best dog kennel in delhi image
A P
Hi i live in Delhi and would like to keep a dog as a pet. Which variety of dog can i keep considering delhi's weather??
Answer
You should seek this information from local sources, as much as possible, since they will be most familiar with your local conditions and with which breeds do well there. I suggest you start with either the Kennel Club of India
www.thekci.org
or Dogs India (www.dogsindia.com)
Good luck!
You should seek this information from local sources, as much as possible, since they will be most familiar with your local conditions and with which breeds do well there. I suggest you start with either the Kennel Club of India
www.thekci.org
or Dogs India (www.dogsindia.com)
Good luck!
Answer the following question for the dog kennel that breed dogs given below:? Labradoodles are heterozygous (?
Lilly
Answer the following question for the dog kennel that breed dogs given below:?
Labradoodles are heterozygous (Aa)
Poodles are homozygous recessive (aa)
Labradors are homozygous dominant (AA)
Answer the following question for the dog kennel that breed dogs given below:
Use this table for parents and next (offspring) generation
Frequency of AA
Frequency of Aa
Frequency of aa
P= frequency of A
Q= frequency of a
What are frequencies of genotypes?
What are frequencies of alleles?
Is this kennel right now in Hardy-Weinberg balance? why or why not?
What will be the frequencies of genotypes in next generation under random breeding conditions and other Hardy-Weinberg assumptions?
If you could show me how to do the calculation in the following problem
5) a kennel which has 50 poodles, 100 labradoodles, and 50 labradors
Answer
I have absolutely no idea what you're even asking, so I'm just going to do my own thing.
> 5) a kennel which has 50 poodles, 100 labradoodles, and 50 labradors
Okay, now, that's really interesting.
Notice that you have 200 dogs total. That might come in handy in some of our calculations.
> Frequency of AA
Well, that's the labradors, so you want labradors as a portion of the total population, so
50 / 200 = 0.25
> Frequency of Aa
Well, that's the labradoodles, so you want labradoodles as a portion of the total population, so
100 / 200 = 0.5
> Frequency of aa
Well, that's the poodles, so you want poodles as a portion of the total population, so
50 / 200 = 0.25
Well okay!
> P= frequency of A
We can kind of streamline that for this problem since both the homozygotes and the heterozygotes have distinctive phenotypes.
So you have 200 dogs and thus 400 alleles 'cause each dog has two.
a poodle has 2 a alleles
a labrador has 2 A alleles
A labradoodle has 1 a allele and 1 A allele
With me so far?
So to get your A allele frequency, count up the number of A alleles in your population and divide by the total number of alleles.
P = (2 * 50 + 1 * 100) / 400 = 0.5
> Q= frequency of a
Do the same analysis we did for the A allele above.
You end up with
Q = (1 * 100 + 2 * 50) / 400 = 0.5
> What are frequencies of genotypes?
> What are frequencies of alleles?
Um, we already did those.
> Is this kennel right now in Hardy-Weinberg balance? why or why not?
Um, not a great question. To be in H-W equilibrium you need a huge population to prevent genetic drift from happening. If you have a huge population, you don't get random mating (you are more likely to marry someone from your home town than you are to marry someone from New Delhi). So no population is ever really in H-W equilibrium in real life.
But that's not what your teacher wants you to do.
Your teacher wants you to plug in the p and q you found above into the Hardy-Weinberg equations.
So let's do that.
p + q = 1
0.5 + 0.5 = 1, Check!
p^2 + 2pq + q^2 = 1
0.25 + 0.5 + 0.25 = 1 Check!
Using these equations, it does indeed appear that your population is in H-W equilibrium for the gene you're interested in.
> What will be the frequencies of genotypes in next generation under random breeding conditions and other Hardy-Weinberg assumptions?
I think you'll still have
25% AA, 50% Aa, and 25% aa
since you know that in H-W equilibrium, allele frequencies do not change in the population, and random mating should give you this. I think.
I have absolutely no idea what you're even asking, so I'm just going to do my own thing.
> 5) a kennel which has 50 poodles, 100 labradoodles, and 50 labradors
Okay, now, that's really interesting.
Notice that you have 200 dogs total. That might come in handy in some of our calculations.
> Frequency of AA
Well, that's the labradors, so you want labradors as a portion of the total population, so
50 / 200 = 0.25
> Frequency of Aa
Well, that's the labradoodles, so you want labradoodles as a portion of the total population, so
100 / 200 = 0.5
> Frequency of aa
Well, that's the poodles, so you want poodles as a portion of the total population, so
50 / 200 = 0.25
Well okay!
> P= frequency of A
We can kind of streamline that for this problem since both the homozygotes and the heterozygotes have distinctive phenotypes.
So you have 200 dogs and thus 400 alleles 'cause each dog has two.
a poodle has 2 a alleles
a labrador has 2 A alleles
A labradoodle has 1 a allele and 1 A allele
With me so far?
So to get your A allele frequency, count up the number of A alleles in your population and divide by the total number of alleles.
P = (2 * 50 + 1 * 100) / 400 = 0.5
> Q= frequency of a
Do the same analysis we did for the A allele above.
You end up with
Q = (1 * 100 + 2 * 50) / 400 = 0.5
> What are frequencies of genotypes?
> What are frequencies of alleles?
Um, we already did those.
> Is this kennel right now in Hardy-Weinberg balance? why or why not?
Um, not a great question. To be in H-W equilibrium you need a huge population to prevent genetic drift from happening. If you have a huge population, you don't get random mating (you are more likely to marry someone from your home town than you are to marry someone from New Delhi). So no population is ever really in H-W equilibrium in real life.
But that's not what your teacher wants you to do.
Your teacher wants you to plug in the p and q you found above into the Hardy-Weinberg equations.
So let's do that.
p + q = 1
0.5 + 0.5 = 1, Check!
p^2 + 2pq + q^2 = 1
0.25 + 0.5 + 0.25 = 1 Check!
Using these equations, it does indeed appear that your population is in H-W equilibrium for the gene you're interested in.
> What will be the frequencies of genotypes in next generation under random breeding conditions and other Hardy-Weinberg assumptions?
I think you'll still have
25% AA, 50% Aa, and 25% aa
since you know that in H-W equilibrium, allele frequencies do not change in the population, and random mating should give you this. I think.
Powered by Yahoo! Answers
Title Post: Which kind of dog can be kept as a pet in Delhi's weather??
Rating: 100% based on 9998 ratings. 5 user reviews.
Author: Unknown
Thanks For Coming To My Blog
Rating: 100% based on 9998 ratings. 5 user reviews.
Author: Unknown
Thanks For Coming To My Blog
Tidak ada komentar:
Posting Komentar