal, and no facial weakness was noted. Her throat and mouth were dry, and her voice was hoarse. The chest was normal to auscultation and to percussion. The abdomen was soft and nontender, with decreased bowel sounds. Deep tendon reflexes were normal. Because of a history of mild hypertension, the findings of tachycardia and hypotension and the history of substernal distress suggested the possibility of a myocardial infarction. The patient was hospitalized. Lab findings included normal results for a complete blood cell count and urinalysis. Values of serum electrolytes, bilirubin, amylase, and protein determined on serum from blood drawn the morning after admission were all within normal limits. Chest and abdominal X-ray films were interpreted as normal. An EKG was unchanged from previous tracings. The patient was treated symptomatically with antacids, nasogastric suction, and intravenous administration of fluids. Three days later the patient had a cardiopulmonary arrest and was resuscitated. Spontaneous respiration did not occur, and breathing was maintained on a mechanical respirator. No apparent benefit resulted from 80,000 units of bivalent (types A & B) and 10,000 units of type E antitoxin, which were administered intravenously. The next day the patient died. Autopsy showed generalized ischemic changes in the central nervous system and moderate arteriosclerosis of the coronary arteries with slight hypertrophy of the left ventricle. The liver and the spleen were enlarged and hyperemic. The lungs showed pulmonary edema with focal acute bronchopneumonia. The other two patients survived the infection.

Regarding DNA extraction:

1) Describe the purpose of each of the following steps or reagents used in DNA isolation (in detail):

- Using fresh vs. dried specimens

- Grinding or tearing tissue

- Lysis solution

- TE buffer

Describe the unique qualities of cyanobacteria with respect to energy production.

If an enzyme found in human cells would be affected by Sofosbuvir, which eukaryotic enzyme would it most likely affect?

a) DNA Polymerase

b) RNA Polymerase

c) Telomerase

d) Reverse Transcriptase

Thank you!

for the Biodiesel Production process

1. Describe the biochemical pathway that occurs in the biodiesel production process
2. Which enzymes participate in the selected industrial process
3. Classify enzymes according to their activity
4. Establish optimal pH and temperature conditions for the functioning of these enzymes.
5. Consult the factors that inhibit the activity of the enzymes that participate in the selected industrial process.

a 46 lb. postoperative patient needs fluids at 75 ml/kg/day with lidocaine (2%) at 30ug/kg/min, ketamine (100 mg/ml) at 5 ug/kg/min, and morpine (10 mg/ml) at 2.5 (ug/kg/min).

a: what is the rate of fluids per hour

b: how many militers of medication to add to 1 L bag:

of lidocaine ? of Ketamine? of morphine??

PLEASE ANSWER ALL PARTS AND SHOW WORK

Describe or draw the endocrine control in the ovary beginning with the primary oocyte and ending with the degeneration of the corpus luteum. Be sure to include the hormones involved.

o carry out 'blue-white screening', the E.coli plasmid pUC18 carries:

** Prior, which I forgot to include:

All Vulcans have pointed ears (P) and a right-sided heart (R) which are dominant over Earthling alleles of rounded ears (p) and a left-sided heart (r). Both Vulcans and Earthlings are homozygous for each allele.

Mr. Spock has a Vulcan father and an Earthling mother.

Gametes produced by an individual that are the same as the gametes contributed by each parent to produce that individual are called parental types.

Gametes that have a combination of parental and maternal alleles are called recombinant types.

Recombinant gametes are the result of independent assortment during Metaphase I if genes are on different chromosomes or crossing over during prophase I if genes are on the same chromosome.

In which I got the following: PR, PP, pR,Pr

_______

In Vulcans, the absence of adrenal glands (A) is dominant over the Earthling allele for adrenal glands (a). All Earthlings have adrenal glands.

A gene map of Mr. Spock shows that the genes for ear shape (P/p) and heart sidedness (R/r) are 20 map units apart on chromosome 7 while the gene for adrenal glands (A/a) is on chromosome 3.

If Mr. Spock fathers a child with an Earthling female, what is the chance that the child will have pointed ears and adrenal glands?

Select one:

a. 10%

b. 20%

c. 25%

d. 50%

Can someone please show me how to get the proper answer? I keep getting myself confused on this self-check question, and I don't know where I keep messing up.

Thank you

Under anaerobic conditions, cells must sacrifice the pyruvate they made from glycolysis. Why?

You have three genes on the same chromosome - A, B and C. Each gene has two alleles in a dominant/recessive relationship. For these genes the homozygous recessive has the mutant phenotype for that trait, the dominant phenotype = wild type for that trait.

• allele A is dominant to a; phenotype a = mutant for trait a; phenotype A = wild type for trait A
• allele B is dominant to b; phenotype b = mutant for trait b; phenotype B = wild type for trait B
• allele C is dominant to c; phenotype c = mutant for trait c; phenotype C = wild type for trait C

Note: phenotypes can be represented by single letters. For example phenotype A = genotypes Aa or AA; phenotype a = genotype aa. Assume that phenotype ab = mutant phenotype for traits a and b, and wild type phenotype for trait C.

You cross an individual heterozygote for all three genes, with an individual who is homozygote recessive for all three. Out of 10,000 offspring you get the following phenotypes and amounts:

• phenotype a - 98
• phenotype b - 150
• phenotype c - 4751
• phenotype ab - 4749
• phenotype ac - 147
• phenotype bc - 99
• phenotype abc - 3
• wild type - 3

Use this information to answer the following questions.

Based on the information given and your answer to the previous questions, what was the genotype of the heterozygous parent, in the correct order?

Select one:

a. BAc/baC

b. ABC/abc

c. AcB/aCb

d. ACB/acb

e. ABc/abC

Where do processed mRNA’s go after leaving the nucleus and how? Explain

You have three genes on the same chromosome - A, B and C. Each gene has two alleles in a dominant/recessive relationship. For these genes the homozygous recessive has the mutant phenotype for that trait, the dominant phenotype = wild type for that trait.

• allele A is dominant to a; phenotype a = mutant for trait a; phenotype A = wild type for trait A
• allele B is dominant to b; phenotype b = mutant for trait b; phenotype B = wild type for trait B
• allele C is dominant to c; phenotype c = mutant for trait c; phenotype C = wild type for trait C

Note: phenotypes can be represented by single letters. For example phenotype A = genotypes Aa or AA; phenotype a = genotype aa. Assume that phenotype ab = mutant phenotype for traits a and b, and wild type phenotype for trait C.

You cross an individual heterozygote for all three genes, with an individual who is homozygote recessive for all three. Out of 10,000 offspring you get the following phenotypes and amounts:

• phenotype a - 98
• phenotype b - 150
• phenotype c - 4751
• phenotype ab - 4749
• phenotype ac - 147
• phenotype bc - 99
• phenotype abc - 3
• wild type - 3

Use this information to answer the following questions.

You cross an individual heterozygote for genes A, B and C, with an individual who is homozygote recessive for all three. Assuming independent assortment for all three genes what do you expect to see out of 10,000 offspring?

Remember:

• for all three traits wild type phenotype is dominant to mutant phenotype
• allele A is dominant to a; phenotype a = mutant for trait a; phenotype A = wild type for trait A
• allele B is dominant to b; phenotype b = mutant for trait b; phenotype B = wild type for trait B
• allele C is dominant to c; phenotype c = mutant for trait c; phenotype C = wild type for trait C

Select one:

a. Three different phenotypes among the offspring; approximately 3333 offspring of each phenotype.

b. Approximately 156 offspring will have the recessive phenotype for all three traits.

c. Approximately 2963 offspring will be wild type (dominant phenotype for all three traits).

d. Eight different phenotypes among the offspring; approximately 1250 offspring of each phenotype.

Complete the tables below by indicating whether the physical, chemical and biological parameters change with the course of a river.

Please include a valid reason (NO PLAGIARISM!)

What is NOT a major characteristic found in the Kingdom Fungi ?
A. they produce spores B. they have rigid cell walls C. they are multinucleate D. they contain chlorophyll E. None of the previous