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a and p lab sem. 2

blood pressure, typing, and enzyme activity

QuestionAnswer
pulse pressure equation PP=Systolic blood pressure- diastolic blood pressure
mean arterial pressure MAP=Diastolic blood presure+ (1/3) Pulse Pressure
what is blood pressure measured in mmHg
systolic blood pressure pressure of blood against walls of larger arteries when ventricles contract
diastolic blood pressure pressure of blood against walls of larger arteries when ventricles are relaxed
when preparing a blood transfusion what specific factors must be considered in order for the transfusion to be successful you do not want to give the recipient any antigens he/she does not already have. the antibodies of the donor are in the plasma and can be removed prior to the transfusion so the reaction of the donors antibodies with the recipients antigen is not an issue
explain the difference between dominant, recessive, and co-dominant inheritance with respect to ABO and Rh blood types dominant traits (A,B,Rh+) are always expressed if the gene is present recessive traits (O, Rh-) are expressed only in the abscence of any dominant traits since A and B are both dominant both will be expressed if present (AB blood) called codominance
a woman with type A+ blood gives birth to a child with type O- blood her husband who has type B- blood states that the child could not be his. Could he be the father of this child? defend the answer without any additional information we must assume the woman and her husband both possess recessive alleles(i,r)therefore the womans genotype could be IAiRr and her husbands genotype could be IAirr possible phenotypes for the children are -A+AB+AB+AB-O+O-
a woman has type B+ blood and her husband has type A- blood based on this information alone list all possible blood types that their children could have again without and additional information we must assume that the woman and her husband both posses recessive alleles therefore the woman's genotype could be I^biRr and her husbands genotype could be I^Airr the possible phenotypes are A+A-B+B-AB+AB-O+O-
why is it necessary to include 2 controls (positive and negative) in the study the-controlconfirmsthereisnocontaminationinthereagentsinmostexperimentsthisshouldyeildnoproduct(-result)thiscaseitalsoindicateshowmuchjuicecanseperatedfromplainapplesaucethe+controlconfirmsthatthereagentsareworking(resultsfrom2experimentsverydifferent)
what influence did the quantity of enzyme have on the amount of filtrate produced as the amount of enzyme increased so did the amount of product all incubated at 22 degrees C (room temp)
what temp condition was best for the activity of pectinase incubation at 22 degrees C resulted in more product compared to incubation at 4 degrees or 37 degrees C all had 1 mL of normal pectinase
did boiling have any effect on the activity of pectinase yes the results using enzyme that had been previously boiled were considerablyless than the positive control (1.0 mL enzyme at 22 degrees C) and were like the negative control
did freezing have any effect on the activity of pectinase some the results using enzyme that had previously fozen resulted in slightly less product than using the same quantity of normal pectinase both incubated at 22 degrees C
did the lack of the acidic preservative have any effect on the activity of pectinase yes there was way more product compared to the control applesauce both with 1.0 mL enzyme incubated at 22 degrees C
what temperature condition represented above is best for the activity of amylase 37 degrees C because all all of the complex carbohydrates had been digested into disaccarides 22 degrees C is an acceptable temperature condition but not the best because some complex carbohydrates remain
Created by: allijeli
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