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jenyasd209 [6]
2 years ago
13

The frequency of sickle-cell anemia, an autosomal recessive condition, is 0.007. Assuming that the population is in H-W equilibr

ium, what is the frequency of the recessive allele that causes the condition? Enter your answer to the nearest 0.001.
Biology
1 answer:
damaskus [11]2 years ago
5 0

Answer:

The correct answer is - 0.084

Explanation:

Sickle cell anemia is a autsmal recessive blood disorder of RBCs where RBC shape altered to the sickle shape or cresecent moon that obstruct and damage the veins and artries.

Normal allele is represented by - HbA - p

sickel cell allele is represented by - HbS - q

genotype frequency of normal allele HbA in the population = p2

genptype frequency of HbAHbS in the population = 2pq

genotype frequency of autosomal recessive HbS = q2 = 0.007 (given)

So, the allelic frequency of the recessive allele =

q =  √ 0.007 = 0.08366 or 0.084

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A new human species was found on a distant planet, whose genetic inheritance and process of reproduction is identical to that of
Vedmedyk [2.9K]

Answer:

  1. E/e ; T/t  ⇒ Square eyes and long tail (Option D)
  2. 3/16 will have round eyes and long tails (Option C)
  3. 1/4 of the progeny will be heterozygous for both traits (Option A)

Explanation:

<u>Available data</u>:

  • Two diallelic unlinked genes
  • Gene E controls eye shape: Dominant allele E expresses square eyes, and recessive allele e expresses round eyes.
  • Gene T controls the tail size. Dominant allele T expresses long trail, and recessive allele t expresses short tail.

<u>Genotypes                                          Phenotypes</u>

EETT, EeTT, EETt, EeTt                     Square eyes and Long tail

eeTT, eeTt                                          Round eyes and Long tail

EEtt, Eett                                             Square eyes and Short tail

eett                                                      Rounf eyes and short tail.

<em>1. An individual of this new species is heterozygous for gene E and heterozygous for gene T. What is their genotype and phenotype?</em>

The heterozygous individual is E/e ; T/t, expressing square eyes and a long tail.

<em>2. Two individuals, who are both heterozygous for eye shape and tail size, mate. Which of the following is a correct statement about the phenotype ratios expected for their offspring?</em>

Cross: between two heterozygous individuals

Parentals) EeTt    x     EeTt

Gametes) ET, Et, eT, et

                ET, Et, eT, et

Punnett square)  .   ET        Et        eT       et

                  ET     EETT    EETt    EeTT    EeTt

                  Et      EETt     EEtt     EeTt      Eett

                  eT    EeTT    EeTt    eeTT     eeTt

                   et     EeTt      Eett     eeTt      eett

F1) Genotype:

  • 1/16 EETT
  • 2/16 EETt
  • 1/16 EEtt
  • 2/16 EeTT
  • 4/16 EeTt ⇒ 1/4 EeTt
  • 2/16 Eett
  • 1/16 eeTT
  • 2/16 eeTt
  • 1/16 eett

      Phenotype

  • 9/16 E-T-, Square eyes and Long tails
  • 3/16 E-tt, Square eyes and short tails
  • 3/16 eeT-, Round eyes and Long tails
  • 1/16 eett, round eyes and short tail

Phenotypic ratio 9:3:3:1

<em>3. From the mating described in question ABOVE, what proportion of ALL of the offspring in will be heterozygous for both traits?</em>

4/16 = 1/4 = 25% of the progeny are expected to be heterozygous for both traits, EeTt.

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