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Question
Given below are some statements related to the dihybrid cross performed by Morgan (considering the body colour and eye colour as the two characters for study). Which of the following would you choose to find out the location of the gene for body colour?
I. Deviation of F2 progeny from the usual 9:3:3:1 ratio
II. Appearance of more parental combinations in F2 generation
III. Appearance of more non-parental combinations in F2 generation
IV. Linked inheritance of body colour and eye colour in Drosophila
Given below are some statements related to the dihybrid cross performed by Morgan (considering the body colour and eye colour as the two characters for study). Which of the following would you choose to find out the location of the gene for body colour?
I. Deviation of F2 progeny from the usual 9:3:3:1 ratio
II. Appearance of more parental combinations in F2 generation
III. Appearance of more non-parental combinations in F2 generation
IV. Linked inheritance of body colour and eye colour in Drosophila
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Solution
The correct option is C I, II, and IV
After conducting monohybrid crosses on Drosophila, Morgan then proceeded to conduct dihybrid cross on them. For this, he chose 2 characters- one of them was eye colour (red colour being dominant over white) and the other was body colour.
In the parental generation, he chose white- eyed yellow bodied female (homozygous recessive for both characters) and red- eyed brown bodied male (dominant for both characters). In the F1 generation he got all brown bodied red eyed females and white eyed yellow bodied males. When he selfed the F1 progenies, he obtained both parental combinations as well as non-parental combinations. But the parental combinations were much greater in percentage than the non parental types. The phenotypes were not in the ratio of 9:3:3:1.
This shows some kind of physical association between the two genes because of which they are largely being inherited together with a little chance of them being inherited separately. This physical association of two genes is termed as “linkage”.
This means that the two genes must be located on the same chromosome to be treated as a unit and inherited together.
From Morgan’s monohybrid cross, it was established that the gene for eye colour is located on the X chromosome. This brings us to the conclusion that the gene for body colour is also located on the X chromosome in Drosophila.
Later, Morgan observed that the genes of both the characters which he chose for his experiment were located on the same chromosome i.e. X chromosome.
After conducting monohybrid crosses on Drosophila, Morgan then proceeded to conduct dihybrid cross on them. For this, he chose 2 characters- one of them was eye colour (red colour being dominant over white) and the other was body colour.
In the parental generation, he chose white- eyed yellow bodied female (homozygous recessive for both characters) and red- eyed brown bodied male (dominant for both characters). In the F1 generation he got all brown bodied red eyed females and white eyed yellow bodied males. When he selfed the F1 progenies, he obtained both parental combinations as well as non-parental combinations. But the parental combinations were much greater in percentage than the non parental types. The phenotypes were not in the ratio of 9:3:3:1.This shows some kind of physical association between the two genes because of which they are largely being inherited together with a little chance of them being inherited separately. This physical association of two genes is termed as “linkage”.
This means that the two genes must be located on the same chromosome to be treated as a unit and inherited together.
From Morgan’s monohybrid cross, it was established that the gene for eye colour is located on the X chromosome. This brings us to the conclusion that the gene for body colour is also located on the X chromosome in Drosophila.
Later, Morgan observed that the genes of both the characters which he chose for his experiment were located on the same chromosome i.e. X chromosome.
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