Mendelian Genetics

"a" is a dominant mutant allele of the A gene, and "b" is the recessive mutant allele of the B gene. To see the mutant phenotype, one requires either an "a" mutant phenotype or a "b" mutant phenotype (both mutations are not necessary). What would the F2 segregation ratio be (wild type to mutant) of the cross of an aabb mutant with AABB wild type? (Note that Arabidopsis nomenclature is used. Mutant alleles are lower-case whether they are dominant or recessive, wild type alleles are upper-case.)

Epistasis, Pedigree Analysis and the Genetics of Sex

Outline of types of epistasis.

Double Cross-overs

Hi. Can you help me with the following genetics problem, please: "Consider a pair of homologous chromosomes heterozygous for three genes (e.g., ABC/abc) during prophase I of meiosis. Let the sister chromatids of one homolog be numbered 1 and 2 (ABC); and the sister chromatids of the other homolog be numbered 3 and 4 (abc). "Assume a double crossover occurs in this pair of chromosomes that results in chromatids of the genotype Abc, ABC, aBc and abC. If the first crossover (the one between A and B) involves chromatids 1 & 4, which chromatids could be involved in the second crossover?" A. 1 & 2 B. 1 & 3 C. 1 & 4 D. 1 & 3 or 1 & 4 or 2 & 3 or 2 & 4 E. any two of the four chromatids ... click for more

Genetics--map units

Hi. Here's the problem: In Zebrafish, 3 genes have been mapped on the third chromosome, in the order fiz--6 m.u.---myob----14 m.u.-------sufi The wild-type allele of each gene is dominant. A strain that is fiz myob sufi/+++ is testcrossed, and there are 1000 offspring fish. 1. How many fiz + + offspring would you expect to see in these 1000 offspring? 2. How many + + sufi offspring would you expect to see in these 1000 offspring? I got two other questions regarding this problem correct: that there would be 400 wild-type offspring (a parental class) and only 4 "+ myob +" offspring (a double crossover class). But I can't seem to figure these t... click for more

Genetics -- map units

EDIT to the problem below: I just realized that the various numbers of offspring add up to 9900, not 10,000, which is the supposed total. So for problem #2, I randomly added 100 to my recombinant numbers and got the answer, 30.1, which turned out to be correct. But the same thing did not work for problem #1. Adding 100 to the recombinant numbers resulted in an answer of 1.2, which is not correct. Nor is the correct answer 1.1, which would make up the distance between 30.1 -- 29 (the distane from sp-cn minus the distance from px-cn.) So here are my questions about this new twist: 1) Why would 100 offspring be left out of the various classes of offspring. How is this a reflection of what... click for more