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Consider a cross of beetles with exoskeleton alleles B and b and leg size alleles N and n. The maleparent is heterozygous in both traits. The female parent is homozygous recessive in both traits.a.Write the four letter genotype for each parent.b.Use the Punnet square below to find the possible allele combinations for an offspring.Possible allele combinations from the maleBNBnbNbnPossible allelecombinationsfrom the femalebnBbNnBbnnbbNnbbnnbnBbNBbnnbbNnbbnnbnBbNnBbnnbbNnbbnnbnBbNnBbnnbbNnbbnnc.What is the probability that the offspring will have a genotype of Bbnn?4/16d.Show how statistics could be used to calculate the probability you found in part c withoutusing a Punnett square.

Respuesta :

Answer:

a. Genotype of male parent= BbNn

Genotype of female parent: bbnn

b: Progeny ratio= 1: 1: 1: 1

c: 1/4

Explanation:

a. Genotype of male parent= BbNn

Genotype of female parent: bbnn

b. Gamete formed by the male parent: BN: Bn: bN: bn in 1:1:1:1 ratio.

Gametes formed by female parent: All gametes with "bn" alleles.

c.  The probability that the offspring will have a genotype of Bbnn=

Bb x bb= 1/2 Bb: 1/2 bb

Nn x nn= 1/2 Nn: 1/2 nn

Therefore, probability that the offspring will have a genotype of Bbnn= 1/2 Bb x 1/2 nn= 1/2.

Ver imagen ArnimZola
marianaegarciaperedo

Assuming diallelic genes that assort independently, a) BbNn and bbnn / b) 25% BbNn, 25% Bbnn, 25% bbNn, 25% bbnn / c) genotype Bbnn = 1/4 = 25% / d) 0.25 x 1 = 0.25 = 25%

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Available data:

  • Two diallelic genes, B and N
  • Gene B codes for exoskeleton
  • Gene N codes for leg size
  • B is the dominant allele
  • b is the recessive allele
  • N is the dominant allele
  • n is the recessive allele
  • Male heter0zyg0us for both traits ⇒ BbNn
  • Female h0m0zyg0us recessive for both traits ⇒ bbnn

Cross) male x female

Parentals)        BbNn              x         bbnn

Gametes) BN, Bn, bN, bn            bn, bn, bn, bn

Punnett square)       BN           Bn           bN           bn

                   bn        BbNn      Bbnn       bbNn       bbnn

                   bn        BbNn      Bbnn       bbNn       bbnn

                   bn        BbNn      Bbnn       bbNn       bbnn

                   bn        BbNn      Bbnn       bbNn       bbnn

F1) Genotype

  • 4/16 = 1/4 = 25% BbNn
  • 4/16 = 1/4 = 25% Bbnn
  • 4/16 = 1/4 = 25% bbNn
  • 4/16 = 1/4 = 25% bbnn

Genotypic ratio ⇒ 1:1:1:1

To calculate the probabilities of getting a certain genotype without the Punnett square, we can just perform a multiplication of the probability of these gametes meeting during fertilization. This is,

The male provides four types of gametes, each with the same probabilities of meeting the female gamete.

Gametes:

  • NB = 0.25
  • Nb = 0.25
  • nB = 0.25
  • nb = 0.25

The female can only provide one type of gamete because she is h0m0zyg0us for both traits, so these gametes will have a 100% probability of meeting a male's gamete.

Gamete:

  • nb = 1

To calculate the probabilities of getting the genotype Bbnn, we just need to multiply the probabilities of gamete Bn (male) meeting gamete bn (female).

Bn  x  bn  = 0.25 x 1 = 0.25 = 25%

a) Male ⇒ BbNn

  Female ⇒ bbnn

b) Offspring

  • 25% BbNn
  • 25% Bbnn
  • 25% bbNn
  • 25% bbnn

c)  Probability of getting the genotype Bbnn = 1/4 = 25%

d) Bn  x  bn  = 0.25 x 1 = 0.25 = 25%

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