Essay about Genetics The Inheritance of Color Blindness Worksheet

The hypothesis was that the majority of males would prefer blue, while females were evenly distributed between

4. Clear wing, Black eye, and Hairless (c, b, and h) are linked, recessive traits carried on

Suppose the feather color of a bird is controlled by two alleles, D and d. The D allele results in dark feathers, while the d allele results in lighter feathers.

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)

This Punnet Square represents the F1 offspring breeding with each other to create more offspring. This second set of offspring is the F2 generation. If both parents are heterozygous dominant, then the offspring expected would be: 50% heterozygous dominant, 25% homozygous dominant and 25% homozygous recessive.

Calculate the ratios of the genotypes and phenotypes of the offspring in the F1 generation.

While child 3 cannot be classified as either Mr.Johnson or Mr.Wilson’s child. When looking at the control blood samples, A blood type forms “clumps” when mixed with the A antigen, B blood type “clumps” when mixed with the B antigen, AB blood type “clumps” with both the A and B antigen, and O blood type does not “clump” with either antigen. To the right are the individuals’ sample trays. Looking at each individual’s blood samples, Mr.Johnson had A blood, Mr.Wilson had B blood, Child 1 had A blood, Child 2 had B blood, and Child 3 had O blood. This supports the claim because knowing Mrs.Johnson’s blood type is O means Mrs.Johnson has the genotype OO. Due to the individuals’ blood types it is known that Mr.Johnson has either the genotype AO or AA, Mr.Wilson has either the genotype BO or BB, Child 1 must have the genotype AO because Child 1 must have had an O trait from Mrs.Johnson, Child 2 must have the genotype BO because Child 1 must have had an O trait from Mrs.Johnson, and Child 3 must have the genotype OO because that is the only genotype for type O blood. When looking at the children’s genotypes it can be concluded that Child A must be Mr.Johnson’s child because it inherited an A trait which can only come from Mr.Johnson, Child 2 must be Mr.Wilson’s child because it had to inherit a B trait which can only come from Mr.Wilson, however Child 3 could come from either Mr.Johnson or Mr.Wilson depending on which individual has a genotype that can pass an O trait. It is not possible from this test to tell whether it is Mr.Wilson or Mr.Johnson that passed on the O trait. Every other research group agreed with our claim. Each group had come to the conclusion that Mr.Johnson was the father of one of the three children

Now mate a mutant F1 female fly with a mutant F1 male fly. Out of the 50 F2 progeny, what percentage of flies are wild type and what percentage are mutant

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.

33. What is the phenotype ratio of the offspring in the Punnett square shown in Figure 11-2?

Using the Hardy-Weinberg equation, calculate the predicted genotype frequencies for each population scenario below. Place your calculations and data in the space provided below. Once you have calculated the frequencies, answer the conclusion questions for each one. Please remember that all calculations must be shown for full credit.

From the cross white eye males with wild females, our results were we got both phenotypes in the males as well as the females to be wild type.

This table helps show all the possible genotypes from one set of parents. The table shows that the genotypes purple and starchy are dominant, and the genotypes yellow and sweet are recessive.(stallsmith)

The pairs of alternative traits examined segregated among the progeny of a particular cross, some individuals exhibiting one traits, some the other