Homozygous dominant would be all Capital letters.
Homozygous recessive would be all lower case letters.
So...in basic color genetics for horses:
A homozygous recessive horse would be aaee
A homozygous dominant horse would be AAEE
You get one homozygous dominant (TT), one homozygous recessive (tt), and two heterozygous (Tt).
Presuming tall is a dominant allele (the 2nd parent is heterozygous and "tall" is it's phenotype), then the square would be as follows: Let T be the dominant gene for tall, and thus every genotype containing this (TT, or Tt) would produce a tall plant. Let t be recessive, and in the absence of T (tt) causes a plant to be small. So the homozygous tall parent would be TT, and the heterozygous tall parent would be Tt. So now you just have to cross them. During meiosis, every gamete from the TT genotype would contain a T allele. However, for the Tt genotype, there is equal chance each gamete will contain EITHER a T or a t. So the square would be: xxTxxT TxTTxTT txTtxTt As you can see, 2 out of the 4 offspring have a Tt genotype (heterozygous), so this translates to a 1/2 fraction, or 50%. Ignore the Xs in the table- if i just used spaces then all the letters crunched up on top of each other when i pressed save...it was the best i could do...:S
Mendel started out with plants that "bred true". That is, when tall plants were self-pollinated (or cross-pollinated with others like them), plants in following generations were all tall; when the short plants were self-pollinated (or cross- pollinated with others like them) the plants in following generations were all short.
the tall plant must be heterozygous
If two homozygous plants with contrasting traits are crossed, the expected genotypes for the offspring will be heterozygous. The dominant trait would be expressed, but they'd be carriers for the recessive trait.
Using Punnett Squares, you can predict the genotypes and phenotypes of the offspring of a cross between a homozygous (purebred) tall pea plant and a homozygous (purebred) short pea plant.
Using Punnett Squares, you can predict the genotypes and phenotypes of the offspring of a cross between a homozygous (purebred) tall pea plant and a homozygous (purebred) short pea plant.
Using Punnett Squares, you can predict the genotypes and phenotypes of the offspring of a cross between a homozygous (purebred) tall pea plant and a homozygous (purebred) short pea plant.
To determine whether a tall plant is heterozygous, it should be crossed with a homozygous recessive plant (short plant). If any offspring are short, the tall plant must be heterozygous; if all offspring are tall, the tall plant is likely homozygous dominant. This test cross allows for the observation of inheritance patterns in the offspring.
You get one homozygous dominant (TT), one homozygous recessive (tt), and two heterozygous (Tt).
This situation represents Mendel's principle of dominance. In this cross, the tall allele is dominant over the short allele, resulting in all F1 offspring exhibiting the tall phenotype. Since both parent plants are homozygous, the F1 generation inherits one tall allele from the tall parent and one short allele from the short parent, but only the dominant tall trait is expressed.
Homozygous for tall is TT Homozygous for short is tt All F1 offspring from this cross are Tt which makes them genotypically heterozygous and phenotypically tall.
TT for the homozygous tall parent, tt for the homozygous short parent and Tt for the heterozygous offspring.
0 (there is no chance it will be short since tall is dominant over short). Hope this helps! - Biology Student
The offspring will look more like the mom but they will have mid hair.
When a true breeding tall pea plant (homozygous for the tall trait, TT) is crossed with a tall pea plant of unknown genotype, the offspring's phenotypes can help determine the genotype of the second plant. If all offspring are tall, the unknown plant is likely also homozygous tall (TT). However, if some offspring are short, the unknown plant must be heterozygous (Tt), as the short trait (tt) can only appear if the recessive allele is present. In summary, the resulting phenotypes of the offspring will clarify the genotype of the unknown parent.
Presuming tall is a dominant allele (the 2nd parent is heterozygous and "tall" is it's phenotype), then the square would be as follows: Let T be the dominant gene for tall, and thus every genotype containing this (TT, or Tt) would produce a tall plant. Let t be recessive, and in the absence of T (tt) causes a plant to be small. So the homozygous tall parent would be TT, and the heterozygous tall parent would be Tt. So now you just have to cross them. During meiosis, every gamete from the TT genotype would contain a T allele. However, for the Tt genotype, there is equal chance each gamete will contain EITHER a T or a t. So the square would be: xxTxxT TxTTxTT txTtxTt As you can see, 2 out of the 4 offspring have a Tt genotype (heterozygous), so this translates to a 1/2 fraction, or 50%. Ignore the Xs in the table- if i just used spaces then all the letters crunched up on top of each other when i pressed save...it was the best i could do...:S