1. Which of the following statements is incorrect about homologous chromosomes?
a. Homologous chromosomes are similar but not identical
b. Homologous chromosomes have the same size and function
c. Homologous chromosomes contain genes for the same kind of protein products
d. Homologous chromosomes cannot be duplicated
When a human cell undergoes meiosis, the resulting daughter cells
a. Are diploid
b. Have identical genes
c. Differ in their genetic combinations
d. Have genes identical to the parental cells
2. Which of the following cells contains two sets of chromosomes (i.e. diploid cell)?
a. Zygote b. Gamete c. Sperm cell d. Egg cell e. Spermatid
3. In a species of tigers, Y = yellow fur (dominant) B = brown eyes (dominant)
y = black fur (recessive) b = gray eyes (recessive)
Which of the following is a cross between a male that is homozygous dominant for fur color, homozygous recessive for eye color and a female that is heterozygous for both characters?
a. YyBb x YYBb b. YyBb x yybb c. YYbb x YyBb d. YYBB x yybb
4. Birds of one particular species produce pigments that allow them to see only gray color. Birds of another species produce pigments that allow them to see only red color. Offspring of these 2 species, however, can fully produce both types of pigment together and therefore can see both gray and red colors. This is an example of
a. Codominance c. Competitive dominance e. Blending
b. Complete dominance d. Incomplete dominance
5. A cross between a pure-breeding white flower, long stem (rrLL) plant, and a pure-breeding red flower, short stem (RRll) plant would produce (R = dominant red flower, r = recessive white flower, L = dominant long stem, l = recessive short stem).
a. All white flowers and long stem plants c. All red flower and short stem plants
b. All red flower and long stem plants d. All white flower and short stem plants
6. A dihybrid cross between YYSS and yyss would produce offspring with which of the following genotype?
a. YYSS b. YySs c. yyss d. Yyss
7. In horses, B = black coat b = brown coat T = trotter t = pacer
Which of the following is the genotype of a black trotter?
a. BBtt b. Bbtt c. BBTt d. bbTt
8. Mary has type AB blood. Which of the following is true?
a. Mary can only receive type AB blood
b. Her plasma contains both antibody A and B
c. Her red blood cells contain surface antigens A and B
d. If Mary receives type A blood, the antibody A would react with her cells
9. Which of the following represent a normal gamete and which is a normal diploid cell? Please label each clearly.
a. GRr b. GgR c. Gr d. GgRrCC
10. Prior to DNA replication, the DNA molecule is unwound by
a. Ligases c. DNA polymerases
b. Helicases d. Single-strand binding proteins
11. If there is one molecule of DNA at the beginning of DNA replication, how many total molecules of DNA would you have after three replications?
a. Two b. Eight c. Four d. Sixteen
12. The primary RNA transcript is processed before it is transported to the cytoplasm for translation. This processing of RNA
a. Speeds the process of transcription
b. Helps in recognition by the ribosome
c. Protects the molecule from being degraded by cytoplasmic enzymes
d. Both a and b are true
e. Both b and c are true
13. In DNA, A, C, G, and T are abbreviations for the
a. different types of bases on the nucleotides.
b. different types of sugars on the nucleotides.
c. types of genes found on DNA chromosomes.
d. different types of bonds that form between nucleotides.
e. different types of phosphates on the nucleotides.
14. There are between 20,000 and 25,000 genes in the human genome, yet our cells are capable of producing more than 90,000 different proteins. How is this possible?
a. Many genes are actually yet to be discovered.
b. Introns are spliced back together to form mRNAs.
c. Micro-RNAs can also be used to make proteins.
d. Primary transcripts can be edited in different ways by alternative splicing to produce different mRNAs.
Please help! I am unsure about this.
We will be glad to respond to your choices.
Sure! I'd be happy to help you with these questions. Let's go through each one step by step.
1. To determine which statement is incorrect about homologous chromosomes, we need to understand what homologous chromosomes are. Homologous chromosomes are similar but not identical, so statement a is correct. They have the same size and function, so statement b is also correct. They contain genes for the same kind of protein products, so statement c is correct. This means that statement d, "Homologous chromosomes cannot be duplicated," is incorrect. Homologous chromosomes can be duplicated during DNA replication.
2. A diploid cell contains two sets of chromosomes. A zygote is formed by the fusion of two gametes, so it has a single set of chromosomes and is therefore haploid. Gametes, such as sperm cells and egg cells, are haploid cells with a single set of chromosomes. Spermatids are also haploid cells. Therefore, the correct answer is a. Zygote.
3. We need to cross a male that is homozygous dominant for fur color (BB) and homozygous recessive for eye color (bb) with a female that is heterozygous for both characters (Bb). The possible genotypes for the offspring will be YyBb, YYBb, Yybb, and YYbb. Among these options, only option a. YyBb x YYBb matches the given genotypes of the male and female.
4. This situation describes codominance, where alleles from both parents are expressed equally in the offspring. Option a. Codominance is the correct answer.
5. The provided cross involves two traits, flower color (R/r) and stem length (L/l). The male parent is rrLL (recessive white flower and dominant long stem), and the female parent is RRll (dominant red flower and recessive short stem). When we cross these parents, all the offspring will have a heterozygous genotype for both traits, resulting in red flower and long stem plants. Therefore, the correct answer is b. All red flower and long stem plants.
6. A dihybrid cross involves two traits, so we need to consider each trait separately. For the genotype YYSS crossed with yyss, all the offspring will have the genotype YySs since both parents are homozygous for each trait. Therefore, the correct answer is b. YySs.
7. A black trotter has a black coat (BB) and is a trotter (Tt). The correct genotype is c. BBTt.
8. Mary has type AB blood, which means her blood contains both A and B antigens on the red blood cells. This means that both antibody A and antibody B are absent from her plasma. Therefore, the correct answer is c. Her red blood cells contain surface antigens A and B.
9. In this question, we are asked to determine which options represent a normal gamete and a normal diploid cell. A normal gamete is haploid, meaning it has a single set of chromosomes. From the given options, only b. GgR can represent a normal gamete. A normal diploid cell has two sets of chromosomes, so d. GgRrCC represents a normal diploid cell.
10. DNA replication involves the unwinding of the DNA molecule. This is done by helicases, option b.
11. DNA replication follows a semi-conservative model, meaning each original DNA molecule is used as a template for the synthesis of a new complementary DNA strand. After three replications, each original DNA molecule will result in eight total molecules of DNA. Therefore, the correct answer is b. Eight.
12. The primary RNA transcript undergoes several modifications before it is transported to the cytoplasm for translation. These modifications include the addition of a 5' cap, splicing out of introns, and addition of a poly-A tail. These modifications serve several purposes, including protecting the molecule from degradation by cytoplasmic enzymes and helping in the recognition by the ribosome. Therefore, the correct answer is e. Both b and c are true.
13. In DNA, the letters A, C, G, and T represent different types of bases on the nucleotides. They are adenine (A), cytosine (C), guanine (G), and thymine (T). Therefore, the correct answer is a. different types of bases on the nucleotides.
14. The process of alternative splicing allows for the production of multiple mature mRNA molecules from a single primary transcript. This means that different combinations of exons can be included or excluded in the final mRNA, resulting in different protein products. Therefore, option d. Primary transcripts can be edited in different ways by alternative splicing to produce different mRNAs is the correct answer.
I hope this helps! Let me know if you have any more questions.
1. The incorrect statement about homologous chromosomes is d. Homologous chromosomes cannot be duplicated. Homologous chromosomes can undergo DNA replication during the S phase of the cell cycle, resulting in duplicated chromosomes.
When a human cell undergoes meiosis, the resulting daughter cells:
c. Differ in their genetic combinations. Meiosis is a process of cell division that reduces the number of chromosomes in each daughter cell and creates genetic variation through the shuffling of homologous chromosomes.
2. The cell that contains two sets of chromosomes (diploid cell) is:
a. Zygote. A zygote is formed when a sperm cell fertilizes an egg cell, resulting in a diploid cell with two sets of chromosomes, one from each parent.
3. The cross between a male that is homozygous dominant for fur color, homozygous recessive for eye color (YYbb), and a female that is heterozygous for both characters (YyBb) is:
c. YYbb x YyBb. This cross would result in offspring with genotypes YYBb and Yybb.
4. The ability of offspring to produce both types of pigment together and see both gray and red colors is an example of:
d. Incomplete dominance. Incomplete dominance occurs when the heterozygous phenotype is an intermediate blend of the two homozygous phenotypes.
5. A cross between a pure-breeding white flower, long stem (rrLL) plant, and a pure-breeding red flower, short stem (RRll) plant would produce:
b. All red flower and long stem plants. The dominant alleles for both flower color (R) and stem length (L) will determine the phenotype in the offspring.
6. A dihybrid cross between YYSS and yyss would produce offspring with the genotype:
d. Yyss. Each parent contributes one allele for each gene, resulting in offspring with a combination of the two alleles for each gene (Yy for fur color and ss for eye color).
7. The genotype of a black trotter in horses is:
c. BBTt. The black coat color is determined by the dominant allele (B), and the trotter gait is determined by the dominant allele (T).
8. Mary having type AB blood means:
c. Her red blood cells contain surface antigens A and B. Type AB blood contains both antigen A and antigen B on the surface of red blood cells.
9. A normal gamete is represented by:
c. Gr. A normal diploid cell is represented by:
d. GgRrCC. Gametes contain only one allele for each gene, while diploid cells contain both alleles.
10. Prior to DNA replication, the DNA molecule is unwound by:
b. Helicases. Helicases are enzymes that unwind the double-stranded DNA molecule by breaking hydrogen bonds between the base pairs.
11. If there is one molecule of DNA at the beginning of DNA replication, you would have:
b. Eight molecules of DNA after three replications. DNA replication results in the synthesis of two identical copies of each original DNA molecule, resulting in exponential growth.
12. The primary RNA transcript is processed before it is transported to the cytoplasm for translation. This processing of RNA:
e. Both b and c are true. Processing of RNA includes modifications like adding a 5' cap and a poly-A tail, which helps in recognition by the ribosome and protects the molecule from degradation by cytoplasmic enzymes.
13. In DNA, A, C, G, and T are abbreviations for the:
a. Different types of bases on the nucleotides. A represents adenine, C represents cytosine, G represents guanine, and T represents thymine, which are the four different nitrogenous bases found in DNA.
14. The ability of our cells to produce more proteins than the number of genes in the human genome is possible because:
d. Primary transcripts can be edited in different ways by alternative splicing to produce different mRNAs. Alternative splicing allows different exons within primary RNA transcripts to be combined, resulting in different mRNA molecules and ultimately different protein products. This increases the protein diversity from a limited number of genes.