In pea plants round seeds are dominant to wrinkled. So to have round seeds you either will have two dominant alleles (homozygous dominant) or one dominant and one recessive allele (heterozygous)
The round seeds are likely to still have a lot of moisture in them. The wrinkles are what happens when they are dried out for storage until planting in the spring. Also, different varieties may have different looking seeds.
Round is dominant and will be represented by the letter R. A homozygous dominant round pea plant would have the genotype RR.
As stated in mendelian inheritance wrinkled seeds have recessive trait(r),while round seeds are dominant trait(R).
When crosses were made taking into account two dominant charater in a parent with another parent having racessive genes for those characters, in F1 generation all plants had dominant characters but in F2 generation, on self-pollination, the segregation in both the characters were observed independent of each other.
An organism that is heterozygous for two traits means that it contains both the dominant allele and the recessive allele for both of the traits in question. For example, take a plant that produces peas. The gene that produces yellow seeds (denoted Y) is dominant, the gene for green seeds (y) is recessive. The gene that produces round seeds is dominant (R), the gene for wrinkled seeds (r) is recessive. So if this organism was heterozygous for both traits its genotype would be: Yy/ Rr Which means it has the allele for yellow seeds and the allele for green seeds (trait one), and also has the allele for round seeds and the allele for wrinkled seeds (trait two). Its phenotype (the traits it displays) will be yellow round seeds, as these are the dominant traits.
True.
depends on the genetic composition of the parents
He allowed plants whose seeds were round or wrinkled in shape to self pollinate. This trait has two variations-either round or wrinkled seeds.
R represents the dominant round allele, and rrepresents the recessive wrinkled allele. :D
As stated in mendelian inheritance wrinkled seeds have recessive trait(r),while round seeds are dominant trait(R).
Suppose that, in a pea plant, round seeds are dominant over wrinkled seeds in the "texture" gene. If you were to take two plants heterozygous for the texture gene and cross them together to make eight new plants, how many of those plants should express the smooth phenotype? Use a Punnett's Square to determine the results.
Mendel described two seed shapes among the pea seeds in his study: Smooth and Wrinkled.
He allowed plants whose seeds were round or wrinkled in shape to self pollinate. This trait has two variations-either round or wrinkled seeds.
He allowed plants whose seeds were round or wrinkled in shape to self pollinate. This trait has two variations-either round or wrinkled seeds.
It helps if you know how to set up the problem in a Punnett square and look at the results. For now, let's just say that for a dihybrid cross is between a plant with green, wrinkled seeds and one with yellow round seeds. The traits of green (G) is dominant over yellow(g) and round (R) is dominant over wrinkled (r). If the traits were "connected" then the offspring should be one or the other of the parental types. But in reality, you'll get plants with green, smooth seeds as well, since a plant with one "R" allele is all that's necessary to produce a plant with the dominant round seeds. And if the cross was between two heterozygous plants (GgRr x GgRr all with the phenotype of green round seeds), the offspring will be a mix of plants with a ratio of 9 green and round to 3 green and wrinkled, to 3 yellow and smooth, to 1 yellow and wrinkled. So if 1600 offspring were produced, there would be approximately 900 with both characteristics showing the dominant phenotype, 300 wit just one trait being dominant, 300 with the opposite trait being dominant, and 100 with neither dominant characteristic instead of all being the dominant characteristics like the parents. Or 50/50. So this shows that the chromosomes that carry these characteristics are independent of each other.
It helps if you know how to set up the problem in a Punnett square and look at the results. For now, let's just say that for a dihybrid cross is between a plant with green, wrinkled seeds and one with yellow round seeds. The traits of green (G) is dominant over yellow(g) and round (R) is dominant over wrinkled (r). If the traits were "connected" then the offspring should be one or the other of the parental types. But in reality, you'll get plants with green, smooth seeds as well, since a plant with one "R" allele is all that's necessary to produce a plant with the dominant round seeds. And if the cross was between two heterozygous plants (GgRr x GgRr all with the phenotype of green round seeds), the offspring will be a mix of plants with a ratio of 9 green and round to 3 green and wrinkled, to 3 yellow and smooth, to 1 yellow and wrinkled. So if 1600 offspring were produced, there would be approximately 900 with both characteristics showing the dominant phenotype, 300 wit just one trait being dominant, 300 with the opposite trait being dominant, and 100 with neither dominant characteristic instead of all being the dominant characteristics like the parents. Or 50/50. So this shows that the chromosomes that carry these characteristics are independent of each other.
The probability that an offspring will have wrinkled seeds is 2 in 4 or 50%
Ressive genes. Smooth seeds are dominant in the pea plants Mendel used in his experiments.
When crosses were made taking into account two dominant charater in a parent with another parent having racessive genes for those characters, in F1 generation all plants had dominant characters but in F2 generation, on self-pollination, the segregation in both the characters were observed independent of each other.