Teb: Txij li q=0.2, thiab p + q=1, ces p=0.8 (80%). Ntau zaus ntawm cov neeg heterozygous. Teb: Qhov zaus ntawm cov neeg heterozygous yog sib npaug rau 2pq. Hauv qhov no, 2pq sib npaug 0.32, uas txhais tau hais tias qhov zaus ntawm cov tib neeg heterozygous rau cov noob no sib npaug li 32% (piv txwv li 2 (0.8) (0.2)=0.32).
Koj pom qhov zaus ntawm homozygous li cas?
In the equation, p2 sawv cev rau qhov zaus ntawm homozygous genotype AA, q 2sawv cev rau qhov zaus ntawm homozygous genotype aa, thiab 2pq sawv cev rau qhov zaus ntawm heterozygous genotype Aa. Tsis tas li ntawd, cov txiaj ntsig ntawm allele frequencies rau tag nrho cov alleles ntawm qhov chaw yuav tsum yog 1, yog li p + q=1.
Yuav ua li cas koj xam allele frequencies?
Ib qho allele zaus yog xam los ntawm faib cov naj npawb ntawm lub sij hawm qhov kev txaus siab pom nyob rau hauv cov pej xeem los ntawm tag nrho cov ntawv luam ntawm tag nrho cov alleles ntawm cov caj ces locus nyob rau hauv cov pejxeem. Allele frequencies tuaj yeem sawv cev ua tus lej, feem pua, lossis ib feem.
Koj pom li cas Hardy Weinberg allele zaus?
Kev xam cov allelic zaus peb tsuas yog faib cov naj npawb ntawm S lossis F alleles los ntawm tag nrho cov lej: 94/128=0.734=p=zaus ntawm S allele, thiab 34/128=0.266=q=zaus ntawm F allele.
Koj nrhiav homozygous thiabheterozygous?
Yog tias qhov xeem cov qhabnias nyob rau hauv ib qho kev xeeb tub, ces tus niam txiv kab mob yog heterozygous rau cov allele hauv nqe lus nug. Yog hais tias qhov kev xeem hla tau tsuas yog phenotypically dominant xeeb tub, ces tus niam txiv kab mob yog homozygous dominant rau allele nyob rau hauv nqe lus nug.
