at 40:24, you mentioned how D. polylepis is the most similar of all the other species but has the greatest amount of difference between the rest, so should it be the most different of all the other species or most unrelated?
Im kind of confused on your hardy weinberg math, to find the allele frequency of q, cant you just square root q^2, so you would do sqrt of .5, which is .707, and then subtract that from one to get allele frquency of p, or am i doing something wrong.
Yes. Recall for the counting alleles method that you need to count the alleles to get the p and q. If you are referring to using the Hardy-Weinberg equilibrium equations, I believe I noticed the error while we were live and explained what to do but I didn’t fix the calculation on the chart since we were live,
Would you be able to go into more detail with why we have to know that extra formula for Hardy-Weinburg? I could not fully grasp why taking the sqaure root of p^2, etc would not give us the right allele frequencies. I know you meant to, and the diagram was not right.
Usually when you are given the dominant phenotype, that is p^2 + 2pq so you can’t just do the square root of that. If you have p^2 theoretically you could solve for p from that. I find that it’s best to tell you to go straight to q^2 as students make the least errors.
The only other part I am not understanding is that in the Hardy-Weinberg Equation video in the first example we were able to take the square root of q^2 to find q and then then other variables. Is the difference in the number of phenotypes that are present? In the sense where there was the dominant rough feathers and only the recessive smooth feathers vs. in this video we have 3 distinct typings ex: white, red, and pink? So is the difference based on the type of inheritance? Mendelian dominant or recessive vs. codominance and incomplete dominance. I hope my question makes sense and thank you for the information and all of the materials you provide. I just wanted clarity because the exam is so soon! :)
@@dennisont if you are given the number of each of the genotypes, you do not need to estimate, you know the values for all do the variables (counting allele method). If they do not give them all, you have to estimate hence your H-W formula. It’s an estimation tool.
For the example of hardy weinberg I don’t understand how u knew if the 200 300 or 500 of the flowers should be p or q or pq^2, when it wasn’t indicated in the problem.
The formula on the formula sheet said p is allele 1 and q is allele 2. I am arbitrarily calling red allele 1 and white allele 2 (if I am remembering the question correctly). Since it tells me the red flower (and this is incomplete dominance) I knew that was homozygous for allele 1 (or p^2). Same logic for the white flower was homozygous for allele 2 (or q^2). Incomplete dominance means that red allele and white allele (heterozygous) is pink so the pink is 2pq. Hope that helps 🫶🏻
the explanation on the error bars was so helpful! thank you!!
Happy to help! Best wishes studying for May 16th ❤️🤓🐧
18:30 HAHHAHAHA I LOVE YOU SO MUCH. i really hope i do well this thursday‼️‼️ 😭😭😭
I’m confused for 37:30 , since X and Y both have similarities in 4 different traits (not having them) wouldn’t they be the most closely related?
You are looking at most closely related. Having traits is better than not.
at 40:24, you mentioned how D. polylepis is the most similar of all the other species but has the greatest amount of difference between the rest, so should it be the most different of all the other species or most unrelated?
😳 I misspoke. They are LEAST similar because they have MOST differences. Sorry thanks for catching my mistake. ❤️🤓🐧
Im kind of confused on your hardy weinberg math, to find the allele frequency of q, cant you just square root q^2, so you would do sqrt of .5, which is .707, and then subtract that from one to get allele frquency of p, or am i doing something wrong.
Yes. Recall for the counting alleles method that you need to count the alleles to get the p and q. If you are referring to using the Hardy-Weinberg equilibrium equations, I believe I noticed the error while we were live and explained what to do but I didn’t fix the calculation on the chart since we were live,
IM COOKED 😭
You got this! I believe in you,
where are the powerpoints you use in the videos on the website i cant find them?
nvm i foun them
Would you be able to go into more detail with why we have to know that extra formula for Hardy-Weinburg? I could not fully grasp why taking the sqaure root of p^2, etc would not give us the right allele frequencies. I know you meant to, and the diagram was not right.
Counting alleles method: th-cam.com/video/1HUw_oZr6k8/w-d-xo.htmlsi=TpiMCScOzIo5Xheq
Equilibrium Equation: th-cam.com/video/n8jWmhm40xM/w-d-xo.htmlsi=ydbBJcGQ0natBbgO
Usually when you are given the dominant phenotype, that is p^2 + 2pq so you can’t just do the square root of that. If you have p^2 theoretically you could solve for p from that. I find that it’s best to tell you to go straight to q^2 as students make the least errors.
The only other part I am not understanding is that in the Hardy-Weinberg Equation video in the first example we were able to take the square root of q^2 to find q and then then other variables. Is the difference in the number of phenotypes that are present? In the sense where there was the dominant rough feathers and only the recessive smooth feathers vs. in this video we have 3 distinct typings ex: white, red, and pink? So is the difference based on the type of inheritance? Mendelian dominant or recessive vs. codominance and incomplete dominance. I hope my question makes sense and thank you for the information and all of the materials you provide. I just wanted clarity because the exam is so soon! :)
@@dennisont if you are given the number of each of the genotypes, you do not need to estimate, you know the values for all do the variables (counting allele method). If they do not give them all, you have to estimate hence your H-W formula. It’s an estimation tool.
For the example of hardy weinberg I don’t understand how u knew if the 200 300 or 500 of the flowers should be p or q or pq^2, when it wasn’t indicated in the problem.
The formula on the formula sheet said p is allele 1 and q is allele 2. I am arbitrarily calling red allele 1 and white allele 2 (if I am remembering the question correctly). Since it tells me the red flower (and this is incomplete dominance) I knew that was homozygous for allele 1 (or p^2). Same logic for the white flower was homozygous for allele 2 (or q^2). Incomplete dominance means that red allele and white allele (heterozygous) is pink so the pink is 2pq. Hope that helps 🫶🏻