rr, Rr
If ALL offspring are Aa, The parents are AA and aa.
The F1 generation consists of the offspring of a cross between two parents; the F2 generation consists of the offspring of a cross between two individuals in the same F1 generation.
When a red flower crosses with a white flower, it can create a pink flower. Why? Because of incomplete dominance. Incomplete dominance is when the heterozygous phenotype(offspring) is a blend of the two homozygous phenotypes(parents).
1. In the fruit fly Drosophila melanogaster, vestigial wings and hairy body are produced by two recessive genes located on different chromosomes. The normal alleles, long wings and hairless body, are dominant. Give the genotype and phenotype of F1 progeny obtained from a cross between a vestigial-winged, hairy male and a normal, homozygous female. If the F1 from this cross are permitted to mate randomly among themselves, what phenotypic ratio would be expected in the F2generation?
He identified and developed many different true-breeding strains of pea plants, and he studied only those traits that had only two possible outcomes; such as short or tall height, and yellow or green pea color. He also prevented self-fertilization of the pea plants by using a technique called cross-pollination, so that he could control each cross.
The probability that an offspring will have wrinkled seeds is 2 in 4 or 50%
If you cross wwgg x WwGg, each parent contributes one allele for the traits. The genotype for wrinkled (w) and green (g) seeds is wwgg, and the genotype for WwGg is heterozygous for smooth and yellow. Therefore, 0% of the offspring will be wrinkled green.
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.
In a cross between two heterozygous pea plants (Rr x Rr), where "R" represents the dominant allele for smooth seeds and "r" represents the recessive allele for wrinkled seeds, the probability of producing wrinkled seeds (rr) can be determined using a Punnett square. The genotype ratio from this cross is 1 RR : 2 Rr : 1 rr. Therefore, the probability of producing wrinkled seeds (rr) is 1 out of 4, or 25%.
3
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.
The hybrid is the offspring so the probability is 1.
The genotype of the P (parent) generation can be done by crossing an offspring from the F1 (first filial) generation with one of the gametes from the P generation and then calculating the genotypic and phenotypic ratios. Such an experiment is called a back cross
In the cross BB x bb, all offspring in the F1 generation will have the genotype Bb, as they inherit one allele from each parent. None of the offspring will have the same genotype as either parent.
Perform a test cross. Cross the organism with an organism with a homozygous recessive genotype and use the phenotypes of the offspring and a Punnett square to figure out the unknown genotype.
false
half white and half purple