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Llana [10]
3 years ago
8

Your biology teacher has asked you to do the impossible! You must measure the diameter of one algal cell. You are about to turn

in a blank lab report when your friend Mel comes along and says it is an easy task. First he shows you how to compute the diameter of your field of view. At 40X, your field of view is .50 mm or 500 um. If you look carefully you count 15 algal cells spanning the field of view.
The diameter of one algal cell is approximately

A) 0.03um
B)1.50um
C)3.00um
D)33.3um
Biology
2 answers:
Tresset [83]3 years ago
7 0
C.
It is d, becuase you divide 500 by 15, and that is the closest answer
fiasKO [112]3 years ago
7 0

the correct answer is D

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Complete question:

A geneticist crossed fruit flies to determine the phenotypic ratio. The geneticist crossed a fly with blistery wings and spineless bristles (bbss) with a heterozygous fly that had normal wings and normal bristles (BbSs). Which proportion of offspring that are dominant for both traits in would you not expect based on Mendel's law of independent assortment? 1/2 , 4/16, 25% , or 1/4

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<u>Available data</u>:

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Let us say that:

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Parentals)        bbss       x        BbSs

Gametes)  bs, bs, bs, bs     BS, Bs, bS, bs

Punnett square)    BS        Bs         bS        bs

                     bs    BbSs    Bbss     bbSs    bbss

                     bs    BbSs    Bbss     bbSs    bbss

                     bs    BbSs    Bbss     bbSs    bbss

                     bs    BbSs    Bbss     bbSs    bbss

F1)  4/16 = 1/4 = 25%  of the progeny is expected to be BbSs, dyhibrid individuals, expressing normal wings and normal bristles

     4/16 = 1/4 = 25% of the progeny is expected to be Bbss, expressing normal wings and spineless bristles

     4/16 = 1/4 = 25% of the progeny is expected to be bbSs, expressing  blistery wings and normal bristles

     4/16 = 1/4 = 25% of the progeny is expected to be bbss, expressing  blistery wings and spineless bristles    

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