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HH and Hh; Hh & Hh;Hh &hh; HH & hh
The possible genotypes HH and Hh. 50% homozygous for hanging earlobes (HH), and 50% heterozygous for hanging earlobes (Hh).
H is a dominant trait for the hair color red. The trait for white hair is recessiv.The parents genotypes are HH x hh.What will the genotype of the offspring be?
What are the possible genotypes if the offspring of the fathers HH for a trait and the mothers hh
The term "hybrid" in biology means one who carries different alleles for the same trait. For example, a hybrid plant's genotype for height is Hh. When crossing two hybrids, we are performing this operation: Hh x Hh. The offsprings' genotypes can vary from: HH Hh hh
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You could try breeding it with a homozygous recessive partner (hh) Lets assume that you breed the original mystery rabbit with an hh recessive partner, and they have 10 offspring. If the original rabbit is homozygous dominant, it would be HH + hh, which would give all 10 the offspring Hh genotypes, which would give them the dominant hair color. If it was heterozygous dominant, it would be Hh + hh, which would lead to either Hh or hh offspring. This means that in theory, 5 would be dominant colored while the other 5 would not be.
More information is needed. The percent of offspring that will display the recessive trait from parents with Hh and HH will be different than the percent of offspring that will display the recessive trait from parents with hh and Hh.
It's obvious from this problem that short hair is dominant in guinea pigs. In that case, if we have 25 out of 100 offspring that have long hair, then there is no doubt that the parents are heterozygous for short hair. If you use the punnet square, and hypothesize that both parents are heterozygous for short hair and long hair, you will get a 25% chance that the offspring will be long haired. So, the answer to this problem is that both parents are heterozygous. Let H = short hair and h = long hair. Hh dam x Hh sire gives us, according to the Punnet Square: 25% HH 50% Hh 25% hh
Heterozygote X HeterozygoteEg.Hh X Hh- produces HH, 2 Hh, hh = 3 different genotypesWhereas:HH X hh- produces all HhAnd HH X HH- produces all HH
Yes to the first one, no to the second one. Firstly, widow's peak is caused by a dominant gene while a straight hairline is caused by a recessive gene. Let 'H' be dominant and 'h' be recessive. So for a dominant gene, the trait still will be expressed whether the genotype of the person is homozygous dominant(HH) or heterozygous(Hh). For a recessive gene however, the trait will only be expressed when the genotype of the person is homozygous recessive (hh). So two people with a widow's peak CAN have a child with a stright hairline, provided both of their genotypes are Hh. This is because by crossing their genotypes, they can have a possibility of having child with the genotype HH(widow's peak), Hh(widow's peak), and also hh(straight hairline). The ratio of these three possibilities however, are 1:2:1. So the odds of having a child with a straight hairline in this case is 1:3. If the two people who have widow's peak have the genotype HH or one of them HH and the other Hh, then the possibility of having a child with a straight hairline is 0. This is because by crossing their genotypes together, the genotype of the child will either be HH for the first case, and HH or Hh for the second case. For two people who have a straight hairline, the genotype of both will definitely be homozygous recessive (hh), thus child will definitely have the recessive gene. Therefore, it is not possible for two people with straight hairline to have a child with widow's peak.
Make a punnett square! 50% Hh 50% hh