Final answer:
Using a Punnett square to determine the phenotypic ratio for the offspring of an elephant cross shows a 3:1 ratio, where there are 3 long trunk, small ears offspring for every 1 long trunk, big ears offspring.
Step-by-step explanation:
The phenotype ratio of the offspring from a cross between an elephant that is homozygous recessive for a short trunk and big ears (ssbb) and an elephant that is heterozygous for a long trunk and small ears (SsBb) can be determined using a Punnett square. Two traits are being considered here: trunk length (S for long trunk is dominant, s for short is recessive) and ear size (B for small ears is dominant, b for big is recessive). Since one parent is homozygous recessive for both traits (ssbb), all its gametes will carry 'sb'. The heterozygous parent can produce four types of gametes: 'SB', 'Sb', 'sB', and 'sb'.
After drawing the Punnett square and filling in the potential combinations of alleles from both parents, we observe that all offspring will have one recessive allele for each trait from the homozygous recessive parent. Since a single dominant allele is enough to express the dominant phenotype, the presence of a dominant allele from the heterozygous parent for each trait will determine the phenotype.
The offspring will have the following phenotypes:
- Long trunk, small ears (SB/sb or Sb/sb): which is the result of the long trunk allele (S) and small ears allele (B) being dominant.
- Long trunk, big ears (sB/sb): the long trunk phenotype is due to the presence of at least one dominant 'S' allele, while the big ears are due to the presence of two recessive 'b' alleles.
Since there is no possible combination for a short trunk and small ears phenotype because the homozygous recessive parent cannot contribute a dominant allele, the phenotypic ratio for these traits will be:
- 3 long trunk, small ears
- 1 long trunk, big ears
Therefore, the correct answer is C) 3:1.