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On the planet Feline, there are only cats. For one particular phenotype, Myles, there are two manifestations: either the cats like jazz music (the wild type phenotype) or they do not (the mutant phenotype). Under the assumption this phenotype is determined by a single gene with two alleles, and given that a mating of two wild-type cats produces the following number of offspring: Female, wild-type: 341; Male, wild-type: 369; Female, Myles: 121; Male, Myles: 123. The mode of inheritance most consistent with these data is:________
A. Autosomal dominant.
B. Two-locus inheritance.
C. Sex-linked recessive.
D. Autosomal recessive.
E. Two-locus dominant.

Respuesta :

marianaegarciaperedo

Answer:

D. Autosomal recessive.

Explanation:

Available data:

  • One diallelic gene codes for the phenotype Myles
  • One allele expresses the wild type phenotype → like jazz music
  • The other allele expresses the mutated phenotype → don´t like jazz
  • Cross: Between two wild cats
  • F1) Female, wild-type: 341; Male, wild-type: 369; Female, Myles: 121; Male, Myles: 123    

Autosomal recessive inheritance characterizes because the mutated gene locates in one of the 22 non-sexual, somatic chromosomes. To express the phenotype, an individual must carry two copies of the mutated gene. For this reason, the recessive allele must be present in both the male/father and the female/mother.  

In the exposed example, the numbers among the progeny are 341:369:121:123. These values approximate a phenotypic ratio 3:1 for females, and for males. This ratio suggests that the cross was performed between two heterozygous individuals.

Let us name the alleles M and m, dominant and recessive respectively. The cross would be like follows

Cross:  wild-type male    x     wild-type female

Parentals)   Mm XY         x                MmXX

Gametes)  MX, MY, mX, mY       MX, MX, mX, mX

Punnett Square)     MX         MX         mX           mX

                  MX    MMXX    MMXX    MmXX     MmXX

                  mX    MmXX    MmXX    mmXX     mmXX

                  MY    MMXY    MMXY     MmXY     MmXY

                  mY    MmXY    MmXY     mmXY     mmXY

F1) Whole progeny

  • 12/16 = 3/4 = 75% Wild-type cats → MMXX + MmXX + MMXY + MmXY
  • 4/16 = 1/4 = 25% Mutated cats → mmXX + mmXY

     Females

  • 6/8 = 3/4 = 75% Wild-type cats → MMXX + MmXX
  • 2/8 = 1/4 = 25% Mutated cats → mmXX

     Males

  • 6/8 = 3/4 = 75% Wild-type cats → MMXY + MmXY
  • 2/8 = 1/4 = 25% Mutated cats → mmXY

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