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Ad libitum [116K]
3 years ago
15

Some fruit flies are born with curly wings which makes them unable to fly. It has been shown that curly wings are caused by a do

minant gene which can be represented by the symbol W, while the recessive normal wing gene is represented by the symbol w. If a parent fly with the genotype WW is crossed with a parent of the genotype Ww, what percentage the offspring will have curly wings?
Some fruit flies are born with curly wings which makes them unable to fly. It has been shown that curly wings are caused by a dominant gene which can be represented by the symbol W, while the recessive normal wing gene is represented by the symbol w. If a parent fly with the genotype WW is crossed with a parent of the genotype Ww, what percentage the offspring will have curly wings?
A) 100%
B) 75%
C) 50%

D) 25%

Biology
2 answers:
Sergeeva-Olga [200]3 years ago
8 0
It would be 100%. As you can see in my punnet square, there are no boxes with ww, which means that ll of the flies would have curled wings. (this is because you have to have two recessive genes for that gene to show) 
So it would be A. 100%

Nostrana [21]3 years ago
3 0

Answer:

A

Explanation:

100% is the answer

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The Offspring produced by a cross between two given types of plants can be any of the three genotypes denoted by A, B, and C. A
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Complete question:

The offspring produced between two given types of plants can be any of the three genotypes, denoted by A, B and C. A theoretical model of gene inheritance suggests that the offspring of types A, B and C should be in a 1:2:1 ratio (meaning 25% A, 50% B, and 25% C). For experimental verification, 100 plants are bred by crossing the two given types. Their genetic classifications are recorded in the table below.

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    A             →        18 individuals

    B             →        55 individuals

    C             →        27 individuals

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Use a 0.05 level of significance.

Determine the chi-square test statistic.

Answer:

Do these contradict the genetic model? No, according to the chi-square test, there is not enough evidence to reject the null hypothesis of the population being in equilibrium.  

Explanation:

<u>Available data</u>:

  • Crossed genotypes: two
  • Genotypes among the offspring; Three → A, B, and C
  • Expected phenotypic ratio → 1:2:1
  • Total number of individuals, N = 100
  • A = 18 individuals
  • B = 55 individuals
  • C = 27 individuals

So, let us first state the hypothesis:

  • H₀= the population is equilibrium for this locus → F(A) = 25%,  F(B) = 50%, F(C) = 25%  
  • H₁ = the population is not in equilibrium

Now, let us calculate the number of expected individuals, according to their expected ratio.

4 -------------- 100% -------------100 individuals

1 ---------------  25% -------------X = 25 individuals A

2 --------------  50% -------------X = 50 individuals B

1----------------- 25%--------------X = 25 individuals C

<u>                                                   A                             B                           C</u>

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  • Expected                         25                           50                         25
  • (Obs-Exp)²/Exp                1.96                        0.5                        0.16

<u>(Obs-Exp)²/Exp</u>

A)  (18 - 25)²/25 = 49/25 = 1.96

B)  (55 - 50)² / 50 = 25/50 = 0.5

C)  (27 - 25)²/25 = 4/25 = 0.16

Chi square = X² = Σ(Obs-Exp)²/Exp  

  • ∑ is the sum of the terms
  • O are the Observed individuals: 2 in chamber B, and 18 in chamber A.  
  • E are the Expected individuals: 10 in each chamber  

X² = ∑ ((O-E)²/E) = 1.96 + 0.5 + 0.16 = 2.62

Freedom degrees = 2

Significance level, 5% = 0.05  

Table value/Critical value = 5.99

X² < Critical value

2.62 < 5.99    

<em>These results suggest that there is </em><u><em>not enough evidence to reject</em></u><em> the null hypothesis. We can assume that </em><u><em>the locus under study in this population is in equilibrium H-W.  </em></u>

 

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