Question

what is the chance of the offspring having a green seed?
0%
25%
50%
75%

Explanation:

To determine the chance of offspring having a green seed, we assume a typical Mendelian genetics scenario (e.g., a cross like \( Yy\times Yy \), where \( Y \) is yellow (dominant) and \( y \) is green (recessive)). The Punnett square for this cross would have genotypes: \( YY \), \( Yy \), \( Yy \), \( yy \). The phenotype for green seed is associated with the \( yy \) genotype (since yellow is dominant). Out of 4 possible genotypes, 1 is \( yy \) (green). So the probability is \( \frac{1}{4} = 25\% \) only if the parents are heterozygous for yellow (dominant) and green (recessive). But wait, if we consider a cross like \( YY\times yy \), all offspring would be \( Yy \) (yellow), 0% green. Or \( Yy\times yy \), the Punnett square gives \( Yy \) and \( yy \) in a 1:1 ratio, so 50% green. However, the most common classic example (Mendel's pea plants, seed color: yellow dominant, green recessive) with two heterozygous parents (\( Yy\times Yy \)) gives 25% green. But maybe the intended scenario here (since 25% is an option and it's a common question) is the \( Yy\times Yy \) cross. Wait, no—wait, maybe I mixed up. Wait, in Mendel's experiments, when crossing two heterozygous yellow-seeded plants (\( Yy \)), the offspring genotypes are \( YY \) (yellow), \( Yy \) (yellow), \( Yy \) (yellow), \( yy \) (green). So the ratio of green (yy) is 1 out of 4, which is 25%? Wait, no—wait, \( YY \), \( Yy \), \( Yy \), \( yy \): three yellow (YY, Yy, Yy) and one green (yy). So green is 25%? Wait, no, 1/4 is 25%, yes. But wait, maybe the question is about a different cross. Wait, the options include 25%, 50%, 75%, 0%. Let's think again. If the parents are, say, \( Yy \) (yellow) and \( yy \) (green), then the Punnett square is \( Yy \) and \( yy \) for each parent. So possible offspring: \( Yy \) (yellow), \( yy \) (green), \( Yy \) (yellow), \( yy \) (green). So that's 2 out of 4, 50% green. But the classic \( Yy\times Yy \) gives 25% green. Since the question is likely referring to the classic Mendelian ratio (as it's a common question), the answer is 25%? Wait, no—wait, maybe I made a mistake. Wait, in the classic monohybrid cross (two heterozygous parents), the phenotypic ratio is 3:1 (yellow:green), so green is 1/4 = 25%. So the chance is 25%.

Assuming a classic Mendelian monohybrid cross (e.g., two heterozygous yellow - seeded parents, \( Yy\times Yy \)), the Punnett square shows genotypes \( YY \), \( Yy \), \( Yy \), \( yy \). The green seed phenotype is associated with the \( yy \) genotype. Out of 4 possible genotypes, 1 is \( yy \), so the probability is \( \frac{1}{4}=25\% \).

Answer:

25% (the option with "25%")