2) Use a Punnett square to help answer the question.
2 answers:
Answer:
2) 75% of the children would have free earlobes
3) 25% of the children would be homozygous dominant.
4) 50% of the seeds would be smooth.
Explanation:
2) The genotype of a parent is given as heterozygous dominant, that is, Ee.
E is the dominant allele which codes for free earlobes and e is the recessive allele which codes for attached earlobes.
Each parent would produce two types of gametes, one containing E and the other would contain e.
The cross would produce offspring with three possible types of genotypes EE, Ee, and ee in the ratio 1:2:1.
Thus, three out of four children would have free earlobes.
Hence, 75% of the children would have free earlobes
3) From the above cross, the genotype ratio obtained was (1) EE: 2 (Ee): 1 (ee).
Thus, one out of four children would have homozygous dominant genotype.
Hence, 25% of the children would be homozygous dominant.
4) Let S and s be the alleles of the gene responsible for the shape of the seed.
S is the dominant allele which codes for smooth shape and s is the recessive allele which codes for wrinkled shape.
The genotype of one parent is homozygous recessive, that is, ss.
The genotype of one parent is heterozygous dominant, that is, Ss.
The cross would produce offspring with two possible types of genotypes; Ss and ss in the ratio 1:1.
Thus, 50% of the seeds would be smooth.
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