Genetics Quiz and Homework Questions

In pea plants, plant height is controlled by a single autosomal dominant gene. Tall plants (H) are dominant to short plants (h). In a cross of two tall heterozygous plants, which phenotype ratio is expected from the resulting offspring?

3:1 once again do a punnett Square.

In pea plants, the allele for red flower color, F, is completely dominant to the allele for white flower color, f. Complete the Punnett square showing the genotypes possible among the offspring when two heterozygous individuals are crossed. Use the information from the Punnett square to answer the second question. In this cross between two heterozygous pea plants, what are the chances that an offspring with red flowers will be produced?

75% Do a Punnett Square to get the answer. Look for all combinations that include the dominant allele F

Which statements describe Y‑linked traits? a) All male offspring of affected fathers will express the trait. b) A carrier female has a 25% chance of having a carrier daughter. c) The trait is passed down to a son by his father. d) A male with the trait will pass the trait to 100% of his female offspring. e) At least 50% of the male and female offspring will inherit the trait.

A and C) -all male offspring of affected fathers will express the trait. -The trait is passed down to a son by his father. The male genotype is XY, which means that all traits linked to the Y chromosome will only relate to males. Since males and fathers are also the only ones who hold the Y trait, any traits linked to the Y chromosome will go to the son.

In butterflies, sex is determined by the ZW sex‑determination system. Female butterflies are heterogametic and have both a Z sex chromosome and a W sex chromosome for sex determination. In contrast, male butterflies are homogametic and have two Z sex chromosomes. Select all of the relatives from which a female butterfly could have inherited her Z sex chromosome. a) father b) maternal grandmother c) mother d) paternal grandfather

A and D) father and paternal grandfather Think of it like this: a female butterfly requires it to have the Z and W chromosome in order to fulfill the condition of being "female". This means that it must be donated at least one W and one Z chromosome. Only mothers have access to the W chromosome, thus to create a "female" they absolutely must donate it. The father can only donate Z chromosomes, but it shows that it is the only one that does it if the mother donates the W. If the mother doesn't donate a W then no daughter is created. do the punnet square as well.

What are the functions of mitotic cell division? a) asexual reproduction b) production of eggs or sperm c) growth of multicellular organisms d) cell regeneration

A, b, and c Mitotic cell division produces two daughter cells that are genetically identical to both the parent cell and to each other. This method of cell division has several functions. Multicellular organisms use mitotic cell division to grow. All multicellular organisms start as a single cell and undergo multiple rounds of mitotic cell division to become multicellular. These types of cell divisions multiply the number of cells in the organism, thereby increasing the size of the organism. In bacteria, cell division is called binary fission. Single‑celled eukaryotes, such as yeast, use mitotic cell division to reproduce. This type of reproduction is called asexual reproduction. Mitotic cell division also functions to replace or regenerate cells that may be old or wounded. For example, when you get a cut on your skin, mitotic cell division produces the cells that repair the wound. Mitotic cell division is not used to produce eggs or sperm. Gametes are produced by a different type of cell division called meiosis.

Cell division by mitosis is a mechanism of asexual cell replication. Some single‑celled organisms reproduce by cell division, and cell division enables multicellular organisms to grow and repair damaged cells. Which is a product of cell division by mitosis? a) daughter cells with half the number of chromosomes b) two cells genetically identical to the original cell c) daughter cells that are genetically variable d) four daughter cells that are genetically the same

B) Two cells genetically identical to the original cell Mitosis involves one division of the parent cell. During interphase of the cell cycle, a cell's DNA is duplicated and the cell prepares for mitosis. During mitosis, the nucleus and cytoplasm of the parent cell divide, and each daughter cell receives an identical set of the copied DNA. The genetically identical daughter cells have the same number of chromosomes as the parent cell. Special cells in sexually reproducing organisms undergo meiosis, which involves two divisions of a parent cell. Meiosis gives rise to four daughter cells that have half the number of chromosomes as the parent cell. These daughter cells become gametes, or sex cells, and do not divide again. Cells produced from meiosis are genetically different from one another and the parent cell.

Mendel's principle of segregation can be explained by what process? a) mitosis b) meiosis c) fertilization d) testcross e) none of these

B) meiosis

A man, Joe, has classic hemophilia, an X‑linked recessive disease. Classify each person depending on whether or not Joe could have inherited the hemophilia gene from him or her. -maternal grandmother -paternal grandmother -maternal grandfather -paternal grandfather

Could have inherited: -maternal grandmother -maternal grandfather Could not inherited: -paternal grandmother -paternal grandfather The X-linked trait is from Joe's mother, the father does not pass down the X trait unless he bore a female child, which Joe is not. Thus any paternal ancestors could not have passed the trait down Which means any maternal ancestors could have passed it down to his mother and further down to him.

When Mendel crossed a plant homozygous for round seeds to another plant homozygous for wrinkled seeds, he found that all the progeny had round seeds. How is this explained? a) The parent that was homozygous for round seeds underwent self‑pollination. b) Segregation of alleles in the two parents produced gametes with both alleles. c) The progeny were homozygous for the allele for round seeds. d) The allele for round seeds is dominant to the allele for wrinkled seeds. e) The allele for round seeds is recessive to the allele for wrinkled seeds.

D) the allele for round seeds is dominant to the allele for wrinkled seeds Homozygous is one trait: Ex) BB or bb

Inheritance Patterns

For each inheritance pattern, first determine the possible genotypes of the parents. When the genotypes of the parents have been determined, place the genotypes in a Punnett square to predict the phenotypes of the children. In the examples, T will be used for the dominant allele, and t for the recessive allele. For an autosomal recessive trait, an affected individual will be homozygous and recessive. The father exhibits the disorder and will have the genotype tt. The mother does not have the disorder and could be either heterozygous, Tt, or homozygous dominant, TT. Because the disorder is rare, the mother is unlikely to have an allele for the disorder and be heterozygous, Tt. Her genotype is more likely to be homozygous dominant, TT. Place these genotypes in a Punnett square. All of the children are heterozygous, so none of the children will express the disorder. For the autosomal dominant trait, the mother must be homozygous and recessive, tt, to not express the disorder. The father, though, could be heterozygous, Tt, or homozygous dominant, TT. Because the allele is rare, it is unlikely the father will be homozygous for the allele. The father more likely possesses only one allele for the disorder and is heterozygous, Tt. Place these genotypes in a Punnett square. One half of the children express the disorder. These phenotypes are distributed equally between the girls and boys. For an X‑linked, recessive disorder, the father will have only one allele since he has only one X chromosome. The father has the recessive allele and the genotype, Xt Y. The mother has two alleles, one of which will be a dominant allele, and is either homozygous dominant or heterozygous. Because the allele that causes the disorder is rare, the mother is more likely homozygous, XTXT, than heterozygous, XTXt. Place these genotypes in a Punnett square. The only children who have the X‑linked, recessive allele are female, but because they also have a dominant allele, they do not have the disorder. The females are carriers for the disorder. The male children have only the dominant allele and do not have the disorder. For an X‑linked dominant disorder, the father has the disorder and has a single dominant allele, XTY. The mother does not have the disorder and thus must have the homozygous, recessive genotype, XtXt. Place these genotypes in a Punnett square. All of the male children have the recessive allele, Xt Y, and do not have the disorder. All of the female children are heterozygous for the trait and exhibit the disorder. For a Y‑linked disorder, the males have the gene for the disorder on their Y chromosome. Females cannot express this disorder. Whether or not a male exhibits the disorder depends entirely on whether his father had the disorder. The father has the allele that leads to the disorder on his Y chromosome, XYT. Place the parental genotypes in a Punnett square. All of the male children have inherited the allele for the disorder from their father and exhibit the disorder. None of the female children have the disorder.

Cells are categorized as either prokaryotic or eukaryotic. Classify the characteristics as being unique to prokaryotic cells, common to both cell types, or unique to eukaryotic cells. Move each description into the correct category. -relatively small, typically less than 10 um in size -contain DNA -lack a true cytoskeleton -nuclear membrane surrounds DNA -use ribosomes to synthesize proteins

For only prokaryotic cells: -relatively small, typically less than 10um in size -lack a true cytoskeleton For only eukaryotic cells: -nuclear membrane surrounds DNA For both types: -contain DNA -use ribosomes to synthesize proteins

Determine the diploid number of chromosomes (2⁢n) in this species of plant.

In order to determine the diploid number of chromosomes in this species of plant, first determine the number of centromeres present in a cell that also contains homologous pairs of chromosomes. Because each chromosome possesses a single centromere, the centromere is particularly useful in determining the number of chromosomes. The images depict the spindle fibers having already attached to the centromeres just prior to cytokinesis. The second image (middle image) depicts the cell during mitosis, after DNA duplication, but just prior to cell division. Thus, each daughter cell will be diploid and so there are six chromosomes in a diploid cell of this plant species. The first image depicts the cell during anaphase I of meiosis, due to the separated homologous pairs of chromosomes. Thus, there are six centromeres and also six chromosomes accordingly. A sister chromatid is one of the two identical chromosomes formed during replication. Each chromatid is considered a DNA molecule, so there are 12 DNA molecules in total. The second image depicts the cell with the sister chromatids already separated after the DNA has been duplicated from six to 12 chromosomes. Accordingly, there are also 12 centromeres present, indicating that there are 12 chromatids and, thus, 12 total DNA molecules. Because of the number of chromosomes, the separating sister chromatids indicate that the cell is depicted during anaphase of mitosis. The third image also depicts the cell with the sister chromatids already separated. However, there are only six total chromosomes present in the cell. There are also six centromeres and, therefore, six chromatids and six total DNA molecules. The presence of six chromosomes indicates that no homologs are present and, thus, the cell is depicted during anaphase II of meiosis. As the cell completes cytokinesis, the resulting two daughter cells will each contain 3 chromosomes and, thus, will be gametes (𝑛).

Suppose Melissa started growing four o'clock plants, Mirabilis jalapa, in her garden. She noticed that the plants have white, green, or patchy white and green (variegated) leaves. Melissa would like to have only four o'clock plants with variegated leaves, so she crosses a few of her plants to see which crosses produce offspring with variegated leaves. She knows that leaf color is determined by the color of the chloroplasts. For each cross, determine the offspring phenotypes that could be observed. Each cross may produce plants with one or more different phenotypes. Parent Crosses: -green female x green male -white female x variegated male -variegated female x green male -variegated female x white male

Offspring phenotypes for each cross: green female x green male --green white female x variegated male --white variegated female x green male --variegated, green, and white variegated female x white male --variegated, green, and white This is a maternal effect question. Four o'clock plants have a variety of possible flower and leaf colors and patterns. In particular, their leaves can be white, green, or patches of white and green, which is called variegated. Leaf coloration is maternally inherited because the genes that code for leaf color are located in the chloroplast DNA, and plastid DNA is maternally inherited. Therefore, the offspring will express the same phenotype as the mother's chloroplasts, and the father does not contribute to the leaf‑color phenotype. If the mother has more than one chloroplast phenotype, then the offspring could be variegated, green, or white, depending on how the chloroplasts segregated. Green leaves are produced when chloroplasts within the leaf cells contain the green pigment chlorophyll, whereas white leaves are produced when the chloroplasts are devoid of chlorophyll. Variegated leaves are produced when leaves contain both chloroplast types. In this example, a green female crossed with a green male would produce offspring that have green leaves because the mother has white leaves. Similarily, a white female crossed with a variegated male would produce offspring that have white leaves. A variegated female crossed with a green male, or a variegated female crossed with a white male, would produce offspring that could have white, green, or variegated leaves.

In humans, there are four blood types: A, B, AB and O. Each parent can give their offspring one of three blood type alleles, IA (coding for phenotype A), IB (coding for phenotype B), or neither, which is expressed as i (coding for phenotype O). This trait displays codominance. Tina has type O blood and her sister Rosa has type AB blood. What are the blood genotypes their parents must have? Select the blood genotype for Parent One a) ii b) IAi c) IBIB d) IAIA e) IAIB f) IBi Select the blood genotype for Parent Two. a) IAIA b) IAi c) IBIB d) IAIB e) IBi f) ii

Parent one: b) IAi Parent two: e) IBi

Coat color in cats is determined by genes at several different loci. At one locus on the X chromosome, one allele (X+) encodes black fur and another allele (Xo) encodes orange fur. Females can be black (X+X+), orange (XoXo), or a mixture of orange and black called tortoiseshell (X+Xo). Males are either black (X+Y) or orange (XoY). Bill has a female tortoiseshell cat named Patches. One night, Patches escapes from Bill's house, spends the night out, and mates with a stray male. Patches later gives birth to the following kittens: one orange male, one black male, two tortoiseshell females, and one orange female. What are the genotypes of Patches, the stray male, and the kittens?

Patches: X+Xo stray male: XoY orange male kitten: XoY black male kitten: X+Y tortoiseshell female kitten: X+Xo orange female kitten: XoXo Patches is a tortoiseshell female, which means she must be heterozygous, and each of the kitten's genotypes are known based on their given phenotype. The genotypes of the male kittens (X+Y and XoY) do not provide information about the stray male's genotype, because they get their X alleles from Patches. The genotype of the tortoiseshell female kitten (X+Xo) also does not provide information because either one of her X alleles could have come from Patches. However, the only way to produce an orange female kitten (XoXo) is if the kitten's father had the orange allele. Therefore, the stray male cat must be XoY.

Use the pedigree to determine the mode of inheritance of the disorder.

Pedigree mapping is a long-standing tool in genetics used to determine information about the mode of inheritance of important traits and disorders. The pedigree provided shows a family with the autosomal recessive disorder Phenylketonuria (PKU). Individuals with PKU cannot metabolize the amino acid phenylalanine. Whereas a strictly regimented diet can control the disorder, individuals who do not avoid phenylalanine in their diets can develop mental and cognitive disabilities. Couples who have relatives with the disorder may wish to do pedigree mapping to assess the likelihood their prospective offspring will have PKU. The type of dominance is determined by looking at the pattern of affected individuals. Generations I, individuals 1 and 2, and II, individuals 3, 4, 5, and 6, show no affected individuals, so the type of dominance must be recessive. Only recessive disorders can skip generations. The chromosomal type is determined by looking at the relationship between the sexes in the pedigree. If the trait were sex-linked recessive, then both parents of affected daughters would have to have an affected allele, thus making the father also affected due to hemizygosity. Since individual 7 is affected but her father is not, the chromosomal type cannot be sex-linked.

Suppose that allele D produces big ears and allele d produces normal ears. A heterozygous male with the Dd genotype mates with a homozygous female with the dd genotype. Big ears is an autosomal dominant trait expressed with 40% penetrance. Determine the probability that their offspring will have big ears. Report your answer as a percentage without decimal points.

Probability: 20% penetrance=probability of showing abnormal trait / predicted ratio of genotype for abnormal trait You will have to rearrange the equation to solve for the abnormal trait probability. In order to identify the genotype that makes big ears, use a punnet square. Square with the D chromosome are the only ones that can make big ears and thus be exposed to abnormalities. There are 2 Dd's out of 4 total so 2/4 Multiply: 2/4 * 40% = 20%

Identify each stage of M phase.

Prophase: Spindles start to form while chromosomes rest in their nucleus Metaphase: Chromosomes lined up in the middle in an "X" formation Anaphase: Chromosomes pulled apart into triangles and dragged to opposite ends of cells Telophase: Cells about to split Cytokinesis: Cells split

In beetles, the gene for color and the gene for pattern are located on the same chromosome and exhibit recombination. The lined pattern allele (l) is recessive to the solid unpatterned allele (L), and the aqua color allele (b) is recessive to the blue color allele (B). Geneticists cross a homozygous lined aqua beetle with a homozygous solid blue beetle. Then, they carry out a testcross using the F1 progeny. The parental chromosomes for the testcross are given. Identify all possible gametic haplotypes each parent can form.

The Heterozygous parent has two different alleles, thus you can mix and match (Punnet square) all the combinations The Homozygous male has the same allele on both sides, only allowing for one combination and one combination only.

Label each phase of the cell cycle with the appropriate name or description.

The cell cycle is the process by which cells grow, replicate their DNA, and undergo cell division. The cycle is divided into four main stages called G1 (gap 1), S (synthesis), G2 (gap 2), and M (mitosis). Some cells also enter into a special phase called G0 . The G1 , S, and G2 phases are together known as interphase. G1 phase occurs after the formation of a new cell and is the first phase of interphase. During G1 , the cell grows in size and synthesizes mRNA and protein. During G1 , the cell must pass a checkpoint called the restriction point (R), after which the cell becomes committed to DNA replication and cell division. If the cell passes R, it eventually enters S phase, during which it replicates its DNA. After S phase, the cell enters G2 phase, during which the cell grows further, synthesizes more mRNA and protein, and prepares for mitosis. G2 is the last phase of interphase. After passing another checkpoint, the cell enters M phase, the cell division phase comprising both mitosis and cytokinesis. If a cell does not pass R, it instead enters G0 phase, also known as a resting or quiescent phase. Cells in G0 do not divide, prepare to divide, or replicate their DNA, although they are still alive and functional. Some cells in G0 can reenter the cell cycle at G1 when stimulated by injury, starvation, or signals from other cells.

Know all organelles and their function in a eukaryotic cell

The plasma membrane is a living barrier between the interior of the cell and the outside environment. It regulates the movement of water and solutes into and out of the cell. Genetic information is stored in the nucleus, which has a double membrane. Ribosomes are complexes of RNA and proteins on which translation from RNA to protein takes place. They are not enclosed by membranes within the cell. The endoplasmic reticulum, both smooth and rough, is responsible for protein transport and some protein processing. Ribosomes bind to rough endoplasmic reticulum, giving it a bumpy appearance. The Golgi apparatus modifies and "packages" proteins for transport out of the cell or to other parts of the cell. Transport vesicles containing the packaged proteins pinch off from the Golgi apparatus and move to lysosomes, storage vesicles, or the plasma membrane. Note the vesicles that pinch off the Golgi apparatus in the diagram. Lysosomes are organelles that contain digestive enzymes, which can digest food particles taken into the cell or break down worn‑out cellular components. Lysosomes can also release their enzymes outside of the cell, where the enzymes break down extracellular material. Much of the cell's energy metabolism takes place in the mitochondrion (plural: mitochondria), which is another organelle with a double membrane. A single eukaryotic cell contains many mitochondria. Mitochondria are usually depicted as cylinders with a highly folded inner membrane. They are often found in association with cytoskeletal microtubules. The cytoskeleton consists of fibers that enable movement and transport within the cell.

Match each of the examples with the corresponding type of sex‑related inheritance. -The trait for pattern baldness is expressed more prominently in males than in females. -One form of heritable icthyosis, a skin disorder, is trasmitted via the X chromosome and shows unequal expression in males and females. -Duchenne muscular dystrophy reduces survival, and expression is determined by sex chromosome genes. -Only female humans lactate. sex‑linked nonlethal sex‑influenced sex‑lethal sex‑limited

The trait for pattern baldness is expressed more prominently in males than in females. --sex-influenced One form of heritable icthyosis, a skin disorder, is trasmitted via the X chromosome and shows unequal expression in males and females. --sex-linked nonlethal -Duchenne muscular dystrophy reduces survival, and expression is determined by sex chromosome genes. --sex-linked lethal -Only female humans lactate. --sex-limited Sex‑linked traits, which refer to characteristics that are determined by genes that are present on the sex chromosomes, can be dominant or recessive. X-linked ichthyosis is an example of a sex-linked trait. The gene for ichthyosis is located on the X chromosome. It is caused by mutations in the steroid sulfatase enzyme that helps in the formation of a healthy skin. Individuals with ichthyosis have dry and scaly skin. X-linked ichthyosis does not affect survival and is therefore nonlethal. Some sex‑linked traits are lethal. Duchenne muscular dystrophy is an example of a sex‑linked trait that frequently results in death. The recessive, mutant allele is found on the X chromosome and therefore, a greater number of males are affected compared to females. Effects of Duchene muscular dystrophy include loss of skeletal muscle function, inability to walk, and heart failure. Most individuals who are affected die by age 20. Sex‑limited and sex‑influenced traits affect features that are specific to the sex of an individual. In contrast to sex‑linked traits, these traits are expressed by genes that are present on autosomes. A sex‑limited trait is limited to one sex. Both sexes may carry the gene for the trait, but only one sex expresses the phenotype. For example, lactation or milk production is only seen in female mammals, such as humans. A sex‑influenced trait is a trait that is seen in both sexes. Sex‑influenced traits are often influenced by sex hormones that alter the extent of expression between the sexes. Pattern baldness is a trait seen in both males and females. Males who are heterozygous for baldness become bald at an early age, whereas heterozygous females do not show much hair loss. Both males and females who are homozygous for the baldness allele become bald, but baldness appears at an earlier age in males than in females.

Match each phenotype description to its corresponding sex chromosome genotype in humans. Categories: XX with SRY on X XY with SRY deleted XXX XXY XYY Options: -phenotypically male but karyotype indicates presence of only X chromosomes -phenotypically female but karyotype indicates presence of both sex chromosomes -phenotypically female with some abnormalities and overexpression of X chromosomal genes -phenotypically male with sterility and hypogonadism -phenotypically male with an increase in average stature

XX with SRY on X: -phenotypically male but karyotype indicates presence of only X chromosomes XY with SRY deleted: -phenotypically female but karyotype indicates presence of both sex chromosomes XXX: -phenotypically female with some abnormalities and overexpression of X chromosomal genes XXY: -phenotypically male with sterility and hypogonadism XYY: -phenotypically male with an increase in average stature The presence of sex chromosomes in humans is essential for normal development. A typical female has two X chromosomes in each cell, whereas a male has one X chromosome and one Y chromosome. Normally, through a process called X‑inactivation, one of the X chromosomes in a female somatic cell is inactivated and forms a Barr body. This process ensures that only one X chromosome is active at any time. Therefore, a female with three or more X chromosomes does not exhibit a highly abnormal phenotype since X‑inactivation inactivates all X chromosomes but one. There are regions of the X chromosome that do not become inactivated, have the ability to recombine, and are functional throughout the lifetime of an individual. These regions are called pseudo‑autosomal regions. Individuals with Klinefelter syndrome have two X chromosomes and one Y chromosome. Even though one of the X chromosomes is inactive, the genes in the pseudo‑autosomal regions of both X chromosomes will interact with genes on the Y chromosome causing a decrease in testosterone levels and sterility due to the increased dosage of X‑linked genes. One of the most important genes on the Y chromosome is the SRY, or sex‑determining region Y, gene, which dictates the formation of testes as well as other male characteristics. The SRY gene is located in the pseudo‑autosomal region of the Y chromosome. If an XY individual is missing the SRY gene, the individual will be phenotypically female. The SRY gene has also been known to translocate to X chromosomes as well as to autosomes during recombination. Therefore, XX individuals that have the SRY gene on one of their X chromosomes or an autosome will develop testes and become phenotypically male. Although the Y chromosome is essential to male development, it does not contain many genes important to the overall health of an individual. Therefore, males whose cells contain an extra Y chromosome are phenotypically normal with a slight increase in stature thought to be due to interactions to the pseudo‑autosomal regions. Individuals with Turner syndrome have an XO genotype and do not contain a Y chromosome. These individuals are also viable and will develop as a female. If there is an SRY gene on an autosomal chromosome in an XO individual, however, this individual will develop testes and male phenotypic characteristics.

Hemophilia is called "the royal disease" because many European royal families had members with the condition. Hemophilia is a recessive, X‑linked disorder. Queen Victoria was unaffected by hemophilia, but was a carrier of the hemophilia gene (XHXh). Suppose Queen Victoria's husband, Prince Albert, was affected with hemophilia (XhY). What is the percent probability that a son of Queen Victoria and Prince Albert would be unaffected by hemophilia? What is the percent probability that a daughter or son of Queen Victoria and Prince Albert would be affected by hemophilia? What is the percent probability Queen Victoria and Prince Albert would have two affected females?

a) 50% b) 50% c) 6.25% Make a Punnet Square. For a), look for males (XY) that bear the Xh chromosome and divide them over the total males. 1/2 = 50% For b), look for any children bearing the Xh chromosome as the only and main X chromosome. If they have the dominant XH chromosome they will not inherit hemophilia. Divide that number by total children. 2/4 = 50% For c), find females with only the sex linked recessive trait Xh. The genotype for this problem would be XhXh. Divide that by total number of children, so 1/4. Do this again for the second child, 1/4. To calculate probability multiply the 1/4 and 1/4 together, then multiply by 100.

Which statement explains why the recombination frequency between two genes is always less than 50%? a) Genes with a recombination frequency near 50% are unlinked and have an equal likelihood of being inherited together or separately. b) Each F1 gamete always has one set of alleles from each parental gamete. c) Recombination cannot be more than 50% because chromosomes are only 50 map units in length. d) There is always an equal ratio of both the parental and recombinant gametes in an F1 generation.

a) Genes with a recombination frequency near 50% are unlinked and have an equal likelihood of being inherited together or separately Recombination is the crossing over of genes during meiosis. For example, in the cross PPRR × pprr, the parental gametes are PR and pr. The outcome of this cross will be F1 progeny that all have the genotype PpRr, with the alleles PR on one chromosome and pr on the other chromosome. The possible F1 gametes are PR, pr, Pr, and pR. In the gametes PR and pr, no recombination has occurred, and these gametes have the same genotype as the parental gametes. However, in the gametes Pr and pR, meiotic recombination has occurred, and thus, the genotype of these gametes are different from the parental gametes. In this example, if an equal ratio of F1 gametes are generated, then the two genes assort independently. Independent assortment leads to equal ratios of both the parental gametes, PR and pr, and recombinant gametes, Pr and pR. However, only nonlinked genes, which include genes that are far apart on the same chromosome or located on separate chromosomes, assort independently. Taken together, only nonlinked genes will produce equal ratios of each gamete type and result in 50% recombinant and 50% parental gametes. Thus, the recombination frequency between two genes is always less than 50% because genes that assort independently will always generate an equal ratio of 50% parental and 50% recombinant gametes. When the recombination frequency between two genes is less than 50%, the two genes are found on the same chromosome and display genetic linkage. When two genes are linked, they are more likely to be inherited together and more parental gametes are produced than recombinant gametes. Recombination frequencies can be used to generate recombination-based maps, which provide an arbitrary map that illustrates the order of genes on a chromosome and the approximate distance between two genes. To generate recombination-based maps, recombination frequencies are converted to map units. For example, two genes with a recombination frequency of 4% would be designated as being four map units apart. Recombination frequencies provide a relative measure of gene distance but do not provide physical gene distance. Because map units are arbitrary units, the length of chromosomes is not restricted to 50 map units.

In fruit flies, long wings (W) are dominant over short wings (w), and red eyes (R) are dominant over orange eyes (r). Each individual possesses two alleles for each trait. If a fly that is homozygous dominant for both traits is crossed with a fly that is homozygous recessive for both traits, what is the predicted genotype of the offspring? a) WwRr b) short wings and orange eyes c) long wings and red eyes d) WwRr, wwRr, Wwrr, and wwrr e) WWRR

a) WwRr Note that we are asking for genotype not phenotype so want the allelic configuration not the physical description. Homozygous dominant = WR Homozygous Recessive = wr Use a punnet square to identify the only obtainable genotype from this dihybrid cross

In Drosophila melanogaster, white eye is an X‑linked recessive trait, and red eye is an X‑linked dominant trait. Assume that the flies have a diploid set of autosomes (2A). Allele Xw codes for white eye, and allele X codes for red eye. Determine each of the genotypes with the expected sex and eye color. -XX -XwX -XwY -XO -XwXY -XwXwX

a) XX - female - red b) XwX - female - red c) XwY - male - white d) XO - male - red e) XwXY - female - red f) XwXwX - metafemale - red In essentiality count the # of X and Xw chromosomes and divide them by 2 (for the 2A autosomes). 1.0 = female 0.5 = male over 1.0 = metafemale under 0.5 = metamale between 0.5 and 1.0 = intersex In more detail: Sex in Drosophila melanogaster is determined by the ratio of X chromosomes to autosomes (A). Unlike humans, the presence of a Y chromosome in Drosophila does not produce males flies. Drosophila can carry diploid (2⁢n) or triploid (3⁢n) sets of chromosomes. X chromosome numbers also can vary due to nondisjunction of X chromosomes during meiosis. This produces a wide range of X to A chromosome ratios. An X to A ratio of 1.0 produces females, 0.5 produces males, above 1.0 produces metafemales, below 0.5 produces metamales, and between 0.5 and 1.0 produces intersexes with features of both sexes. In this scenario, there are diploid (2A) sets of autosomes. The XX genotype has an X to A ratio of 2:2, or 1.0, and has no alleles for white eyes. Therefore, XX flies are females with red eyes. For the XwX genotype, the X to A ratio is 2:2, or 1.0, and the recessive Xw allele for white eyes is masked by the dominant X allele for red eyes, resulting in females with red eyes. Similarly, the XwXY genotype has an X to A ratio of 2:2, or 1.0. Since the Y chromosome does not impart maleness, the XwXY flies are also females with red eyes. For the XwXwX genotype, the ratio of X to A is 3:2, or 1.5, and they carry one dominant allele for red eyes. Therefore, the XwXwX genotype produces metafemales with red eyes. Metafemales are rarely viable. For the XwY genotype, the X to A ratio is 1:2, or 0.5, and the flies carry only the Xw allele for white eyes. Therefore, the XwY flies are males with white eyes. For the XO genotype, the X to A ratio is also 0.5. Since these flies carry the dominant red allele, XO males have red eyes. XO males are sterile, whereas XY males are fertile. The difference in fertility is due to the absence of a Y chromosome in XO males. The Y chromosome is not required for determining sex, but it is required to produce fertile males.

The platypus is a sexually reproducing mammal in which the sexes are separate. Unlike other mammals, the male has five X and five Y chromosomes. The female has 10 X chromosomes. Select the term that describes the organization of the sexual organs in the platypus. a) dioecious b) monoecious c) intersex d) hermaphroditic

a) dioecious The terms dioecious, monoecious, and hermaphroditic refer to the distribution of the reproductive organs between different individuals in sexually reproducing species. A dioecious individual is either male or female, but not both. Dioecious individuals belong to species in which the sexes are always separate. In monoecious individuals, male and female reproductive structures occur in the same individual. In animals, the terms monoecious and hermaphroditic are synonymous. In plants, there is a subtle difference between these terms. A hermaphroditic plant has flowers with both male and female reproductive structures on the same flower or cone, whereas a monoecious plant has flowers with only male reproductive structures on certain parts of the plant and flowers with only female reproductive structures on other parts of the same individual.

Consider the set of crosses for hypothetical genes that control eye color and tail length in mice. Diane is working with mice with recessive mutations for the genes that control light eye color (b) and short tail length (t). She knows that these genes display genetic linkage and are found on chromosome III. In her work, she crosses a true‑breeding male with light eyes and a long tail (bbTT) and a true‑breeding female with dark eyes and a short tail (BBtt). She then crosses the resulting heterozygous progeny (BbTt) together in a dihybrid cross. The number of animals of each phenotype of this second cross is shown. dark‑eyed, short‑tailed (𝐵𝐵𝑡𝑡)=24 dark‑eyed, long‑tailed (BbTt or BBTT)=55 light‑eyed, long‑tailed (bbTT)=25 light‑eyed, short‑tailed (bbtt)=2 What is the most likely explanation for why Diane sees light‑eyed, short‑tailed (bbtt) progeny in this cross? a) recombination b) random mutagenesis c) horizontal gene transfer d) independent assortment

a) recombination Do the Punnet Square to reveal the possible progenys and notice the lack of bbtt progeny. As can be seen in the figure, the gametes produced by independent assortment will lead to expected phenotypes where 25% of the offspring are dark‑eyed, short‑tailed (BBtt), 50% are dark‑eyed, long‑tailed (BbTt), and 25% of the offspring are light‑eyed, long‑tailed (bbTT). However, in Diane's cross, she sees some progeny that are light‑eyed, short‑tailed (bbtt). These progeny are produced by recombination between homologous chromosomes. Recombination occurs during prophase I of meiosis and results in two homologous chromosomes exchanging genetic information. This process generates new allele combinations along the same chromosome. During Diane's cross, some of the gametes made by the heterozygotes were the product of recombination (bt or BT). These gametes resulted in the light‑eyed, short‑tailed (bbtt) progeny she saw in her cross. Recombination would have also produced progeny homozygous for both dominant alleles (BBTT), but these progeny would have had the same phenotype as heterozygotes (BbTt). Although random mutagenesis could have also produced new alleles, the rate of random mutagenesis is very low and would not explain why Diane sees several light‑eyed, short‑tailed (bbtt) progeny. To isolate events that would have occurred due to random mutagenesis, Diane may have needed to screen over a thousand offspring. Horizontal gene transfer is a means of exchanging genetic information used by bacteria and would not explain the result of the cross.

The principle of independent assortment involves at least how many different gene pairs? a) 3 b) 2 c) 5 d) 1 e) 4

b) 2

Consider two cells, where one cell is haploid and the other is diploid. Which of the following differences between the two cells is observed during prophase of mitosis? a) In the haploid cell, only one pair of homologous chromosomes are present, whereas in the diploid cell, two pairs of homologous chromosomes are present. b) In the haploid cell, only one pair of DNA molecules are present, whereas in the diploid cell, two pairs of DNA molecules are present. c) In the haploid cell, only one pair of centrosomes are present, whereas in the diploid cell, two pairs of centrosomes are present. d) In the haploid cell, only one pair of telomeres are present, whereas in the diploid cell, two pairs of telomeres are present.

b) In the haploid cell, only one pair of DNA molecules are present, whereas in the diploid cell, two pairs of DNA molecules are present. The haploid chromosome number, represented by n, is equal to the number of unique chromosomes in the nucleus of a cell. A diploid cell contains 2n chromosomes, or a pair for each n. During interphase, each chromosome is duplicated to form a pair of DNA molecules attached together at the centromere. Chromosomes can be visually observed after condensation occurs during prophase. A haploid cell contains only one copy of each unique chromosome. In the question above, the haploid cell (n=1) contains only one unique chromosome. A diploid cell (2n=2) contains a pair of homologous chromosomes, with one homolog derived from each parent. Thus, the haploid cell contains one pair of DNA molecules, whereas the diploid cell contains two pairs of DNA molecules in prophase. Mitosis ensures that the duplicated chromosomes, in the form of sister chromatids, are distributed evenly to the daughter cells. Additional cellular structures are needed to facilitate the mitotic process. In both haploid and diploid cells, one of the early events in prophase involves the migration of two pairs of centrioles to opposite ends of the cell. Centrioles are situated outside the nuclear envelope in an area of the cytoplasm called the centrosome. The centrosome facilitates the formation of the mitotic spindle. Both haploid and diploid cells contain the same number of centrosomes. Maintaining the integrity of the chromosomes, generation to generation, requires specialized sequences called telomeres. Each chromosome has a pair of telomeres, with one at each end of the chromosome. Therefore, each chromosome contains two pairs of telomeres. The haploid cell contains two pairs of telomeres, whereas the diploid cell contains four pairs of telomeres.

What information can the chi‑square goodness‑of‑fit test provide? a) the genotypes of the progeny in the test cross b) how well the observed results of a genetic cross fit the expected values c) the genotypes of the two parents in the test cross d) how well the genetic cross was carried out e) that the results of the genetic cross are correct

b) how well the observed results of a genetic cross fit the expected values

In a hypothetical mouse species, brown fur (B) is completely dominant to white fur (b), and long fur (L) is completely dominant to short fur (l). If two mice heterozygous for both traits mate and produce a litter of pups, what is the probability that an individual pup will have white, short fur? a) 3/16 b) 1/4 c) 1/16 d) 9/16 e) 3/4

c) 1/16 Use a Punnet Square to determine the answer. in a dihybrid cross between two heterozygous parents (two traits ex - Bb or Ll) result in a ratio of: 9:3:3:1

A maternal effect can cause the offspring phenotype ratio to depart from that of classic Mendelian inheritance. In a species of moth, the dominant allele N codes for brown eyes and recessive allele n codes for red eyes. If an Nn female with brown eyes mates with an Nn male, what is the eye color phenotypic ratio of their offspring? a) 0, brown eyes : 4, red eyes b) 3, orange eyes : 1, red eyes c) 4, brown eyes : 0, red eyes d) 3, brown eyes : 1, red eyes

c) 4 brown eyes: 0 red eyes The term maternal effect means that offspring will express the phenotype determined by their mother's genotype regardless of their own genotype. In this example, N is the dominant allele, and the Nn female has brown eyes. Therefore, even if she mates with an Nn male, all of their offspring will have brown eyes, and the phenotypic ratio of their offspring would be 4, brown eyes : 0, red eyes. If the mother had the genotype nn and expressed red eyes, all of her offspring would have red eyes and the offspring phenotypic ratio would be 0, brown eyes : 4, red eyes. However, if this gene did not exhibit maternal effect, the phenotypic ratio of their offspring would be 3, brown eyes : 1, red eyes. If this gene had co-dominant alleles and did not exhibit maternal effect, the phenotypic ratio of their offspring may be 3, orange eyes : 1, red eyes.

Fur color in a species of mouse is controlled by a single gene pair. BB animals are black and bb animals are white. Bb animals have gray fur and each hair is gray. What type of interaction is being shown by the two alleles in heterozygous animals? a) codominance b) complementation c) incomplete dominance d) complete dominance e) epistasis

c) incomplete dominance

A boy has blood type MN with a genotype of LMLN. His red blood cells possess both the M antigen and the N antigen. What is the relationship between his two alleles for this gene? a) incomplete dominance b) codominance c) complete dominance d) epistasis e) complementation

codominance

Which statement is the definition of a map unit (centimorgan)? a) It is 1/100 (0.01) of the percent chance of a crossover between two loci. b) It is the physical distance between neighboring genes on a chromosome. c) It is the distance equal to 1000 base pairs of DNA. d) It is the percent chance of a crossover between two locations on a chromosome. e) It is the distance equal to 1/1000 of the length of the chromosome.

d) It is the percent chance of a crossover between two locations on a chromosome The map unit (m.u.), also known as the centimorgan (cM) in honor of the geneticist Thomas Hunt Morgan, is a unit of recombination frequency. A map unit is defined as a one percent probability of crossing over between two loci on a chromosome. For example, if genes A and B recombine 20% of the time, then genes A and B are separated by 20 m.u. or 20 cM. Despite its name, a centimorgan is not 1/100 of the percent chance of recombination, but instead, it is derived from a one percent (1%, or 0.01) chance of recombination. Importantly, the map unit is not a unit of physical distance. Recombination frequency does not directly correlate to physical distances between genes, the numbers of base pairs of DNA, or fractions of a chromosome's length. Although the relationship between recombination frequency and physical distance can be estimated, the conversion differs between species, chromosomes, and even regions within the same chromosome.

What information about recombination frequencies enables scientists to create linkage maps? a) The recombination frequency between two genes reveals the nucleotide sequence of the two genes. b) The recombination frequency between two genes is equal to the distance in nanometers between the two genes. c) The higher the recombination frequency, the closer two genes are on a chromosome. d) The recombination frequency is proportional to the distance between the two genes.

d) The recombination frequency is proportional to the distance between the two genes.

Suppose a group of researchers analyzed a new species of flowering plant. In this species, flower color is determined by two genes. They crossed a truebreeding red‑flowered stock with a truebreeding white‑flowered stock. They then crossed the F1 offspring together and analyzed the next generation. Upon examining the flower color, they found that the F2 generation contained 97 purple, 24 red, and 8 white flowers. What genetic term best describes the inheritance pattern of flower color for this species? a) genes in one pathway b) recessive epistasis c) no gene interaction d) dominant epistasis

d) dominant epistasis Make sure you know ur ratios, this ones 12:3:1

What is the physical appearance or manifestation of a characteristic called in genetics? a) independent assortment b) alleles c) genotype d) phenotype e) wild type

d) phenotype i stg if you miss this bro

Suppose a group of researchers analyzed a new species of flowering plant. In this species, flower color is determined by two genes. They crossed a truebreeding red‑flowered stock with a truebreeding white‑flowered stock. They then crossed the F1 offspring together and analyzed the next generation. Upon examining the flower color, they found that the F2 generation contained 178178 purple, 60 red, and 80 white flowers. What genetic term best describes the inheritance pattern of flower color for this species? a) no gene interaction b) dominant epistasis c) genes in one pathway d) recessive epistasis

d) recessive epistasis In this case, flower color displays a recessive epistasis inheritance pattern. Recessive epistasis typically displays a 9:3:4 phenotype ratio. The phenotypic classes that would normally make up on of the 3/16 and the 1/16 components of the ratio are combined in recessive epistasis. The 3/16 class appears phenotypically identical to the 1/16 class because the 3/16 class is epistatically masked. In this case, the white gene is recessive epistatic to the red gene. Epistasis is a genetic interaction wherein one gene masks another gene so that a double mutant displays the phenotype of one gene. The gene that masks the effects of the other gene is called epistatic. If the F2 ratio is the standard Mendelian F2 ratio of 9:3:3:1, there is no genetic interaction. Different types of epistasis result in different modifications to the expected 9:3:3:1 F2 phenotype ratio. Dominant epistasis typically displays a 12:3:1 phenotype ratio. Dominant epistasis has a 12/16 class that is comprised of the 9/16 and one of the 3/16 classes. The 3/16 class appears phenotypically identical to the 9/16 class because the 3/16 class is epistatically masked. Two genes in one pathway will typically display a 9:7 phenotype ratio. This ratio occurs when the double mutant shares the same phenotype as both of the individual mutants. Therefore, the 9/16 class are the wildtype individuals and the 7/16 class contains two different single mutant classes and the double mutant class.

In humans, mitochondrial genetic disorders are inherited from only the mother. The severity of such diseases can vary greatly, even within a single family. What form of inheritance does this represent? a) sex‑limited inheritance b) sex‑influenced inheritance c) genetic maternal effect d) genomic imprinting e) cytoplasmic inheritance

e) cytoplasmic inheritance

An individual possesses two alleles at a locus and these two alleles separate when gametes are formed, one allele going into each gamete. This genetic concept is known as the: a) concept of dominance. b) chromosome theory of heredity. c) reciprocal cross. d) principle of independent assortment. e) principle of segregation.

e) principle of segregation

Ann's family has a history of cystic fibrosis, a recessive genetic disease. In the pedigree, family members who are afflicted with the disease are shown in red. Members who are unafflicted may or may not be carriers. Which of the given family members can be identified definitively as unafflicted carriers of cystic fibrosis?

grandmother, sister, and aunt's husband In a recessive genetic disease, an afflicted individual has a homozygous recessive genotype (aa). An individual with a heterozygous genotype (Aa) is a carrier who is not afflicted by the disease. All definitive carriers can be identified by looking at the parents and children of afflicted individuals. Individuals with the disease (aa) must have inherited a recessive allele from each parent. Therefore, each parent must have at least one recessive allele. If a parent does not have the disease, then the parent must be a heterozygous carrier (Aa). Similarly, if an afflicted individual has a child, then a recessive allele must have been donated to the child. If the child is not afflicted, then the child must be a carrier. In Ann's family, the afflicted individuals (aa) are Ann's father, aunt, and one of her male cousins. Each unafflicted parent of these individuals must have donated a recessive allele. Hence, Ann's grandparents and aunt's husband must all be carriers (Aa). Now, look at the children of the afflicted individuals to identify the remainder of the carriers. Of the afflicted individuals, Ann's father and aunt have children. Ann's father can only donate a recessive allele to Ann and her sister. Ann's aunt can only donate a recessive allele to each of her children. Therefore Ann, her sister, and her two unafflicted cousins must be carriers. Some other family members may be carriers, but it is not definitive. Both Ann's mother and one uncle are unafflicted, so their genotypes may be Aa or AA. However, because neither individual has parents nor children who are afflicted, there is not enough information to definitively identify whether either individual is a carrier. Lastly, Ann's father, aunt, and afflicted male cousin are not carriers because they have cystic fibrosis (aa).