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The possible genotypes HH and Hh. 50% homozygous for hanging earlobes (HH), and 50% heterozygous for hanging earlobes (Hh).
if a man were non- hemophiliac and he marries a woman whois homozygus for nan-hemophilia,give the possible genotypes of the children
Haemophilia is a recessive, X-based disorder. The woman in your question is a carrier, meaning she has the defective gene, but isn't bothered by it. Therefor, the woman is of the type 'Xx'. The man is of the type 'XY', not carrying the defective gene. Their children can then be: XX, xX, XY, xY. This means that their daughters won't be affected by it, but might carry it, and their sons either not carry it at all, or carry it and be haemophilic.
Tongue rolling is homozygous dominant and all issue from this pairing will be tongue rollers. This is the only result that is allowable with a standard Punnett square or branch diagram representation. T = tongue roller t = non-tongue roller TT X tt = 4 Tt ======With tongue rolling expressed.
The woman could be AA or AO and the man could be BB or BO. The children would all be AB.
Yes. It's a matter of phenotypes, not genotypes.
CC is the phenotype for sickle cell anemia. They can also have SS or SC genotypes. These people can marry only some other genotypes and not have their children develop the disease.
reguardless of genetype, that's just not right. Genotypes are not pertinent in a same-sex relationship .
A wise man knows when to hold his tongue.
The front man is the vocalist
because the man doesnt want a tongue on his body but using it to insert in her party box .
She wants to lick him
homer (: ?
The possible genotypes HH and Hh. 50% homozygous for hanging earlobes (HH), and 50% heterozygous for hanging earlobes (Hh).
to rev roll to the platform
by giving him lots of tongue
the man you had last night