No, a heterozygous genotype, which consists of one dominant and one recessive allele (e.g., Aa), will typically express the dominant trait, not the recessive one. However, if both parents are heterozygous (Aa), there's a possibility for offspring to inherit two recessive alleles (aa), which would express the recessive trait. Thus, while a heterozygous individual cannot express a recessive trait, such traits can appear in their offspring if the right allele combinations occur.
The genotype of the offspring with yellow pods is likely homozygous dominant (YY) or heterozygous (Yy), assuming yellow pods are dominant over green. The possible genotypes of the offspring with green pods would be homozygous recessive (yy), as green is the recessive trait. If both yellow-podded parents are heterozygous (Yy), some offspring could also be yellow (YY or Yy) while others could be green (yy).
If both parents have the same phenotype, but the offspring did not share that phenotype, then it is likely that the parents have a dominant phenotype, but the offspring has a recessive phenotype, which means that the offpring's genotype would be homozygous recessive, and it's parents' genotypes would be heterozygous. For example, the parents may both have the genotype Bb, which gives them black fur. Approximately 25% of their offspring should have the genotype bb, which gives them the phenotype of white fur.
If the gene for a trait has two alleles, one dominant (D) and one recessive (d) there are three possible combinations in the genotype: DD (homozygous dominant) Dd (heterozygous) dd (homozygous recessive)
In a cross between Parent 1 (Tt) and Parent 2 (tt), the possible genotypes of the offspring are Tt and tt. The Tt offspring will be heterozygous and display the dominant trait, while the tt offspring will be homozygous recessive and display the recessive trait. There is a 50% chance (2 out of 4 possibilities) that the offspring will be tt and show the recessive trait. Therefore, 50% of the offspring will display the recessive trait.
The probable phenotype of the offspring would be black, as black color is dominant over chestnut color. The genotype of the offspring would be heterozygous for black (Bb), since one parent is homozygous black (BB) and the other is homozygous chestnut (bb).
The genotype of the offspring with yellow pods is likely homozygous dominant (YY) or heterozygous (Yy), assuming yellow pods are dominant over green. The possible genotypes of the offspring with green pods would be homozygous recessive (yy), as green is the recessive trait. If both yellow-podded parents are heterozygous (Yy), some offspring could also be yellow (YY or Yy) while others could be green (yy).
In a cross between a homozygous recessive parent (AA) and a heterozygous parent (Aa), the possible genotypes of the offspring are 50% homozygous recessive (AA) and 50% heterozygous (Aa). Therefore, the probability that an offspring will be homozygous recessive is 50%.
If they both are heterozygous and the recessive gene is blue it can happen
If both parents have the same phenotype, but the offspring did not share that phenotype, then it is likely that the parents have a dominant phenotype, but the offspring has a recessive phenotype, which means that the offpring's genotype would be homozygous recessive, and it's parents' genotypes would be heterozygous. For example, the parents may both have the genotype Bb, which gives them black fur. Approximately 25% of their offspring should have the genotype bb, which gives them the phenotype of white fur.
They perform a test cross. A test cross takes the unknown genotype and crosses it with a known homozygous recessive. If the F1 generation is all dominant, then they know the organism was a homozygous recessive. If recessive offspring appear, then the organism was a heterozygote. As an example, consider a gene with two alleles, A and a, with Adominant. Now consider the test cross. The unknown genotype can only be one of two possibilities: AA (homozgous dominant) Aa (heterozygous) In a test cross,the unknown genotype is crossed with a known homozygous recessive. Since there are only two possible unknown genotypes, there can be only two possible results. First, consider the case of the unknown genotype being a homozygous dominant. The cross looks like this: AA X aa Remember that a homozygote for an allele can only produce one kind of gamete. In this case the homozygous dominant can only produce gametes with the allele A in them, while the homozygous recessive can only produce gametes with the recessive allele a in them. This means the F1 offspring can only be ONE genotype; Aa. Therefore, all of the offspring would have the dominant phenotype. Now consider the other possible cross, where the unknown genotype is heterozygous: AaX aa Remember that a heterozygote can produce two types of gametes. In this case, the unknown would produce gametes with the dominant allele A or the recessive allele a. The homozygous recessive would still only produce one kind gamete, with the recessive a allele. Therefore, we expect to see only two genotypes in the F1, Aa and aa, in equal proportions. In either case, only one test cross is needed to tell one immediately the nature of the unknown genotype. If all of the F1 are of the dominant phenotype, then the unknown genotype must be homozygous dominant; if a mixture of phenotypes appears in equal proportion, then the unknown genotype must be a heterozygote.
depends how it self fertilises. If the gametes are produced with random genes in them the offspring could have recessive charateristics not displayed on the parent. However it could only have charateristics that were present in the parents genotype to begin with. If the offspring has the same genotype as tha parent then it would be the exact same.
So, if one parent is Aa (heterozygous) and the other parent is aa (homozygous recessive) the punnett square would look like this: ___|_A__|__a_ _a_|_Aa_|_aa_ _a_|_Aa_|_aa_ The genotypes of the offspring 50% heterozygous and 50% homozygous recessive
Both of the parents were heterozygous with the blonde hair allele, which is recessive. When there are two parents that are heterozygous, there is a 25% chance their offspring will get two of the recessive alleles. A punnett square can be useful when determining the different phenotypes and genotypes possible in offspring
50%. There are four possible outcomes of the cross:dominant trait from "dad", dominant trait from "mom"recessive trait from "dad", dominant trait from "mom"dominant trait from "dad", recessive trait from "mom"recessive trait from "dad", recessive trait from "mom"Therefore, to get hybrid offspring (one dominant, one recessive), you have a 2 out of 4 chance.
If the gene for a trait has two alleles, one dominant (D) and one recessive (d) there are three possible combinations in the genotype: DD (homozygous dominant) Dd (heterozygous) dd (homozygous recessive)
that is what im figuring out, sorry :(
In a heterozygous cross (e.g., Aa x Aa), the possible genotypes of the offspring are AA, Aa, and aa. The probability of having two offspring with the same genotype can be calculated as follows: the probabilities of each genotype are 1/4 for AA, 1/2 for Aa, and 1/4 for aa. Thus, the probability that both offspring have the same genotype is the sum of the probabilities of each genotype occurring twice: (1/4 * 1/4) + (1/2 * 1/2) + (1/4 * 1/4) = 1/16 + 1/4 + 1/16 = 5/16. Therefore, there is a 5/16 chance that both offspring will have the same genotype.