Official DAT Destroyer Q&A Thread

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densaugeo

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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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But without substrates, the enzymes will not be able to function and thus would no longer be active. Enzyme activity is measured by the rate of reaction, and Vmax is simply the maximum rate at which a reaction can proceed, when all enzymes are saturated with substrate. Hope this helps :)
Thats true, thank you! A very simple question that I somehow got confused by!
 
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Hello Dr Romano,

For Q.77 DAT Destroyer (2017) Orgo Chem...So Im a little confused as to why the aldol addition didnt lead to a condensation? The solution says when in dilute acid or base, do the aldol condensation. However, in this example it says propanal is treated with dilute sodium hydroxide, so wouldnt that be a dilute base? Thanks!
 
Hello Dr Romano,

For Q.77 DAT Destroyer (2017) Orgo Chem...So Im a little confused as to why the aldol addition didnt lead to a condensation? The solution says when in dilute acid or base, do the aldol condensation. However, in this example it says propanal is treated with dilute sodium hydroxide, so wouldnt that be a dilute base? Thanks!
This reaction is indeed the Aldol condensation. Propanal reacts with sodium hydroxide and the aldol condensation results. I am not clear on what you are actually asking. The aldol product is shown and named in choice D. Hope this helps.....
 
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This reaction is indeed the Aldol condensation. Propanal reacts with sodium hydroxide and the aldol condensation results. I am not clear on what you are actually asking. The aldol product is shown and named in choice D. Hope this helps.....
Oh okay, I was just confused why the correct answer choice was the Aldol product rather than the condensation product, but I'm assuming it's because the other choices weere completely wrong?
 
Hello Dr. Romano,
Question 269 on Organic Chemistry Destroyer. Why is the acyl halide added to the amine portion and not the benzene. I thought NH2 would have directed O/P. Thank you!
 
Hello Dr. Romano,
Question 269 on Organic Chemistry Destroyer. Why is the acyl halide added to the amine portion and not the benzene. I thought NH2 would have directed O/P. Thank you!

raysoflight
CAREFUL !!! This reaction is NOT a Friedel Crafts reaction. An NH2 group as well as a meta direction will interfere with a Friedel Crafts alkylation as well as an acylation reaction. The NH2 group is indeed an o/p director BUT not in this reaction. The NH2 group would react with the Lewis acid catalyst and deactivate the molecule. As you can see, no Lewis acid was employed. The acyl halide reacts with the NH2 in a nucleophilic acyl substitution reaction to form the amide as shown. Keep hammering away !

Hope this helps.

Dr. Jim Romano
 
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raysoflight
CAREFUL !!! This reaction is NOT a Friedel Crafts reaction. An NH2 group as well as a meta direction will interfere with a Friedel Crafts alkylation as well as an acylation reaction. The NH2 group is indeed an o/p director BUT not in this reaction. The NH2 group would react with the Lewis acid catalyst and deactivate the molecule. As you can see, no Lewis acid was employed. The acyl halide reacts with the NH2 in a nucleophilic acyl substitution reaction to form the amide as shown. Keep hammering away !

Hope this helps.

Dr. Jim Romano
Oh ok, that makes sense. Thank you!
 
Hello Dr Romano,

For Q189 DAT Destroyer 2017 Biology; if a person if born with a genetic defect that produces abnormal microtubules, why would sperm cells, cells of the larynx and trachea be affected specifically?
Thanks!
 
Hello Dr Romano,

For Q189 DAT Destroyer 2017 Biology; if a person if born with a genetic defect that produces abnormal microtubules, why would sperm cells, cells of the larynx and trachea be affected specifically?
Thanks!
Tissues of the trachea and larynx have cilia, which contain microtubules. Since microtubule polymerization does not occur, those tissues will be affected.

Sperm cells use flagella, which also contain microtubules, therefore lack of microtubule polymerization would make the sperm cell unable to swim towards the egg.

Hope this helps.
 
Tissues of the trachea and larynx have cilia, which contain microtubules. Since microtubule polymerization does not occur, those tissues will be affected.

Sperm cells use flagella, which also contain microtubules, therefore lack of microtubule polymerization would make the sperm cell unable to swim towards the egg.

Hope this helps.
ahhh I see!! I didnt think of that, thank you!!
 
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For GC Destroyer 2017 Q.352: isn't Sr++ bigger than S-- because it is a row down (Sr becomes Kr and S becomes Ar right?), also in the same question, I am confused why answer choice C is true, I understand how London dispersion forces increase with mass but aren't electron masses negligible anyway?
 
For GC Destroyer 2017 Q.352: isn't Sr++ bigger than S-- because it is a row down (Sr becomes Kr and S becomes Ar right?), also in the same question, I am confused why answer choice C is true, I understand how London dispersion forces increase with mass but aren't electron masses negligible anyway?
All non-metals (except for the noble gases which do not form ions) form anions which become larger down a group. For non-metals, a subtle trend of decreasing ionic radii is found across a period . Anions are almost always larger than cations, although there are some exceptions . I looked up the literature values for you....Sr++ is 113 pm and S -2 is 184 pm. A great general rule for the DAT exam is that the more - signs you see,,,,the larger the ion !

Hope this helps.

Dr. Jim Romano
 
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Dr. Romano,

#266 OC.. Can you explain why the answer isn't E? I thought that the methyl hydrogens were unequivalent because they are diastereotopic.
Thanks for you help!
 
Dr. Romano,

#266 OC.. Can you explain why the answer isn't E? I thought that the methyl hydrogens were unequivalent because they are diastereotopic.
Thanks for you help!


@1229ko Hey! I am unsure about what your question is asking, but let me see if I can help you understand the concept behind this question.

The question is obviously testing your knowledge about HNMR's, but what makes this question a bit harder is that it asks for the HNMR signal after the reactions has occurred. Thus, you also have to understand reactions, a radical reaction, in this case. Be aware that Br2/hv is NOT the same as HBR/ROOR. In this radical reaction, it will take places through the most stable radical intermediate, which would be C1 from the benzene (the CH2), because it is secondary. And then you add the Br to that carbon, so 1 of the 2H's is replaced by a Br. Now you go to the secondary part of the question which asks what the HNMR signal would be by the methyl protons. Now I am assuming you understand the basics of HNMR and the 3 bond length rule. There is only 1H left on C1. The you apply the N+1 rule, which gets you the ultimate answer or 1H+1=2, thus a doublet by the methyl protons.

Please let me know if that helps.
 
Dr. Romano,

#266 OC.. Can you explain why the answer isn't E? I thought that the methyl hydrogens were unequivalent because they are diastereotopic.
Thanks for you help!
This is a very easy problem. As shown in the solution, simply remove the benzylic hydrogen and replace it with a Br. Now simply look at the methyl hydrogens.......ALL METHYL HYDROGENS ARE EQUIVALENT>>>>>>>ALWAYS..........they are adjacent to a single hydrogen.....hence the signal is a doublet.

Hope this helps.

Dr. Romano
 
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I have an orgo question on one of the problems from DAT Destroyer.

Can this reaction occur? I was under the impression that SN1 reactions couldn't occur on a primary carbon, but this reaction would suggest that a carbocation formed followed by a hydride shift. Does that mean this is something I should always be looking out for this?
 

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Dr. Romano,

For a SA/WB titration (HCl with NH3), how could we know to choose the methyl red over the methyl yellow?

GC #345
 
I have an orgo question on one of the problems from DAT Destroyer.

Can this reaction occur? I was under the impression that SN1 reactions couldn't occur on a primary carbon, but this reaction would suggest that a carbocation formed followed by a hydride shift. Does that mean this is something I should always be looking out for this?
Usually a rearrangement is not seen on a primary halide. Undergrads think this is definitive rule. Not so ! Look closely, and you will see BRANCHING........thus it is possible that as soon as protonation occurs, our leaving group departs AND a simultaneous shift occurs to generate a stabilized carbocation. It must be simultaneous since the high energy primary carbocation is too energetically unstable. Usually, a primary halide would prefer SN2, but not here. Using SOCl2 or PCl3 would involve a different mechanism and this simultaneous shift would not occur. I have read virtually every Organic textbook in the USA, England, and India.....and only 3 use this type of problem as a classic example. The David Klein Text, Kalsi text, and Clayden text do discuss this. We often refer to it as a neopentyl or neopentyl-like system.

Hope this helps.

Dr. Jim Romano




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Dr. Romano,

For a SA/WB titration (HCl with NH3), how could we know to choose the methyl red over the methyl yellow?

GC #345
Great question.....We choose an indicator that will change color at a pH value that is near the pKa of the indicator,,,,,, HCl is a strong acid,,,,NH3 is a weak base,,,,,,thus in this type of titration, we would see a pH about 5. Now see which one of these indicators have a pKa equal or close to this value. Methyl red has a pKa value VERY close to the Equivalence Point of a Strong Acid vs Strong Base Titration.,,,namely 5. Methyl yellow has a pKa value of 3.3,,,,,,,quite far from the target value of 5. If we titrated HBr with KOH,,,,,which indicator do you think would be best ? I hope you said Bromothymol Blue !

Hope this helps......

Dr. Romano
 
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Great question.....We choose an indicator that will change color at a pH value that is near the pKa of the indicator,,,,,, HCl is a strong acid,,,,NH3 is a weak base,,,,,,thus in this type of titration, we would see a pH about 5. Now see which one of these indicators have a pKa equal or close to this value. Methyl red has a pKa value VERY close to the Equivalence Point of a Strong Acid vs Strong Base Titration.,,,namely 5. Methyl yellow has a pKa value of 3.3,,,,,,,quite far from the target value of 5. If we titrated HBr with KOH,,,,,which indicator do you think would be best ? I hope you said Bromothymol Blue !

Hope this helps......

Dr. Romano

How could we have known that the equivalence pt of HCl/NH3 was around 5 and not... say around 3.5? I was assuming that a sa/wb could easily reach lower...

Thank you!
 
Usually a rearrangement is not seen on a primary halide. Undergrads think this is definitive rule. Not so ! Look closely, and you will see BRANCHING........thus it is possible that as soon as protonation occurs, our leaving group departs AND a simultaneous shift occurs to generate a stabilized carbocation. It must be simultaneous since the high energy primary carbocation is too energetically unstable. Usually, a primary halide would prefer SN2, but not here. Using SOCl2 or PCl3 would involve a different mechanism and this simultaneous shift would not occur. I have read virtually every Organic textbook in the USA, England, and India.....and only 3 use this type of problem as a classic example. The David Klein Text, Kalsi text, and Clayden text do discuss this. We often refer to it as a neopentyl or neopentyl-like system.

Hope this helps.

Dr. Jim Romano




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Wow that's interesting, and I am surprised so few undergraduate textbooks mention it. Thanks for taking the time to answer this question!
 
How could we have known that the equivalence pt of HCl/NH3 was around 5 and not... say around 3.5? I was assuming that a sa/wb could easily reach lower...

Thank you!
Let's review,,,,,,,A strong acid vs a strong base has an Equivalence Point pH of 7..........A weak acid vs a strong base is about 9......a weak base and a strong acid is about 5.....

Hope this helps

Dr. Romano
 
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Wow that's interesting, and I am surprised so few undergraduate textbooks mention it. Thanks for taking the time to answer this question!
Keep up the great work! If you have any more questions I will be happy to answer them.

Dr. Romano
 
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Hello Dr. Romano,
On Question 364 of Chemistry Destroyer, why aren't we excluding the 3d^7 electrons in addition to the 4s^2 electrons? Thank you!
 
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Hello,

Im wondering, for E2 reactions, do they always undergo a antiperiplanar conformation or is it only when theres a strong bulky base ? THanks!
 
Hello,

Im wondering, for E2 reactions, do they always undergo a antiperiplanar conformation or is it only when theres a strong bulky base ? THanks!
If you read most organic chemistry books they will say yes. For your purposes,,,,,,the E2 needs to have the leaving group and the hydrogen in an antiperiplanar position. The base need not be a big bulky base. Sodium ethoxide works fine.....or sodamide, NaNH2. Why ? The anti bonding orbital is perfectly aligned when 180 degrees apart, and this allows for a lower energy transition state. The electrons from the departing hydrogen fill the anti bonding orbital..... HOWEVER,,,,,,,if you have certain bicyclo compounds the 180 degrees needed for E2 departure can NOT be attained,,,,,and we see zero degree departure,,,,,we call this in advanced organic chemistry a syn elimination. For your purposes,,,,,think anti periplanar and you will be set !!!!!

Hope this helps.

Dr. Jim Romano
 
If you read most organic chemistry books they will say yes. For your purposes,,,,,,the E2 needs to have the leaving group and the hydrogen in an antiperiplanar position. The base need not be a big bulky base. Sodium ethoxide works fine.....or sodamide, NaNH2. Why ? The anti bonding orbital is perfectly aligned when 180 degrees apart, and this allows for a lower energy transition state. The electrons from the departing hydrogen fill the anti bonding orbital..... HOWEVER,,,,,,,if you have certain bicyclo compounds the 180 degrees needed for E2 departure can NOT be attained,,,,,and we see zero degree departure,,,,,we call this in advanced organic chemistry a syn elimination. For your purposes,,,,,think anti periplanar and you will be set !!!!!

Hope this helps.

Dr. Jim Romano
Thank you! :)
 
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Hello Dr. Romano,
On Question 364 of Chemistry Destroyer, why aren't we excluding the 3d^7 electrons in addition to the 4s^2 electrons? Thank you!
The effective nuclear charge is the ATOMIC number minus the non valence electrons. Look closely how I drew cobalt out....The valence shell has the LARGEST principle energy number,,,,,,here it is the 4S2 only !!!! With that said,,,,,count up what is left.....and that is the non valence electrons. Thus we see 25 non valence electrons . Now plug it into our formula,,,,,,,,Effective Nuclear Charge = Atomic number - non valence electrons. We get 27 -25 = 2.

I hope this helps.

Dr. Jim Romano
 
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The effective nuclear charge is the ATOMIC number minus the non valence electrons. Look closely how I drew cobalt out....The valence shell has the LARGEST principle energy number,,,,,,here it is the 4S2 only !!!! With that said,,,,,count up what is left.....and that is the non valence electrons. Thus we see 25 non valence electrons . Now plug it into our formula,,,,,,,,Effective Nuclear Charge = Atomic number - non valence electrons. We get 27 -25 = 2.

I hope this helps.

Dr. Jim Romano
Thank you!!
 
Hi,

For Bio Q 576 the answer is that E is false. When I read solutions it sounds to me from the last sentence that E is true! Can you please explain this? Thanks
 
Hi,

For Bio Q 576 the answer is that E is false. When I read solutions it sounds to me from the last sentence that E is true! Can you please explain this? Thanks
Answer is correct.
Liver does not have cilia, so columnar epithelium is less likely. Liver has cuboidal epithelium.
 
I think what you are answering is for number 526. I am wondering about number 576 which asks about Killer T cells in Destroyer 2017. :)

For 576, he is asking for Killer T-Cell. E refers to Helper T Cell, which is specific for the virus that causes HIV (and I am pretty sure that it mainly targets CD4 Help T Cells). Killer and Help T cells, are not he same, despite both being primarily regulated within the cell mediated response.
 
For 576, he is asking for Killer T-Cell. E refers to Helper T Cell, which is specific for the virus that causes HIV (and I am pretty sure that it mainly targets CD4 Help T Cells). Killer and Help T cells, are not he same, despite both being primarily regulated within the cell mediated response.
I thought they were the same. Thanks!
 
I think what you are answering is for number 526. I am wondering about number 576 which asks about Killer T cells in Destroyer 2017. :)

Sorry I looked at the wrong question.

The answer is correct.

The question asks about the Cytotoxic Killer T-cells. Answer choice E talks about Helper T-cells. That is why choice E is false. The target of the AIDS virus is CD-4 cells, a type of a Helper T-cell. They express CD-4 glycoprotein on their surface.

Hope this helps.
 
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Hello I was wondering if someone could help me on this question I came across on gen chem bootcamp. The Q says "With increasing temperatures, which of the following occurs?" the correct answer is 'an increase in entropy'. This is fairly obvious because heating a solid goes to liquid, liquid to gas, so entropy increases. However, I dont understand it in the context of the gibbs free energy equation. Say for example you have a reaction with negative entropy, well if dG= dH-TdS, then wouldnt increasing the temperature make dS (entropy) more negative??
 
Hi I have a question regarding QR destroyer 2017
Question 38.
I don't know why A is the answer, D gives us the answer
Are we to assume that the square root gives only positive values? Because this is the only way A would be true not D
 
Hi I have a question regarding QR destroyer 2017
Question 38.
I don't know why A is the answer, D gives us the answer
Are we to assume that the square root gives only positive values? Because this is the only way A would be true not D
Yes, you should assume that when it's given with the square root sign, you only get positive numbers.
I had the same question lol, you can see Mickey's explanation here: Bootcamp QR confusion ...
Hope this helps :)
 
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Hi Dr.Romano!

For #14 in the OC section of Destroyer 2017, is it possible for a markovnikov addition of an OH to occur at the end after the grignard is added? (since H3O+ is added after the grignard is added to the aldehyde, and the grignard reagent contains a double bond) I was thinking that the would be E because of this.

Thanks!
 
Hello,

Im wondering, for E2 reactions, do they always undergo a antiperiplanar conformation or is it only when theres a strong bulky base ? THanks!

yes, that is one of the calling cards of an E2 reaction. the hydrogen absolutely must be antiperiplanar to the leaving group =). the bulkiness of a base has more to do with the major product you will get, which in this case will be the Hoffman product (due to steric hindrance).
 
Hello I was wondering if someone could help me on this question I came across on gen chem bootcamp. The Q says "With increasing temperatures, which of the following occurs?" the correct answer is 'an increase in entropy'. This is fairly obvious because heating a solid goes to liquid, liquid to gas, so entropy increases. However, I dont understand it in the context of the gibbs free energy equation. Say for example you have a reaction with negative entropy, well if dG= dH-TdS, then wouldnt increasing the temperature make dS (entropy) more negative??

dH and dS, in the context of gibbs free energy, is actually independent of temperature. this quote from khan academy explains: "Although dG is temperature dependent, it's generally okay to assume that the dH and dS values are independent of temperature as long as the reaction does not involve a phase change." so i wouldn't look at it as "increasing temperature is making entropy more negative". it just makes the product of TdS a more positive or negative value (depending on dS), which in turn dictates what dG is (based on dH).

lets dig deeper. for the gibbs free energy equation, lets look at the scenario you proposed:

negative entropy, so dS is negative. therefore, based on the equation, you will actually be adding the product of temperature and entropy to your enthalpy (the double-negatives cancel out), which in turn dictates whether dG is positive or negative. now lets shift to dH. if dH is negative and dS is negative, the temperature has to be LOW for dG to be negative because the addition of the product of TdS to dH CANNOT make dG positive. that is why you commonly see "dH negative, dS negative, spontaneous at low temperature".

hope this helps!
 
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Hi Dr.Romano!

For #14 in the OC section of Destroyer 2017, is it possible for a markovnikov addition of an OH to occur at the end after the grignard is added? (since H3O+ is added after the grignard is added to the aldehyde, and the grignard reagent contains a double bond) I was thinking that the would be E because of this.

Thanks!
Not a chance in Hell. The final step is called a quench.....The intermediate an O-MgBr.......it is in desperate need for a proton to attain stability !!!!!

Hope this helps.

Dr. Romano

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Hello Dr Romano,

for Q.244 of the DAT Destroyer Bio (2017). Im a little confused as to why the probability of daughters with colour blindness is 50%. In the solutions at the back, it seems like 50% of the daughters inherit colour blindness. But since there is already a 1/2 chance of the gender of the child being male or female, wouldnt that be (1/2)*(1/2) -> 25%?
 
Hello Dr Romano,

for Q.244 of the DAT Destroyer Bio (2017). Im a little confused as to why the probability of daughters with colour blindness is 50%. In the solutions at the back, it seems like 50% of the daughters inherit colour blindness. But since there is already a 1/2 chance of the gender of the child being male or female, wouldnt that be (1/2)*(1/2) -> 25%?
Well, you are already given the fact that it is a daughter, so you don't need to include the probability of it being a girl vs. boy. Since that is given, you only look at the genotypes of the daughters, one will be XcX and the other XX, so only 50% of the daughters will be colorblind! Hope this helps :)
 
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Well, you are already given the fact that it is a daughter, so you don't need to include the probability of it being a girl vs. boy. Since that is given, you only look at the genotypes of the daughters, one will be XcX and the other XX, so only 50% of the daughters will be colorblind! Hope this helps :)
hmm okay yeah i understand that, thanks :)! It just cause the question was worded 'What is the probability that the first daughter is colourblind'. I thought it implied that you need to include the probability of the gender being a female first. Ive seen a similar question where it did require calculating that too, so im confused how to recognise the different type of question! Thank
 
hmm okay yeah i understand that, thanks :)! It just cause the question was worded 'What is the probability that the first daughter is colourblind'. I thought it implied that you need to include the probability of the gender being a female first. Ive seen a similar question where it did require calculating that too, so im confused how to recognise the different type of question! Thank
Well, if it says the gender (what is the prob of a daughter, son being colorblind), then you don't include the (1/2) for M or F. But, if the question says what is the probability of a colorblind male? Then you include both the (1/2) for M and the (1/2) for colorblindness. Hope this helps :)
 
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Well, you are already given the fact that it is a daughter, so you don't need to include the probability of it being a girl vs. boy. Since that is given, you only look at the genotypes of the daughters, one will be XcX and the other XX, so only 50% of the daughters will be colorblind! Hope this helps :)
I actually literally just came across another similar question no.276, but this time the probability of the gender is taken into account... :shrug: @orgoman22 what the difference between 244 and 276 - very confused!
 
I actually literally just came across another similar question no.276, but this time the probability of the gender is taken into account... :shrug: @orgoman22 what the difference between 244 and 276 - very confused!
In 244, they are telling you that the first child is a daughter. In 276, they don't know what the first child is, so they want to know how likely it is to have a normal son.
 
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Well, if it says the gender (what is the prob of a daughter, son being colorblind), then you don't include the (1/2) for M or F. But, if the question says what is the probability of a colorblind male? Then you include both the (1/2) for M and the (1/2) for colorblindness. Hope this helps :)
I think i kinda understand that now, thank you :) i guess its based on the specific wording!
 
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