Which of the following mutations might lead to the formation of a recessive allele?

I. Silent

II. Frameshift

III. Nonsense

IV. Missense

Human chromosomes are divided into two arms, a long q arm and a short p arm. A karyotype is the organization of a human cell’s total genetic complement. A typical karyotype is generated by ordering chromosome 1 to chromosome 23 in order of decreasing size.

When viewing a karyotype, it can often become apparent that changes in chromosome number, arrangement, or structure are present. Among the most common genetic changes are Robertsonian translocations, involving transposition of chromosomal material between long arms of certain chromosomes to form one derivative chromosome. Chromosomes 14 and 21, for example, often undergo a Robertsonian translocation, as below.

A karyotype of this individual for chromosomes 14 and 21 would thus appear as follows:

Though an individual with aberrations such as a Robertsonian translocation may be phenotypically normal, they can generate gametes through meiosis that have atypical organizations of chromosomes, resulting in recurrent fetal abnormalities or miscarriages.

During meiosis, gametes result from the isolation of one chromosome from each of a human’s homologous 23 pairs. In Figure 2, der(14,21) is treated as a homologous pair to either 14 or 21. An egg, then, could be formed that had a normal chromosome 21 and the der(14,21) instead of the normal chromosome 14. If a sperm fertilizes this egg, and the sperm has a normal chromosome 14 and 21, what will result?

Cryptosporidium is a genus of gastrointestinal parasite that infects the intestinal epithelium of mammals. Cryptosporidium is water-borne, and is an apicomplexan parasite. This phylum also includes Plasmodium, Babesia, and Toxoplasma.

Apicomplexans are unique due to their apicoplast, an apical organelle that helps penetrate mammalian epithelium. In the case of cryptosporidium, there is an interaction between the surface proteins of mammalian epithelial tissue and those of the apical portion of the cryptosporidium infective stage, or oocyst. A scientist is conducting an experiment to test the hypothesis that the oocyst secretes a peptide compound that neutralizes intestinal defense cells. These defense cells are resident in the intestinal epithelium, and defend the tissue by phagocytizing the oocysts.

She sets up the following experiment:

As the neutralizing compound was believed to be secreted by the oocyst, the scientist collected oocysts onto growth media. The oocysts were grown among intestinal epithelial cells, and then the media was collected. The media was then added to another plate where Toxoplasma gondii was growing with intestinal epithelial cells. A second plate of Toxoplasma gondii was grown with the same type of intestinal epithelium, but no oocyst-sourced media was added.

You are conducting a study of an isolated tribe in New Guinea, and you find that there is widespread resistance to cryptosporidium infection. Upon examination, you find that the resistance is caused by a change in one nucleotide pair in a gene on chromosome 13. What kind of genetic change does this likely reflect?

An mRNA sequence is supposed to read UAUGGA, but a mutation replaces the second uracil base with guanine. What is the most specific term for this type of mutation?

Which answer choice correctly shows a frameshift mutation?

Type II diabetes results from defective pancreatic beta cells and increased insulin resistance, indicating that peripheral tissues (such as skeletal muscle) do not properly respond to insulin.

Mouse models have been developed to model type II diabetes. In addition to global mutations, tissue-specific mutations can be used to delete genes of interest in precise regions of the body. A group of investigators is interested in characterizing the role of the gene Dia in the onset of diabetes.

Four groups of male mice are compared. Group A is a control group, group B has a global deletion of Dia, group C has a beta cell-specific Dia mutation, and group D has a skeletal muscle-specific Dia mutation.

In order to measure the ability of these mice to respond to a glucose challenge, the mice are fasted overnight. Following the fast, their blood glucose levels are measured (in mg/dL). The mice are then injected with two grams of glucose, and blood glucose levels are measured at 30, 60, 90, and 120 minutes post-injection.

| | 0 min | 30 min | 60 min | 90 min | 120 min | | | ------------ | ---------- | ---------- | ---------- | ----------- | --- | | Group A | 80 | 150 | 120 | 90 | 80 | | Group B | 90 | 220 | 180 | 160 | 140 | | Group C | 100 | 260 | 190 | 150 | 135 | | Group D | 75 | 145 | 110 | 90 | 75 |

Based on the data, what role does Dia play in insulin regulation?

The concept of genomic imprinting is important in human genetics. In genomic imprinting, a certain region of DNA is only expressed by one of the two chromosomes that make up a typical homologous pair. In healthy individuals, genomic imprinting results in the silencing of genes in a certain section of the maternal chromosome 15. The DNA in this part of the chromosome is "turned off" by the addition of methyl groups to the DNA molecule. Healthy people will thus only have expression of this section of chromosome 15 from paternally-derived DNA.

The two classic human diseases that illustrate defects in genomic imprinting are Prader-Willi and Angelman Syndromes. In Prader-Willi Syndrome, the section of paternal chromosome 15 that is usually expressed is disrupted, such as by a chromosomal deletion. In Angelman Syndrome, maternal genes in this section are deleted, while paternal genes are silenced. Prader-Willi Syndrome is thus closely linked to paternal inheritance, while Angelman Syndrome is linked to maternal inheritance.

Figure 1 shows the chromosome 15 homologous pair for a child with Prader-Willi Syndrome. The parental chromosomes are also shown. The genes on the mother’s chromosomes are silenced normally, as represented by the black boxes. At once, there is also a chromosomal deletion on one of the paternal chromosomes. The result is that the child does not have any genes expressed that are normally found on that region of this chromosome.

In addition to the chromosomal deletion on chromosome 15 in the passage, the father is found to have another gene with a mutation, which adds a stop codon prematurely in the base pair sequence. This mutation is best described as a __________.

Cellular division is an essential part of the cell cycle. When a cell divides it passes genetic information to daughter cells. The amount of genetic information passed on to daughter cells depends on whether the cell undergoes mitosis or meiosis. Mitosis is the most common form of cell division. All somatic cells undergo mitosis, whereas only germ cells undergo meiosis. Meiosis is very important because it produces gametes (sperm and eggs) that are required for sexual reproduction. Human germ cells have 46 chromosomes (2n = 46) and undergo meiosis to produce four haploid daughter cells (gametes).

An individual containing three sex chromosomes (XXY) is called a polysomic individual. What is the reason for polysomy?

Which of the following will least likely affect the length of a protein product?

Which of the following mutations will not change the amino acid sequence of the polypeptide created during translation?