Principles of Biology : Homework #7, Genetics 2

This homework assignment consists of a problem in genetics. It is worth 10 points. See Three Point Mapping handout for help.

Assume: Corn Mendelian genetics where a wild type corn kernel is a yellow kernel, with a matte, or dullish, surface finish and a plump, full kernel body. Recessive mutations of colorless, waxy and shrunken kernels exist. The three genes (color, surface, body) should be coded as follows:

gene
 alleles
 Phenotypes
 Dominant Phenotype
 Dominant Genotype
 Recessive Phenotype
 Recessive Genotype
kernel color
 yellow (wild type), colorless
 either yellow kernels or whiteish
yellow kernels
 YY or Yy
 coloreless kernels
 yy
kernel surface
 matte (wild type), waxy
 matte surface or waxy, shiny surface
 matte surface of kernels
 MM or Mm
 waxy, shiny surface
 mm
kernel body
 full (wild type), shrunken
 either full or shrunken appearance of skin of kernel
 full bodied kernels
 FF or Ff
 shrunken kernels
 ff

Mapping Genes on a Chromosome in Corn

This problem involves three genes on a single chromosome in corn plants. The results obtained from crosses can be used to calculate the frequency of crossovers. Frequencies of crossovers can tell us where genes are located relative to each other. The further apart two genes are, the higher the frequency of crossover. The closer the genes are, the lower the frequency of crossover. By examining crossover frequency, we can start to make a chromosome map. Consider the following:

A wild type corn plant (kernels yellow, matte finish, and full-bodied) is crossed with a strain that is homozygous recessive on three genes : kernel color, kernel surface and kernel body. This strain is phenotypically colorless kernel, a waxy surface and a shrunken body. We can indicate this cross as follows:


P1      YYMMFF         X          yymmff

We can see by inspection that each parent can only produce one type of gamete, and the resultant F1 generation will be heterozygous at all three loci (ie. YyMmFf).

F1   all YyMmFf

When we back cross these F1 individuals with their triply homozygous, recessive parental genotype (a type of "test cross"), we expect to see the P1 phenotypes in equal proportion.

F2 = F1 generation "backcrossed" with parental yyffmm
       YyMmFf       X       yymmff

Normal Punnett Square


 ymf
YMF (remember: they are on one chromosome)
 YyMmFf : yellow, matte, full
ymf (they are on the other chromosome)
 yymmff : colorless, waxy, shrunken

The actual results, presented below, do verify this, but the data also shows the effects of crossing over. Use this "three point cross" data to make a map of this chromosome.
In other words, calculate the map distances between these genes and then place the genes in the correct sequence in a diagram of the chromosome.

F2 generation of a cross between the trihybrid yellow, matte, full (YyMmFf) and the triple, homozygous recessive colorless, waxy, shrunken (yymmff).


Phenotype
 Number
yellow, matte, full
 17,959
colorless, waxy, shrunken
 17,699
yellow, waxy, shrunken
 509
coloreless, matte, full
 524
yellow, waxy, full
 4,455
coloreless, matte, shrunken
 4.654
yellow, matte, shrunken
 20
colorless, waxy, full
 12


Problem:

Given this data, compute the distance, measured in map units, between the three genes and construct a chromosome map that includes the order of these three genes. Explain your calculations and logic. You do not need to summarize the genetics of the genes (as presented here) or include the raw data table. You should include a table that shows your calculations of the various crossover types. You will find the discussion of "three point crosses" on page 245 helpful. Also, see the meiosis lab, especially the mapping of the tan gene in Sordaria. These calculations are similar but you will not divide anything by "2" here as you are instructed to do in the lab.