worksheet Essay

In this experiment first the stages of an onion cell undergoing mitosis are going to be observed and every stage is going to be detected and drawn on paper. A brief description to what is going on should be attached to the pictures. This is important to understand the basics of cell division which is necessary growth,repair and asexual reproduction. Second the number of cells undergoing each phase is going to be counted to figure out in which phase the cell remains the most. If interphase is the stage in which the cell grows and prepares for cell division then the

This equation is used to calculate the genotype frequency, so 1 = 100% of the population.

The total number of Pago Fuagans that are carriers of the (g) allele and phenotypically normal =

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

9. No, because each child's 2 sets of chromosomes are rearranged independently, so that there is an equal probability of all of them having the same genotype (1/2^4, or 1/16) as there is of them having all different ones

1. Suppose two Dd birds mate. What percentages of DD, Dd, and dd offspring would you predict? Use the Punnett square at right to help determine your answer.

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)

2. How will the alleles for these traits assort into the gametes that each parent might produce? (Hint: For a reminder on how alleles sort independently into gametes, refer to the illustration in Part 2, Question 2, in the Student Guide.)

Genomic DNA is heterogenrous because it shows 2 fragments on 2% agarose gel which come from parents, mom and dad. Moreover, the tandem repeats(n) is within the standard limit (14-41) of heterogenic DNA. So, the sample is heteregenerous.

Now you have determined some facts about the grounded allele and the trait that it causes. Given what you know, do you expect the mutant F1 flies to be homozygous or heterozygous for the allele that causes the grounded trait? According to your reasoning, if you mated two mutant F1 flies, what percentage of flies would you expect to be wild type versus mutant in the F2 progeny? Draw a Punnett square of this cross to justify your answer.

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.

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.

The above Punnett Square demonstrates the inheritance pattern of phenylketonuria in two heterozygous parents. The inheritance pattern in the squares indicates that there are three possible genetic outcomes. The father and the mother supply the offspring with a dominant P gene in the top left quadrant resulting in a homozygous dominant offspring. The offspring in the top left quadrant receives two dominant alleles from both parents and therefore, does not have phenylketonuria. Also, the offsprings in the second and third quadrant receive dominant (P) and recessive (p) genes from both the father and the mother; hence, they have heterozygous genotypes. Lastly, both the father and mother pass on two recessive genes to the offspring in the fourth quadrant resulting in a homozygous recessive child with phenylketonuria.

In a population you will have a portion of alleles that will be categorized by letters for example p has an allele of A1. Then there will be another portion that will be labeled as q which represents the A2 genotype. When these two genotypes mate and reproduce offspring have the potential to have one of the parents genotypes or essentially a combination of their parents genotypes which would look like A1A2. In the big picture the genotypes are recognized as the following A1 A1, A2 A2, or A1 A2. Even though these frequencies don’t change from generation to generation the proportions of the genotype ratio in the populations do change in the generations afterward.

Figure 1 Relationship between the proportion of abnormal sperm and meant heterozygosity. Graph taken from Slate and Pemberton, 2006. Graph A shows the relationship among individual