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arsen [322]
3 years ago
6

Which non nutritive artificial sweetener should be avoided by someone who has phenylketonuria (pku?)?

Biology
1 answer:
anygoal [31]3 years ago
3 0
Aspartame is very high in the protein, phenylalanine

Individuals with PKU are unable to metabolize this protein from their bodies & it builds up in toxic levels. Anything containing artificial sweeteners, like aspartame, or are very high in phenylalanine should be avoided to minimize the chances of complications.

Hope that helps!
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3 0
3 years ago
In the laboratory, a genetics student crossed flies with normal, long wings to flies with mutant, dumpy wings, which she believe
Marianna [84]

Answer:

a. Chi-Square Value = 0.12

b. Degrees of freedom = 1

c. Reject or fail to reject the hypothesis = fail to reject the hypothesis

Explanation:

<u>Available data:</u>

  • Cross: long wings to flies with mutant, dumpy wings
  • F1) 100% long-winged flies
  • F2)  672 long-winged flies 218 dumpy-winged flies
  • Hypothesis: the dumpy wing is inherited as a recessive trait

Let us name the allele for Long L and the allele for dumpy l

The number of observed individuals:

  • Total number: 890 flies (672 + 218)
  • Long-winged = 672
  • Dump-winged = 218

In the F2 generation, if l was a recessive allele and L a dominant allele, exhibiting complete dominance, we would expect a 3:1 phenotypic ratio.

Cross:          Ll   x   Ll

Gametes)   L   l    L  l

Punnett square)   L     l

                     L    LL    Ll

                      l     lL     ll

F2) 3/4 LL + Ll long-winged individuals

      1/4 ll dump-winged individuals

From this ratio and the total number of individuals in the F2 generation, we can get the expected number of individuals:

4 possible genotypes ------- 890 individuals

3 LL + Ll genotype ------------X = 667.5 long-winged individuals

1 ll genotype ---------------------X = 222.5 dump-winged individuals

 

Chi square = ∑ (O-E)²/E

LL + Ll = (O-E)² /E

LL+ Ll = (672 - 667.5) ² / 667.5

LL + Ll = 0.03

ll= (O-E)²/E        

ll= (218 - 222.5)² / 222.5

ll= 0.09  

  • Chi square= ∑ ((O-E)²/E) = 0.03 + 0.09 = 0.12
  • Freedom Degrees = genotypes - alleles = 3 - 2 = 1
  • Significance level = 0.05    
  • Table value/Critical value = 3.84

0.12 < 3.84 meaning that there is enough evidence to reject the null hypothesis.

Fail to reject the hypothesis.  The genotypes in the population might be in equilibrium H-W. L is the dominant allele, and l is the recessive allele. The dumpy wing is inherited as a recessive trait.

6 0
3 years ago
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