# Problems with a Single Chromosome

1) The genes GHI are on a single chromosome in the following order:

---G--------H-------------------------------------- I ------------------

5 mu 15 mu

The distance between G and H is 5 mu and the distance between genes H an I is 15 mu. What are the numbers of various progeny (assume a total of 4000 progeny are measured) in a GHI x ghi cross when:

1) c.o.c. = 1.0

2) c.o.c.= 1.5

3) c.o.c. = 0.5

Thank you! I really appreciate it!!

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#### Solution Preview

Hello

I have walked you step by step to show you how to answer this question. I have included the calculations as well as explanation about why the calculations are being done. I have completed done the first question for you to show you how to calculate these types of problems. I gave you advice on how to do questions 2 and 3 however I have not done those full calculations out so that you can complete them yourself. This will let you know if you understand the concept.

To answer this question you must understand a few concepts first. The first is to remember that the map units (mu) between two genes is equivalent to the recombination frequency or the percent of time there will be crossing over between two genes. The second thing to understand is that the coc stands for coefficient of coincidence. The coefficient of coincidence can be calculated by the formula c.o.c = actual double crossover frequency / expected double crossover frequency.

The question provides a map of three genes and then asks how many offspring of different types will be produced from a GHI x ghi cross. First off I will assume that the GHI individual is heterozygous so the two chromosomes found in that organism are really GHI/ghi which gives it a phenotype of GHI. If this is the case then there will be 8 different offspring types that could be produced. Those are the two parental types GHI and ghi and then 6 different recombinant types. If you have crossover between the G and H you will produce offspring that are Ghi and gHI, if there is ...

#### Solution Summary

This solution helps answer problems regarding a single chromosomes.