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In rabbits, short hair is due to a dominant allele, S, and long hair to its recessive allele, s. Black hair is due to a dominant allele, B, and white hair to its recessive allele, b. When two rabbits are crossed, they produce 2518 short-haired, black offspring and 817 long-haired, black offspring. What are the possible genotypes of the parents?

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Answer:

SsBB  and SsBB

or

SsBB and SsBb

Explanation:

To produce a ratio of offspring where the number of short haired black offspring is three times that of the long haired black offspring. Since, the long haired offspring can only be inherited in the recessive manner in two copies, thus possible genotypes of the parents could be:

SsBB  and SsBB

or

SsBB and SsBb

SsBB  and SsBB

         SB      sB

SB    SSBB   SsBB

sB     SsBB   ssBB

3/4 short haired black offspring and 1/4 long haired black offspring

SsBB and SsBb

          SB        Sb      sB       sb

SB    SSBB  SSBb  SsBB   SsBb

sB     SsBB  SsBb   ssBB    ssBb

6/8=3/4 short haired black offspring and 2/4=1/4 long haired black offspring

marianaegarciaperedo

According to the phenotypic percentages and ratios expressed by the offspring, we may say that one of the parents is SsBB. The other parent might be either SsBb or Ssbb.

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Available data:

  • Two diallelic genes ⇒ S for hair length and B for color

  • S gene

⇒  S is the dominant allele and codes for short hair

⇒  s is the recessive allele and codes for long hair

  • B gene

⇒  B is the dominant allele and codes for black hair

⇒  b is the recessive allele and codes for white hair

  • N = 2518 short-haired, black offspring + 817 long-haired, black offspring

  • N = 3335

We need to figure out which is the genotype of the parental generation.

A way to do it is by taking the percentages and ratios of the progeny. This is,

3335 individuals -------------------- 100% of the progeny

2518 black, short-haired --------- X = 75.5%

817 black, long-haired -------------X = 24.5%

These percentages approximate to the phenotypic ratio 3:1 from a cross between two heterozygous individuals ⇒ 75% expressing the dominant trait and 25% expressing the recessive phenotype.

The first conclusion is that, knowing that there are two genes involved, if both parents were dihybrid, the expected ratios expressed by the progeny should be 9:3:3:1. But this is not the case. So parents must express a different genotype.

Let us analyze the phenotypes of the progeny.

About the hair length

  • Some of them are long-haired, and most of them are short-haired.

  • This fact, together with the percentages calculated before, suggests that the parents are heter0zyg0us for the hair length trait.

  • If this is the case, then 75% of the progeny should express the dominant phenotype, and only 25% the recessive one.

  • The observed percentages coincides with the expected ones.

  • Both parents must be heter0zyg0us for this trait, Ss.

About the color

  • All of them are black.

  • This result suggests that at least one of the parents is h0m0zyg0us dominant for the trait, BB. The other parent could be either h0m0zyg0us recessive -bb- or heter0zyg0us -Bb-.

  • The BB parent could only provide dominant alleles to the progeny, so the whole offpring expresses black color.

So, there are two possible options for parental genotype:

  • Option 1 ⇒ BBSs   and   BbSs
  • Option 2 ⇒ BBSs   and   bbSs

Let us perform the crosses:

  • Option 1

Parentals) BBSs   x   BbSs

Gametes) BS, BS, Bs, Bs

                 BS, Bs, bS, bs

Punnett square)            BS        BS        Bs       Bs

                        BS      BBSS    BBSS    BBSs    BBSs

                        Bs       BBSs    BBSs    BBss     BBss

                        bS       BbSS    BbSS    BbSs    BbSs

                        bs       BbSs     BbSs     Bbss    Bbss

F1) 12/16 B-S- = 0.75 B-S- = 75% of the progeny black and short-haired

     4/16 B- ss = 0.25 B-ss = 25% of the progeny black and long-haired

  • Option 2

Parentals) BBSs   x   bbSs

Gametes) BS, BS, Bs, Bs

                 bS, bs, bS, bs

Punnett square)            BS        BS          Bs       Bs

                        bS       BbSS    BbSS     BbSs    BbSs

                        bs        BbSs    BbSs     Bbss     Bbss

                        bS       BbSS    BbSS     BbSs    BbSs

                        bs       BbSs     BbSs     Bbss     Bbss

F1) 12/16 BbS- = 0.75 BbS- = 75% of the progeny black and short-haired

     4/16 Bbss = 0.25 Bbss = 25% of the progeny black and long-haired

Both crosses result in the same phenotypic ratios of the progeny, meaning that both options are possible.

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