A nucleotide is the building block of nucleic acids, such as DNA & RNA. They contain a phosphate group, 5-carbon sugar & nitrogen-containing base all covalently bound together.

Pyrimidine bases include cytosine, thymine, and uracil (RNA), and are monocyclic. Purine bases include adenine and guanine, and are bicylic with two rings in the molecular structure. 

The correct answer here is uracil. Remember that in DNA replication, the nucleic acids present are TCGA. When we switch over to RNA production, the thyamine is replaced by uracil to form UCGA. If you chose any of the other options, remember that each one has a pair (AT and CG) and in RNA it switches over to UA and CG. If you remember that, you will always recall that uracil replaces thyamine in RNA sequencing

The correct answer here is fatty acids because lipids are fats. If you look to the other options, we can eliminate nucleic acids because those are what characterize DNA and RNA (thyamine, uracil, guanine, adenine, and cytosine). Proteins are characterized as amino acids, not lipids. You can easily eliminate sugar and carbohydrates because those are the same thing, both characterized as -saccharides. 

This question requires us to do a dihybrid cross. We can represent the gene for beak color with the symbol "A" for dominant yellow and "a" for the recessive orange. Likewise, for feather color, we can use "B" for blue feathers and "b" for black.

The problem states that both birds are heterozygous for each trait, implying that our cross is between two birds with the genotype AaBb.

Now we look at the gametes that can be produced by these parents: AB, Ab, aB, and ab. These gametes can then be used to make a punnet square.

Offspring: 1 AABB, 3 Aabb, 8 AaBa, 3 aaBb, 1 aabb

There are 16 total offspring. 12 of them carry the dominant A allele, giving them the yellow beak phenotype.

Yellow beaks: 1 AABB, 3Aabb, 8 AaBb

Finally, of these 12, 9 carry the dominant B allele for blue feathers.

Yellow beaks and blue feathers: 1 AABB, 8 AaBb

This gives a total of 9 out of the 16 offspring that will express both the yellow beak and blue feather phenotypes.

You should be familiar with the 9:3:3:1 phenotypic ratio resulting from dihybrid crosses.

When dealing with two traits, it helps to approach each trait separately. The question asks for the fraction of plants in the second generation that are heterozygous for both traits.

First, we will look at the first generation. The parents are both pure breeding, meaning they are homozygous for each trait. One is purple (dominant) and wrinkled (recessive), while the other is white (recessive) and round (dominant). The result will be dihybrid offspring.

Parent cross: PPrr x ppRR

Offspring: All offspring will be PpRr and exhibit both dominant phenotypes (purple and round).

Now we will look at the first generation self-cross for each trait.

First generation: PpRr x PpRr

Offspring for color: 1 PP, 2 Pp, 1 pp; 3 purple and 1 white.

Offspring for seeds: 1 RR, 2 Rr, 1 rr; 3 round and 1 wrinkled.

We can see that half of the offspring will be heterozygous for color, and half of the offspring will be heterozygous for seed shape.

Since we are looking for plants that have both of these traits, we multiply these two probabilities together.

The genotype to obtain pointed leaves is rr, while the genotypes to obtain white flowers are WW or Ww. Determine the frequence of obtaining WWrr or Wwrr using a Punnett Square.

You should be familiar with the ratios represented in a dihybrid cross: 9:3:3:1. There will be nine inidividuals with both dominant traits (pointed leaves and white flowers), three individuals dominant for one trait (round leaves and pink flowers), three individuals dominant for the other trait (pointed leaves and white flowers), and only one individual recessive for both traits (pointed leaves and pink flowers).

In Hardy-Weinberg equilibrium, the sum of the dominant allele frequency (p) and the recessive allele frequency (q) is equal to 1.

The question says that 49% of the population consists of mice with the homozygous dominant gene, therefore, the dominant genotype frequency is equal to 0.49.

The question asks us to find the frequency of the recessive allele (q). In order to find the frequency of the recessive allele, we must first find the frequency of the dominant allele (p). According to the Hardy-Weinberg principle, the square root of the homozygous genotype frequency is equal to the allele frequency.

The dominant allele frequency is 0.7. Using this, we can solve for the recessive allele frequency.

We can use the Hardy-Weinberg equilibrium formulas to calculate the allele frequencies.

We know that the frequency of homozygous recessive (pink) rabbits is . This is equal to  in the Hardy-Weinberg calculation. We can use this information to solve for , the recessive allele frequency.

Now that we know the value of , we can solve for the value of .

The frequency of heterozygotes is equal to in the Hardy-Weinberg calculation. Now that we know the frequency of each allele, we can complete this calculation.

For problems of this type, we need to understand the Hardy-Weinberg equations:

Here,  represents the frequency of the dominant allele, while c refers to the frequency of the recessive allele.  and  denote the proportion of homozygous dominant and recessive phenotypes, respectively. Finally, the proportion of heterozygotes is denoted by .

We already know that , and if only two alleles are present in the population,  must be equal to .

Using the values for and , we can solve for the proportion of heterozygotes using the term of the Hardy-Weinberg equation.