F2 Progeny Lab Report

The likelihood of close relatives sharing the same recessive alleles is greater than in the general population, raising the risks that a child would be homozygous recessive for a trait.

What observations can you make regarding the gene pool and gene frequency of the surviving individuals?

---If given traits and parents, be able to use a Punnett square or patterns to predict the probability of offspring for a given cross and express it as a fraction, percent, or ratio.---

Recall from the background information that purple corn kernels are dominant and yellow kernels are recessive. The second ear of corn was the result of crossing two heterozygous ears of male purple corn (Pp x Pp). This is represented by the Punnett square below. Complete the Punnett square by writing the correct letters that correspond to each number indicated in the table. (4 points)

The purpose of Mendelian Genetics: Fast Plant lab is to determine if Mendel’s law of segregation applies to the reproduction of the Brassica Rapa. The law of segregation suggest that allele pairs separate during the production of gametes. Which then the offspring gets one factor from each of the parents. To show this, Mendel suggests that the F2 generation plants will have a three to one ratio between anthocyanin gene (purple) and the absence of the anthocyanin gene (green). The purple stem being the dominant allele and the green stem being the recessive allele. In the lab, we harvest F1 hybrid seeds of the Brassica Rapa and pollinated them so we can have a monohybrid cross between the plants. This will show us the F2 generation of Brassica

parent carried the b allele. The F1 offspring of such a cross would be Bb, and

Imagine that you are crossing two plants that are heterozygous for flower color and seed shape. The dominant and recessive alleles for these traits are:

For one of the monohybrid crosses you performed in this Investigation, describe how to use the phenotype ratios to determine

The “Brassica rapa” is a fast plant known as the field mustard. This plant is well known for its rapid growing rate, which makes it an easy breeding cycle and easy to pollinate. In giving so this makes “Brassica rapa” a great participant for testing Gregor Mendel’s theories of inheritance. The “Brassica rapa” acts like a test subject in testing cross-pollination giving the understanding to the dominant allele of colored stems. There are different colors that are visible on the stem that are above the soil; the colors vary from green to purple. P1 seed was ordered, germinated and cross-pollinated until germination of the next off spring of plants were also done. It was

An expected F2 cross would be as follows: (The expected ratio would be 3:1 in this case – wildtype : mutant).

     There is a chance that there may be different outcomes to the F2 generation due to the possibility that the Line A and the Line B generation were not all homozygous dominant in wing type for females and eye color in males. Such a genotype in the males, vvBb, and such in females, Vvbb, may lead to a different ratio in the F2 offspring.

This lab had 2 exercises. Exercise 9.1 involved observing pictures of 60 F2 offspring and recording the phenotypes for 6 different traits. Exercise 9.2 required us to perform the “chi-square test” to determine whether the data we collected matches the standard Mendelian ratio.

The genotype of the parental genotypes where chosen by looking if there was a difference in the data. If there was a linkage in the sex or in the gender. Since, there were none, the two allele for each mutant was used but for the men we use the apterous to be wild and wild on the eyeless for female. When we perform the dihybrid hypothesis cross shown in Table 4 we obtain a ratio to be 9:3:3:1. The result of the dihybrid cross show that the predicted values where excellent however when the Chi-Square was made the obtain value was high which makes the data not to fall into the

You are also provided with a heterozygous female, and a homozygous recessive male for a genetic cross. In this particular female, all the dominant alleles are on one chromosome, and the recessive counterparts are on the other homologous chromosome. Due to a chromosomal condition, in the female no recombination occurs between the M and N loci. Normal recombination occurs between the L and M loci. Diagram this cross, and show the genotypes and frequencies of all offspring expected from this cross.

Other forms of the genotype, (ygr/ YGR) and (YGR/YGR) will result in green leaves. A third gene in Brassica rapa is the rosette mutant, homozygous recessive. The genotype needed for the short, rosette plant form is (ros/ros). The other two genotypes (ros/ROS) and wild type (ROS/ROS) will result in the normal form of the plant. The phenotypes and genotypes are related in that the phenotypes provide a visible indication of the genotype. This is true in an individual with a homozygous recessive gene. However, in the case of dominant genes, since only one copy is needed for the phenotype to be present, then the second copy is not indicated. The second copy can be identified process where two individuals (P1 and P2) with the same dominant phenotype, called the parental generation, are bred. This produces an F1 or first generation of offspring. The F1 generation can also be bred and produce an F2 generation. Each individual in the F1 and F2 generations receives one copy from each parent of the 3-letter genotype code, called an allele.