Thug Life

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

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

---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.---

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.

_____ In swine, when a pure-breeding red is crossed with a pure-breeding white the F1 are all red. However, the F2 shows 9 red, 1 white and 6 of a new color, sandy. The Sandy phenotype is most likely determined by

It would be expected that the mutant F1 flies would be heterozygous for the allele responsible for the grounded trait. If two F1 flies were mated, the percentage of flies that would be expected to be wildtype in the F2 generation would be 25% mutants given that the mutant allele (ap) is predicted to be recessive and, leaving 75% to be wildtype (ap+).

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.

If 1000 offspring were scored for both characters, what number of offspring would be expected to have each of the four phenotypes (assuming no linkage)? List them below.

the definition of sex-linked genes, and examples of sex-linked genes in humans and other organisms

The results of our monohybrid cross shows that the number of males was higher for having white eyes, however the difference between the recorded number was not significant. So, we cannot definitely say due to sex linked inheritance males should have more white. This result shows our chi square value and we will discuss further what p value that corresponds to that value and what it means to the aforementioned hypothesis and prediction about the cross. The p value we got for X^2-0.362 is between 0.8 and and 0.5 which suggests that the null hypothesis is accepted because we expected males to have more white eye although due to the non-significant difference,the cross is more likely to be a 1:1 ratio which violates mendelian law of equal segregation.

The idea of the project was to experiment breeding Drosophila Melanogaster (fruit fly) to figure out if certain genes of that species were sex linked or not (autosomal). A mono-hybrid cross and di-hybrid cross was performed. For the mono-hybrid cross, white eyed female and red eyed male were placed in one vial for them to reproduce. For the di-hybrid cross, red eyed and normal winged flies and sepia eyed and vestigial winged flies were placed in their vial to reproduce. In the mono-hybrid cross the results expected were within a 1:1:1:1 ratio. Expected results similar to the expected desired null hypothesis proposed with what the F1 parental generation breeds. The potential results would have had

The female subject shows signs of a chromosomal condition that appears to be hereditary through her paternal pedigree. This pedigree shows that her paternal grandfather shows this trait, along with her father. However, this pedigree chart also shows that her grandmother, aunt, uncle, and two brothers show no sign of this trait. Through testing with a pedigree chart, the subject's condition may be narrowed down from certain chromosomes through a process of elimination.

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