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Answer :
Final answer:
To determine the probability of chick phenotypes produced by Farmer Jim's turkeys, we use Mendelian genetics and a Punnett square. An orange female (presumed genotype 'oo') crossed with a heterozygous brown male (genotype 'Bo') would yield a 50% chance of brown phenotype ('Bo') and a 50% chance of orange phenotype ('oo').
Explanation:
The question pertains to genetics and the concept of phenotype probabilities from a cross-breeding scenario. To determine the probability of chick phenotypes produced by Farmer Jim's turkeys, we can apply Mendelian genetics. If an orange female turkey is bred with a heterozygous male turkey, assuming that 'orange' is a recessive trait, we can predict outcomes using a simple Punnett square. Since the question doesn't provide explicit information on genetic dominance or the genotype of the female (just that she is orange), we may presume 'orange' to be recessive (oo), and 'brown' to be the dominant phenotype. Thus, the heterozygous male would have one dominant allele (let's say 'B') for 'brown' and one recessive allele ('o') for 'orange'. The cross would be between a female 'oo' and a male 'Bo'.
To predict the phenotypic ratio, the resulting Punnett square would show that there is a 50% chance that the offspring would be heterozygous ('Bo') displaying the brown phenotype and a 50% chance they would be homozygous recessive ('oo') displaying the orange phenotype. Therefore, the probability of obtaining an orange turkey from this pair is 50%.
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