It will be transported into the cell through the process of endocytosis, which is a form of active transport. Active transport requires energy in the form of ATP. Hope this helps! :)
Answer:
no 1. A is the answer
Explanation:
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Answer:
Homologous structures provide evidence for common ancestry analogous structures show that similar selective pressures can produce similar adaptations beneficial features Similarities and differences among biological molecules in the DNA sequence of genes can be used to determine species relatedness
Answer:
50% of their children are likely to be carriers of cystic fibrosis
Explanation:
Since the normal allele "F" will be the dominant allele while the mutated CFR allele "f" will be the recessive allele, <u>the gene (pair of alleles) of the person that is a CFR carrier will be "Ff" while that of the normal person who isn't a carrier will be "FF"</u>. The attachment shows the crossing between the two parents. From the illustration in the attachment, for every 4 children given birth to, 2 of them will likely be normal, "FF", (not a carrier and doesn't have cystic fibrosis) while 2 others will likely be carriers of cystic fibrosis (Ff). Hence, 50% of their children are likely to be carriers of cystic fibrosis.
Answer;
The correct answer would be Rr.
In a monohybrid cross, the phenotype ratio of 3:1 is obtained only when both the parents are heterozygous for the trait.
In this condition, the heterozygous genotype would be Rr. The phenotype of both the parents would be round seeds (R) as the round is a dominant trait.
The cross of these parents would produce offspring with three types of genotypes RR, Rr, and rr in ratio 1:2:1.
Thus, the phenotypic ratio would come out to be 3 (round seed):1 (wrinkled seed).