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
The offspring in the first column both have two dominant T alleles. Therefore, you know that the father (parent on the left hand side) is TT and the mother (parent on the top) has a dominant T in that column too. The offspring in the second column are Tt. The dominant T came from the father, as figured out in the first part of the problem. The recessive allele t came from mother's second empty box.
Cause the offspring wouln't like know which trait comes the parent. In other words the traits are mixed up. :) <3 hope i helped u !!! ---SMILEY <3 :)
The F1 (first filial) generation is always 100% heterozygous.This is because the F1 generation is defined as the offspring of pure-breeding (homozygous) parents carrying different alleles from each other, e.g. AA x aa. All the offspring must be Aa, in other words heterozygous.
The offspring in the F1 generation are heterozygous for the indicated trait.
(I don't know if this is gonna be good?) The flower,Lilly,have carried heredity.
The offspring in the first column both have two dominant T alleles. Therefore, you know that the father (parent on the left hand side) is TT and the mother (parent on the top) has a dominant T in that column too. The offspring in the second column are Tt. The dominant T came from the father, as figured out in the first part of the problem. The recessive allele t came from mother's second empty box.
In genetics a type of grid is used to show the gametes of each parent and their possible offspring and also called the checkerboard is used. The Punnet Square is named after Reginald Crundall Punnett an English geneticist.
To determine the probability of an offspring having the genotype BBBbbb from a specific cross, we need to know the genotypes of the parents involved in the cross. If we assume one parent is homozygous dominant (BBB) and the other is heterozygous (Bb), the resulting offspring will have a probability of 50% for BBB and 50% for Bb, leading to a combined probability of 0% for BBBbbb, as it requires one of each type of allele that cannot be obtained from these parents. Please provide the parental genotypes for a more accurate calculation.
To determine the probability of offspring having wrinkled seeds, we need to know the genetic makeup of the parents. If we assume that wrinkled seeds are a recessive trait (represented by "r") and smooth seeds are dominant (represented by "R"), then the probability depends on the genotypes of the parents. For example, if both parents are heterozygous (Rr), there is a 25% chance that the offspring will inherit the wrinkled seed trait (rr). Without specific parental genotypes, we cannot provide an exact probability.
To determine the probability of offspring having green seeds, you need to know the genetics involved, such as the parental genotypes and whether green seeds are dominant or recessive. For example, if green seeds are recessive (represented by "g") and both parents are heterozygous (Gg), the probability of obtaining green-seeded offspring (gg) is 25%. If you have more specific information about the parental genotypes, I can provide a more precise calculation.
To determine the percentage of offspring with short beaks, you would need to know the genetic traits involved, including whether the trait is dominant or recessive, and the genotypes of the parents. For example, if short beaks are a recessive trait and both parents are heterozygous, 25% of the offspring would typically express the short beak phenotype. Without specific genetic information, it's impossible to give an accurate percentage.
If both parents carry a recessive gene for a particular trait, there is a 25% chance that their offspring will inherit two copies of the recessive gene and exhibit the corresponding trait. This can be determined through genetic Punnett square analysis to predict the probability of different genetic outcomes in offspring.
well i want someone to answer it i dont know it
No because you don’t know what the speed detmerman’s
To determine the probability of an offspring mouse being born with the genotype ff ee, you need to know the genotypes of the parent mice. If both parents are heterozygous (Ff Ee), the probability of producing ff offspring is 1/4 and for ee offspring is also 1/4. To find the probability of the combined genotype ff ee, you multiply these probabilities: (1/4) x (1/4) = 1/16. Thus, there is a 1/16 probability that an offspring mouse will have the genotype ff ee, assuming both parents are Ff Ee.
Yes. It would take too long to explain genotypes and phenotypes, just know that it is possible.
traits are passed from parents to offspring