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Thepotemich [5.8K]
3 years ago
5

EXAM TOMORROW! PLEASE HELP! The genotype of F1 individuals in a tetrahybrid cross is AaBbCcDd. Assuming independent assortment o

f these four genes, what are the probabilities that F2 offspring would have the following genotypes? Show your work.
(1)aabbccdd
(2) AaBbCcDd
(3) AABBCCDD
(4) AaBBccDd,
Biology
2 answers:
yKpoI14uk [10]3 years ago
4 0

You can use the Punnett square or the shortcut method. Either way, it is given that in any circumstance that the probability of a heterozygote (Aa) is 2/4 or simply 1/2 and the probability of a homozygote (AA) or (aa) is ¼.

a. aabbccdd = (¼)(¼)(¼)(¼) =1/256

b. AaBbCcDd = (1/2) (1/2) (1/2) (1/2) = 1/16

c. AABBCCDD = (¼)(¼)(¼)(¼) = 1/256

<span>d. AaBBccDd =(1/2)(¼)(¼)(1/2) = 1/64</span>

solniwko [45]3 years ago
4 0
Probability of heterozygote is given by 2/4 which is 1/2 (Zz)and Probability of homozygote is given by 1/4 (ZZ or zz)Hence any thing in format AA or aa will be given probability as 1/4 and anything in the format as Aa will be assigned probability as 1/2.Now, 
1) aabbccdd = 1/4*1/4*1/4*1/4= 1/256
2) AaBbCcDd= 1/2*1/2*1/2*1/2= 1/16
3) AABBCCDD= 1/4*1/4*1/4*1/4= 1/256
4) AaBBccDd= 1/2*1/4*1/4*1/2= 1/64
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Explanation:

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Mole (n) = Mass (M) ÷ Molar mass (MM)

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Where atomic mass of C= 12, H=1, O= 16

Molar mass of C6H12O6= 12(6) + 1(12) + 16(6)

= 72 + 12 + 96

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a) In 100g of glucose;

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Mole = 0.56moles

b) In 500g of glucose

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Mole = 2.78moles

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Answer:

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2.

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Explanation:

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