The Given Information:
Bryant was hired by Dr. Smith to work on a maize genome project. His research was to map three linked genes which can cause brown midrib (bm), virescent seedling (v) and purple aleurone (pr) phenotypes respectively in maize. All the above mentioned alleles are recessive ones compared with the dominant wild type phenotype (designated with +).
Dr. Smith gave Bryant a triple heterozygote strain which has both the recessive mutant and dominant wild type alleles. Bryant was asked to test cross the strain and collect the data from the progeny. Here is the data Bryant obtained after one year's hard work in the field. Based on the phenotype, they are categorized into 8 groups in the following table. This is a multiple answer test.
Table 1:
What I was asked to find:
1) The phenotype of the given strain.
2) The genotype of the given strain.
3) The phenotype of the second strain (not given) used for the test cross.
4) The genotype of the second strain (not given) for the test cross.
5) Which progeny phenotype in the table above results from the non-crossover parental gametes?
6) Which progeny phenotype in the table above results from the single-crossover gametes?
7) Which progeny phenotype in the table above results from the double-crossover gametes?
8) What are the other two alleles on the same homologous chromosome with the wild type bm allele of the heterozygous strain used in this study?
9) What are the other two alleles on the same homologous chromosome with the mutant bm allele of the heterozygous strain used in this study?
10) Identify which gene is located int he middle.
11) What is the distance between the v and pr genes?
12) What is the distance between the pr and bm genes?
13) What is the distance between the bm and v genes?
How I got the answers:
First analyze the given. I have highlighted the given information you must pay attention to.
Bryant was hired by Dr. Smith to work on a maize genome project. His research was to map three linked genes which can cause brown midrib (bm), virescent seedling (v) and purple aleurone (pr) phenotypes respectively in maize. All the above mentioned alleles are recessive ones compared with the dominant wild type phenotype (designated with +).
Dr. Smith gave Bryant a triple heterozygote strain which has both the recessive mutant and dominant wild type alleles. Bryant was asked to test cross the strain and collect the data from the progeny. Here is the data Bryant obtained after one year's hard work in the field. Based on the phenotype, they are categorized into 8 groups in the following table. This is a multiple answer test.
The figure out the genotype of the given strain, this is the answer to question 2, but it will help you find the answer to question 1.
The strain that Dr. Smith gave to Bryant is a triple heterozygote "which has both the recessive mutant and dominant wild type alleles. " therefore using the given letters above "brown midrib (bm), virescent seedling (v) and purple aleurone (pr) " you can write the genotype. Heterozygous involves a dominant and a recessive. In here the dominant is "designated with +". Therefore dominant brown midrib is bm+, dominant virescent seedling is v+ and dominant purple aleurone is pr+. Combine the dominant and recessive of each bm+bmv+vpr+pr and you have the answer to question 2.
2) The genotype of the given strain.
bm+bmv+vpr+pr
Now to answer question 1. Remember that the dominant allele will be expressed over the recessive. Therefore this strain bm+bmv+vpr+pr will be wild type midrib, wild type seedling and wild type aleurone.
1) The phenotype of the given strain.
Wild type midrib, wild type seedling and wild type aleurone.
Now you must know that in order to do a test cross one parent is heterozygous where as the other must be homozygous recessive. Therefore the second strain's genotype is all homozygous recessive like so: bmbmvvprpr.
4) The genotype of the second strain (not given) for the test cross.
bmbmvvprpr.
And the phenotype will be what the recessive alleles express which is part of the given:
brown midrib, virescent seedling and purple aleurone.
3) The phenotype of the second strain (not given) used for the test cross.
brown midrib, virescent seedling and purple aleurone.
Now for question five you have to look at the table given. The trick is the parental non-crossover gametes are always the 2 highest numbers from the data. In the table it is the first 2 (710 and 698). So, Brown midrib and virescent seedling, purple aleurone.
Table 2:
5) Which progeny phenotype in the table above results from the non-crossover parental gametes?
Brown midrib and virescent seedling, purple aleurone.
Also from the table, double crossover are the lowest 2 numbers. In this case the last two on the table (5 and 3) therefore purple aleurone and brown midrib, virescent seeding.
7) Which progeny phenotype in the table above results from the double-crossover gametes?
Purple aleurone and brown midrib, virescent seeding.
Then to find the single crossover gametes it is the 4 numbers in between the parental and double crossover. In this case the 4 middle numbers. (170, 150, 42 and 38). So, Wild type, brown midrib-virescent seedling-purple aleurone, virescent seedling and brown midrib-purple aleurone.
6) Which progeny phenotype in the table above results from the single-crossover gametes?
Wild type, brown midrib-virescent seedling-purple aleurone, virescent seedling and brown midrib-purple aleurone.
Since we already figured out the two strains given we can answer question 8 and 9 from it.
8) What are the other two alleles on the same homologous chromosome with the wild type bm allele of the heterozygous strain used in this study?
In the table the given is all mutant, the not given is assumed to be wild type. The parental with the wild type bm+ allele is the second one (698). This is because the given is all recessive virescent seedling - recessive v and purple aleurone is recessive pr. In the homologous chromosome the other two alleles are v and pr. So the answer to this question is v and pr.
9) What are the other two alleles on the same homologous chromosome with the mutant bm allele of the heterozygous strain used in this study?
The answer to this question is similar to question 8. The parent with the mutant (recessive) bm is the first (710). Brown midrib (bm), assuming wild type for the other two phenotypes we say, wild type seedling (v+) and wild type aleurone(pr+).So, the answer is v+ and pr+.
10) Identify which gene is located in the middle.
Here it gets a little tricky so I will try to explain the best way possible. To figure out which gene goes in the middle you must first figure out the genotype of the parental gametes in the table. The first 2 again. I worked out all the genotypes for easier reference.
Step 1: To work them out just read the phenotype, remember it is recessive because it was given, and fill in the rest as wild type. The wild types are heterozygous because one of the parents will always donate a recessive and if it is wild type it must have a dominant. For example wild type seedling will be v+v and not v+v+ because the second parent is homozygous recessive and so it has not v+ to donate.
Example:
Parental Gamete 1: bmbmv+vpr+pr = bm v+ pr+ (dominant alleles will be picked over recessive)
Parental Gamete 2: bm+bmvvprpr = bm+ v pr
Step 3: Now do the same for the first two single crossed gametes.
Example:
Single crossed gamete 1:bm+bmv+vpr+pr = bm+ v+ pr+
Single crossed gamete 2: bmbmvvprpr = bm v pr
Step 4: Now take a guess as to which gene might be in the middle, I guess the v gene, so layout the parental gamete 1 and 2 like shown below. (from step 2):
 |
| Parental Gametes 1 and 2 |
Step 5: Do the same for the single crossed.
 |
| Single crossed gametes 1 and 2 |
Step 6: The single crossed layout is for reference. The point of this is you will do a single cross on your parental gametes and when you do, if my guess of having the v gene in the middle is correct, the answer should look like the single cross. So, lets do the single cross on the parental.
 |
| Single-crossed parental gametes. |
And its correct. The answer matches the given data.
10) Identify which gene is located in the middle. The v gene goes in the middle.
11) What is the distance between the v and pr genes?
Questions 11, 12 and 13 will be answered in this section.
To calculate the distance between any genes you must first add all the numbers that were given to you in the table 1. So lets add:
710 + 698 + 170 + 150 + 42 + 38 + 5 + 3 = 1,816
Know that the distance is being measured in centimorgans (cm). Also, 1 cm = 1%.
To find the distance between v and pr you must add 42 + 38 + 5 + 3 = 88
Now divide 88 by the total and multiply by a 100.
So, 88/1,816 x 100 = 4.84% (4.84cm)
The answer to question 11 is (4.84cm)
13) What is the distance between the bm and v genes?
Lets do this one first as the answer to 12 is the addition of the answers to 11 and 13.
So to calculate this one just add 170 + 150 + 5 and 3 = 328
Then 328/1,816 x 100 = 18.06% (18.06cm)
The answer to question 13 is (18.06cm)
Finally, to find the answer to question 12, What is the distance between the pr and bm genes?,
add 4.84cm + 18.06cm = 22.90cm
The answer to question 12 is 22.90cm
This is the end of this tutorial. I hope I was as helpful as I wanted to be. if you have any questions feel free to ask in the comments section below. If I made any mistakes feel free to correct me too.
If you need more info on this topic check out the pdfs below:
http://www.austincc.edu/asession/biol2316/CH15sp06.pdf
http://www.csun.edu/~cmalone/pdf360/Ch06-1chi%202pt.pdf
http://www.mhhe.com/biosci/cellmicro/tamarin7/information/tam7ch06.pdf
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