Dictionary » H » Heterozygotes


(noun) Plural form of heterozygote.

See heterozygote for definition and additional information.

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Do chromosomses ever get competitive?

... I will call their protein products "A" and "B". A and B are beneficial on their own, but they interfere with eachother in heterozygotes, so heterozygotes are less fit than homozygotes. Since natural selection can only select against alleles, not genotypes, the heterozygous ...

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by wildfunguy
Sun Sep 29, 2013 2:50 pm
Forum: Cell Biology
Topic: Do chromosomses ever get competitive?
Replies: 9
Views: 11118

mutation-selection balance

... mutation and selection against recessive alleles. The Hardy-Weinberg principle showed that most copies of harmful recessive alleles are found in heterozygotes and are thus protected from the effects of natural selection. We would therefore expect their gene frequencies to be higher than those ...

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by Monco
Mon Mar 04, 2013 10:41 pm
Forum: Genetics
Topic: mutation-selection balance
Replies: 0
Views: 4036

I really need help with the Punnett square & genotypic

If they were all heterozygotes, they must be Rr. That should be easy. Now you have to do Punnett: .... R .. r R . RR . Rr r . Rr . rr thus you see that F1 will be genotypically RR : Rr : rr - 1 : 2 : 1. It should be easy to make phenotypic ...

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by JackBean
Thu Jan 24, 2013 11:23 am
Forum: Genetics
Topic: I really need help with the Punnett square & genotypic
Replies: 4
Views: 5212

Re: Genetics problems

... above, just combined 2. Pure breeding means they are homozygous (thus all progeny of self-breeding has the same phenotype, that's not possible for heterozygotes). Since F1 is tall red, the tall and red alleles should be dominant. Now you have to determine the ratio of F2 and compare with numbers ...

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by JackBean
Fri Jan 18, 2013 9:09 am
Forum: Genetics
Topic: Genetics problems
Replies: 3
Views: 4968

Re: Help with Hardy-Weinberg Equation?

... of the type O allele (q): sqrt (.35) = .592 Therefore, q = .592, and p = .408 (1 - .592) In the Hardy-Weinberg equation, the frequency of heterozygotes is equal to 2pq. Therefore: 2 (.408) (.592) = .483 So, the proportion of the population that are heterozygous carriers is equal to .483 ...

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by knorman
Fri Jun 08, 2012 5:58 am
Forum: Genetics
Topic: Help with Hardy-Weinberg Equation?
Replies: 2
Views: 2537
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