Biology Genetics: Mapping and Recombinant Frequency

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The discussion focuses on a genetic cross between two homozygous plants resulting in an F1 generation of yellow, long-stemmed offspring. The F2 generation displays a specific phenotypic ratio, leading to the calculation of recombinant frequency. The calculated recombinant frequency is 14%, indicating that the genes are linked and are 14 map units apart. There is clarification about including all progeny in the denominator for accurate calculations. The conclusion confirms that the genes are indeed linked based on the observed data.
shawonna23
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Homework Statement



Two homozygous plants are crossed and all offspring (F1) are yellow with long stems. A cross of the offspring (F1)produces the following phenotypes in the F2 generation:

1,263 yellow, long stem
1,196 white, short stem
205 yellow, short stem
195 white, long stem

What information can be determined about the genes involved in the cross?

Choose one answer.
a. The genes are linked and are 14 map units apart.
b. The genes are linked but you can tell how far apart they are from each other.
c. The genes are linked and are 7 map units apart.
d. The genes are not linked.



Homework Equations





The Attempt at a Solution



1263 and 1196 are the parental type

205 and 195 are the recombinant offspring

Recombinant frequency= (205+195) / (1263+1196) * 100= 14%

the genes are linked and are 14 map units apart
 
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Are you sure only the # of parental's goes in the denominator? :wink: Be careful.
Your final answer is correct.
 
Thanks...i actually forgot to put all of the progeny in the denominator...(205+195+1263+1196)
 

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