Official DAT Destroyer Q&A Thread

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Hi guys, since I'm currently going through the DAT Destroyer and I'm sure many of you guys are as well I felt this thread would be helpful. I usually have various questions while going through Destroyer regarding why a certain answer choice is correct or other times I am in need of a more detailed explanation.

Instead of making multiple threads each time we have questions, I thought we could just post them here and anyone can chime in with an explanation or further clarification.

Please don't post entire questions as that would be a violation of copyright. You can ask specific questions regarding a particular problem in the Destroyers.

Hopefully others find this a useful thread. Thanks!
 
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May I ask these questions please? I am confused about the stroke volume formula. is that (end diastolic volume - end systolic volume) ? Also is there any formula for heart beat? could you please tell for microscope what I should know for the test because there was just one question about microscope on DAT destroyer?
I would greatly appreciate your help.
 
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ROMI15 said:
May I ask these questions please? I am confused about the stroke volume formula. is that (end diastolic volume - end systolic volume) ? Also is there any formula for heart beat? could you please tell for microscope what I should know for the test because there was just one question about microscope on DAT destroyer?
I would greatly appreciate your help.
Click to expand...
That's blood in the ventricles discharged with each contraction*. So at systolic (when ventricles contract) it's minimal and at diastolic (ventricles fill up with blood when ventricles relax) it's maximal. So when you do end diastolic (most blood) - end systolic (least) that's how much blood is pumped per heart beat (stroke volume) from ventricles. CO=SV*HR. Formula for heart rate is just beats per minute. That's why cardiac output is stroke volume (volume pumped per beat) * Heart rate (beats per minute) to be volume pumped per minute. For the microscopes, know all of the major SEM TEM and Light microscopes.
 
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ROMI15 said:
May I ask these questions please? I am confused about the stroke volume formula. is that (end diastolic volume - end systolic volume) ? Also is there any formula for heart beat? could you please tell for microscope what I should know for the test because there was just one question about microscope on DAT destroyer?
I would greatly appreciate your help.
Click to expand...

You can express "Stroke Volume" = (cardiac output)/(heart rate). Or, you can use "Stroke volume" = (end diastolic volume) - (end systolic volume).
You can therefore calculate "heart rate" if you have the value for a "stroke volume" and the value of the "Cardiac Output". You will simply divide "Cardiac output"/"Stroke Volume".

As for the Microscope, I would know the difference between the "Light Microscope", "Electron Microscope" and "Confocal Microscope"

Hope this helps.
 
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orgoman22 said:
You can express "Stroke Volume" = (cardiac output)/(heart rate). Or, you can use "Stroke volume" = (end diastolic volume) - (end systolic volume).
You can therefore calculate "heart rate" if you have the value for a "stroke volume" and the value of the "Cardiac Output". You will simply divide "Cardiac output"/"Stroke Volume".

As for the Microscope, I would know the difference between the "Light Microscope", "Electron Microscope" and "Confocal Microscope"

Hope this helps.
Click to expand...
Thank you Dr. Romano
 
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orgoman22 said:
First split it to get:

Tan(x)/csc (x) + cot(x)/csc(x)

Which is equal to :
[Sin(x)/cos(x)]/[1/sin(x)]+ [cos(x)/sin(x)]/[1/sin (x)] which simplifies to:

Sin^2(x)/cos(x) + cos(x). With the common denominator cos(x) we get:
(Sin^2(x)+cos^2(x))/cos(x) = 1/cos(x) = sec(x)

Hope this helps!



cleardot.gif
Click to expand...

This is about QR in the dat destroyer 146. I dont see how you go from Sin^2(x)/cos(x) + cos(x) to (Sin^2(x)+cos^2(x))/cos(x)

did you multiply by cosx? and if so why is (Sin^2(x)+cos^2(x)) being divided by cos?
 
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OMG wow, i had to rework this problem soooo many times, now I get what you did. You just put a cosx under cosx/1 to make common denominators so thats why you get cos^2x. trickyyyyyyyyyyyyy, i would never have thought to do that on a test haha. On the dat other than just trying to simplify the problem, are there any strategies to go about these kinds of problems, because I feel like there are so many ways to do these. Should I just look at the answer choices and try to get something around those lines?
 
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Can someone explain why distance isn't the round trip but only from NY to LA in question 27 of math test 15?
 
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Average speed = (Total Distance)/(Total Time)

Distance from NY to LA we can term d, so a round trip would be 2d. Total time is 14 hours. So 2d/14 is the answer reduced to d/7
 
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@orgoman22 Hi Dr, Romaro, For question 160 why is the protected carbonyl a o/p director (Does it act more like a methyl after being protected?) ? Other similar groups like carboxylic acids, esters, ketones act as moderate deactivating groups because the oxygen can pull electron density to make the carbon more positive. My guess as to why a protected carbonyl is an o/p director is because that carbon is no longer sp2 and now that its sp3 it cannot lose electron density to be a meta director.
 
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Graffix said:
@orgoman22 Hi Dr, Romaro, For question 160 why is the protected carbonyl a o/p director (Does it act more like a methyl after being protected?) ? Other similar groups like carboxylic acids, esters, ketones act as moderate deactivating groups because the oxygen can pull electron density to make the carbon more positive. My guess as to why a protected carbonyl is an o/p director is because that carbon is no longer sp2 and now that its sp3 it cannot lose electron density to be a meta director.
Click to expand...

The protected carbonyl group will simply act as an OR group......think methoxy !!!!! This is a great way to convert a meta director into a WICKED WICKED para director......after installing the needed functionality, we can treat the compound with acid and reform the carbonyl group. The group is a monster, thus little or no ortho isomer would be made. This is a vital reaction for the synthetic organic chemist. In the mechanism, we are able to stabilize the intermediate arenium ion by electron DONATION from this group. Draw out the intermediate,,,,,there will be a nice electron -donating effect via a hyperconjugative structure,,,,,,

Hope this helps.

Dr. JIm Romano
 
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Dr. Romano, I am still a bit confused. The entire substituent would act as a -OR group? So then how i've drawn the partial charges in the picture below would be incorrect? I am still unsure why its not a deactivating, electron withdrawing group.
 

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Graffix said:
Dr. Romano, I am still a bit confused. The entire substituent would act as a -OR group? So then how i've drawn the partial charges in the picture below would be incorrect? I am still unsure why its not a deactivating, electron withdrawing group.
Click to expand...

I will be back in my office shortly and I will write it out for you. You will LOVE my answer.

Dr. Romano
 
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i
Graffix said:
Dr. Romano, I am still a bit confused. The entire substituent would act as a -OR group? So then how i've drawn the partial charges in the picture below would be incorrect? I am still unsure why its not a deactivating, electron withdrawing group.
Click to expand...

Hope this helps!

Dr. Romano
sdn orgo.jpg
 
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Graffix said:
OH! I see! Thank you for your explanation! Protecting the ketone changes the substituent from an inductive electron withdrawing group to an electron donation group. Fingers crossed I get it on my DAT- ill be ready!
Click to expand...

You have the idea.....In uncompetitive inhibition, this situation is a very rare class of inhibition. An uncompetitive inhibitor binds to the enzyme and enhances the binding of substrate (so reducing Km), but the resultant enzyme-inhibitor-substrate complex only undergoes reaction to form the product slowly, so that Vmax is also reduced.
 
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Bio question #475 in the 2015 edition asks "What is a promoter site?" Why isn't B (a start site for DNA transcription) a correct answer as well? The answer in the back seems to allude that it would be a start site for transcription. Is it that the start site is near the promoter and not actually on the promoter?
 
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soundthealarm said:
Bio question #475 in the 2015 edition asks "What is a promoter site?" Why isn't B (a start site for DNA transcription) a correct answer as well? The answer in the back seems to allude that it would be a start site for transcription. Is it that the start site is near the promoter and not actually on the promoter?
Click to expand...

You seem to be confused about the function of the promoter site. Promoter, is a sequence on the DNA molecule that is present before, or (upstream of the transcription initiation site). Transcription initiation, or the start site, is where transcription begins. Promoter site, is where the RNA polymerase must bind to, in order to determine where the transcription initiation site will be. That is what the Destroyer implied here.

Hope this helps.

Nancy
 
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Dr. Romano (or Nancy, or whoever has the correct answer),

On the 2014 version of DAT Destroyer, problem #45, you wrote that epinephrine binding to its receptor will lead to the conversion of ATP to cAMP. However, why wouldn't glucagon also be a correct answer?
 
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Frank22 said:
Dr. Romano (or Nancy, or whoever has the correct answer),

On the 2014 version of DAT Destroyer, problem #45, you wrote that epinephrine binding to its receptor will lead to the conversion of ATP to cAMP. However, why wouldn't glucagon also be a correct answer?
Click to expand...

Yes, this question needs clarification. Glucagon can also activate adenylyl cyclase and make cAMP.
 
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Hello, Dr. @orgoman22

I had a question regarding the explanation of Organic Chemistry problem #67 on the 2015 version of DAT Destroyer :
It asks which of the following reagents would be best for the reaction of 2- methyl-propanal to 1-Chloro-2-methylpropane. The answer is SOCL2, which I understand why (Doesn't rearrange, no backside attack possible). However, in the explanation it says that with HCl, it would undergo a shift.

I thought primary alcohols (With H-X) undergo SN2 where you can't form a carbocation, and with secondary and tertiary, rearrangements are possible (SN1). Am I just confused atm?
 
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SirT said:
Hello, Dr. @orgoman22

I had a question regarding the explanation of Organic Chemistry problem #67 on the 2015 version of DAT Destroyer :
It asks which of the following reagents would be best for the reaction of 2- methyl-propanal to 1-Chloro-2-methylpropane. The answer is SOCL2, which I understand why (Doesn't rearrange, no backside attack possible). However, in the explanation it says that with HCl, it would undergo a shift.

I thought primary alcohols (With H-X) undergo SN2 where you can't form a carbocation, and with secondary and tertiary, rearrangements are possible (SN1). Am I just confused atm?
Click to expand...

You are indeed A perspicacious student ! Usually, a primary alcohol will not rearrange, and do the SN2 reaction process when treated with reagents such as HCl. HOWEVER,,,,,if the primary alcohol is branched adjacent to the alcohol group....the Sn1 can occur. How ? First you protonate,,,,,as soon as the water leaves, a shift SIMULTANEOUSLY occurs. Most text books never even mention this....but it is wonderfully shown in the 2nd edition book by Dr. David Klein. Professor Paula Bruice also shows this.

Hope this helps.

Dr. Romano
 
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orgoman22 said:
You are indeed A perspicacious student ! Usually, a primary alcohol will not rearrange, and do the SN2 reaction process when treated with reagents such as HCl. HOWEVER,,,,,if the primary alcohol is branched adjacent to the alcohol group....the Sn1 can occur. How ? First you protonate,,,,,as soon as the water leaves, a shift SIMULTANEOUSLY occurs. Most text books never even mention this....but it is wonderfully shown in the 2nd edition book by Dr. David Klein. Professor Paula Bruice also shows this.

Hope this helps.

Dr. Romano
Click to expand...

Awesome. This definitely helps!

Thank you!
 
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Dr. Romano, @orgoman22

When asked about the most acidic hydrogen in a compound, I know that if there was a hydrogen bonded to an oxygen, you would choose that one over the one bonded to a carbon because the negative charge is more stabilized by the more electronegative oxygen. However, if the carbon was resonance stabilized but the oxygen was not, is the most acidic hydrogen still on the oxygen or would it now be on the carbon since it is resonance stabilized?
 
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m145491 said:
Dr. Romano, @orgoman22

When asked about the most acidic hydrogen in a compound, I know that if there was a hydrogen bonded to an oxygen, you would choose that one over the one bonded to a carbon because the negative charge is more stabilized by the more electronegative oxygen. However, if the carbon was resonance stabilized but the oxygen was not, is the most acidic hydrogen still on the oxygen or would it now be on the carbon since it is resonance stabilized?
Click to expand...
I am not clear on your question.....Can you please provide a specific example, and I would be happy to answer this for you.
Dr. Romano
 
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On number 55 in the biology section of the 2015 book, cold blooded is a blanket term that can include poikilotherms, so how does one know the poikilotherms is a better answer than cold blooded when cold blooded animals are also described as "an animal whose body temperature varies with the external environment?"

Thank you
 
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Hey guys!! I hope all is well. I have a question as well. Question# 51 on the QR section of the Destroyer. ( Ruth has gotten 52 hits in 150 at bats. How many hits does she need in her next 100 at bats to have a total avg of .325?)
I did 0.325= (x+52)/250 and got x=29.25. Answer choices are 29,30,31,32 and 33. So according to my work, the answer should be A) 29. However, the book says that it is B) 30. Isn't 29.25 closer to 29 than it is to 30? What I am missing?
 
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Lostmoonofpoosh said:
On number 55 in the biology section of the 2015 book, cold blooded is a blanket term that can include poikilotherms, so how does one know the poikilotherms is a better answer than cold blooded when cold blooded animals are also described as "an animal whose body temperature varies with the external environment?"

Thank you
Click to expand...
Both answers can be correct. However, Poikilotherms is a better term because, those animals can adapt to the temperature of the environment as well as Maintain a constant body temperature if they have been in the constant-temperature environment for a long period of time.
Sometimes they are also referred to Ectotherms or Ectothermic Poikilotherms.

Hope this helps.
 
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FrenchyM.D/D.O said:
Hey guys!! I hope all is well. I have a question as well. Question# 51 on the QR section of the Destroyer. ( Ruth has gotten 52 hits in 150 at bats. How many hits does she need in her next 100 at bats to have a total avg of .325?)
I did 0.325= (x+52)/250 and got x=29.25. Answer choices are 29,30,31,32 and 33. So according to my work, the answer should be A) 29. However, the book says that it is B) 30. Isn't 29.25 closer to 29 than it is to 30? What I am missing?
Click to expand...

The correct answer is 30. If you pick 29 the total average ( hits/ at bat) will be less than 0.325. Therefore we pick the higher number.

This is not a question of rounding up or down depending on the number right after the decimal.

Hope this helps!
 
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orgoman22 said:
The correct answer is 30. If you pick 29 the total average ( hits/ at bat) will be less than 0.325. Therefore we pick the higher number.

This is not a question of rounding up or down depending on the number right after the decimal.

Hope this helps!
Click to expand...

Thank you very much Ms.Steen and Dr.Romano.
 
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Hi Dr. Romano! I was trying to message you privately but I couldn't. Anyway just wanted to say I was one of your students in Summer of 2013 (I was scammed by someone who was teaching DAT course with your material in Houston) and you allowed me to take your course 🙂 Thank you so much!!! I just wanted to say I am attending dental school now after I got the scores that I did from your help. If it wasn't for you I wouldn't have been here. Thanks again!
 
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podYI said:
Orgoman22,

I am having trouble figuring out how
2x=1 was obtained for Destroyer 2015, gchem problem 273
Click to expand...


In the balanced reaction you produce 2 moles of Ammonia. The equilibrium concentration (as stated in the problem) is 2 moles. You have a 2 Liter container and you need to figure out Molarity.
So 2moles/2Liters = 1.

Using the ICE table you obtain 2X = 1.


Hope this helps.
 
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orgoman22 said:
The correct answer is 30. If you pick 29 the total average ( hits/ at bat) will be less than 0.325. Therefore we pick the higher number.

This is not a question of rounding up or down depending on the number right after the decimal.

Hope this helps!
Click to expand...
I'm kinda confused bu this. Aren't you rounding up to the nearest whole number? Because you can't have 29.25 hits?
 
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GC #174

I feel like this is something I should know because I havent seen anyone ask this, but I can't seem to get it.
I get when you balance H's with H+ and o's with H2O for acids.

so Cr2O7(-2) ---> 2Cr3(+) (good)
14H(+) + Cr2O7(-2) ----> 2Cr3(+) + 7H2O (good)

I dont get why it is 6e- when balancing the charges
6e(-) + 14H(+) + Cr2O7(-2) ----> 2Cr3(+) + 7H2O (Whaaaaa....lost me there)

Please help....thanks!!!
 
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Denevistas said:
GC #174

I feel like this is something I should know because I havent seen anyone ask this, but I can't seem to get it.
I get when you balance H's with H+ and o's with H2O for acids.

so Cr2O7(-2) ---> 2Cr3(+) (good)
14H(+) + Cr2O7(-2) ----> 2Cr3(+) + 7H2O (good)

I dont get why it is 6e- when balancing the charges
6e(-) + 14H(+) + Cr2O7(-2) ----> 2Cr3(+) + 7H2O (Whaaaaa....lost me there)

Please help....thanks!!!.
Click to expand...

If you look at the right side of the equation, the charge is 6+, because
(2Cr3+)

On the left side, the charge is 12+. So to equal out the charges, you need to place 6 electrons on the left. This will contribute to 6- charge and you get 6+ on the left side. The charges must be balanced.

Hope this helps.
 
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