The reactants for fermentation are:
1.) Pyruvate
2.) NADH
3.) A proton
The products are a lactate and NAD+.
The process of fermentation results in the reduction of pyruvate to form lactate acid and the oxidation of NADH to form NAD+
Answer:
Explanation:
Fertilization occurs when the nucleus of both a sperm and an egg fuse to form a diploid cell, known as zygote. The successful fusion of gametes forms a new organism.
<span>Endoplasmic Reticulum
ER Golgi apparatus, because it packages proteins received from the ER cytoplasm </span>
<span>The Golgi body are the ones that slightly alter, organize and prepare so-called parcels to be delivered for all the organelles in the cell. They receive these packages mainly in the rough endoplasmic reticulum. These packages that set out by Golgi body are macromolecules that used and synthesized by cells in many operations. If ER is absent then it would only mean that Golgi body would have no use other than simply lysosomes but these macromolecules plays a dynamic role in many organelles –nutrients, ATP and cell metabolism. It'll have a ripple effect if ER is absent in the cell.<span>
</span></span>
Answer:
0.0177
Explanation:
Cystic fibrosis is an autosomal recessive disease, thereby an individual must have both copies of the CFTR mutant alleles to have this disease. The Hardy-Weinberg equilibrium states that p² + 2pq + q² = 1, where p² represents the frequency of the homo-zygous dominant genotype (normal phenotype), q² represents the frequency of the homo-zygous recessive genotype (cystic fibrosis phenotype), and 2pq represents the frequency of the heterozygous genotype (individuals that carry one copy of the CFTR mutant allele). Moreover, under Hardy-Weinberg equilibrium, the sum of the dominant 'p' allele frequency and the recessive 'q' allele frequency is equal to 1. In this case, we can observe that the frequency of the homo-zygous recessive condition for cystic fibrosis (q²) is 1/3200. In consequence, the frequency of the recessive allele for cystic fibrosis can be calculated as follows:
1/3200 = q² (have two CFTR mutant alleles) >>
q = √ (1/3200) = 1/56.57 >>
- Frequency of the CFTR allele q = 1/56.57 = 0.0177
- Frequency of the dominant 'normal' allele p = 1 - q = 1 - 0.0177 = 0.9823
Condominance is the answer hope it helps
