GENETICS FINAL!!!! BIO 2300 Pt. 2

A cross was made between a homozygous wild-type Drosophila female and yellow-bodied Drosophila male. All of the resulting offspring were phenotypically wild-type. Offspring of the F2 generation had the following phenotypes: Sex: M ; M ; F Pheno: W ; Y ; W Num: 96 ; 99 ; 197 What mode of inheritance for the yellow body mutant allele?

X-linked recessive

Which is the term for the double-stranded circular extrachromosomal DNA element found in F+ bacterial strains.

- Plasmid

From a liquid bacteria culture, 1 mL was transfer to 9 mL of bacteria-free liquid. Then 1 mL of this first dilution was transferred to 9 mL of bacteria-free liquid. Then 1 mL of this second dilution was transferred to 9 mL bacteria-free liquid. Then 1 mL of this third dilution was transferred to an agar plate. From this plate, 100 colonies appeared. In bacteria/mL, what was the concentration of bacteria in the original culture?

( 1st dilution: 1/10 fold 2nd dilution: 1/100 fold 3rd dilution: 1/1000 etc. ) (You take the number of colonies and divide it by the fold number) - 100 colonies / (1/1000) == 100,000 colonies in the original solution.

The white eye and miniature wing genes are separated by about 34 map units. If flies from the above F1 progeny are crossed, how many of the 500 progeny are expected to have both wild-type eyes and miniature wings?

(Cross the females genes ++/wm --> w+/+m and then determine how many offspring have wild-type eyes and miniature wings when the female is crossed with the fully mutant male (wm/Y)) The punnet square - 0.17 X 500 = 85 ***

In a particular species of plant, locus A_ produces tall stems, while aa produces short stems, and locus B_ produces broad leaves, while locus bb produces thin leaves. A true breeding plant with tall stems and broad leaves is crossed with a true-breeding plant with short stems and thin leaves. As you would expect, all the F1 plants are tall/broad. An F1 plant is then crossed with the parental plant exhibiting the recessive phenotypes to produce the following progeny: 36 tall/thin, 164 tall/broad, 44 short/broad and 156 short/thin. These results indicate that the a and b loci are linked. What is the distance between the two loci?

(To determine the distance between a & b loci, add the smallest numbers, divide by the total and multiply by 100%.) 36 + 44 / 400 = 0.2 * 100 = 20% == 20 mu.

The genes for vermilion eyes and forked bristles are approximately 25 map units apart on chromosome 1 in Drosophila. Assume that a vermilion-eyed female was mated to forked-bristle male and that the resulting F1 phenotypically wild-type females were mated to vermilion, forked-bristle males. Of 1000 offspring, what would be the expected phenotypes, and in what numbers would they be expected?

(To determine the distance between the two loci, add the smallest numbers and divide by the total and finally multiply by 100% to find the map units) (To solve this problem, go through each answer choice and follow the instructions above until you find the answer choice that adds up to be 25% or 25mu.) D) Vermilion = 375; forked-bristle = 375; wild-type = 125; vermilion-forked-bristle = 125. 125 + 125 = 250 --> 250/1000 = 0.25 --> 0.25 x 100 = 25% or 25mu.

Assume that the a and b loci in the plant species above are completely linked. What are all the possible phenotypes of progeny plants that would be obtained by crossing an F1 plant with the parental plant exhibiting the recessive phenotypes?

(completely linked = no crossover = only parental phenotypes show) Tall/broad (200) and Short/thin (200)

Assume that you are able to microscopically observe the vermillion and forked-bristle loci on meiotic Drosophila chromosomes. If you examined 200 primary oocytes, in how many would you expect to see a chiasma between these two loci mentioned above?

(map units x 2 == chiasmata) 25 mu x 2 == 50 chiasmata = 0.50 --> 200 x 0.50 == 100

Assume that there are chromosome markers for the above loci that allow them to be visualized/identified under the microscope. Among 180 sporangia undergoing meiosis, how many would display chiasmata the a and b loci?

(mapunits x 2 == chiasmata) 20 mu x 2 = 40 chiasmata = 0.40 180 x 0.40 == 72 display chiasmata.

Which of the following events is occurring?

- Positive Interference

What process can lead to a bacterial cell's acquiring resistance to a particular phage

* A spontaneous mutation that occurs in the presence or absence of phage is considered the primary source of genetic variation in bacteria. * Selection is the growth of the organism under conditions in which only the mutant of interest grows well, whereas the wild type does not.

How are DNA markers used to identify loci associated with certain human conditions (e.g. analysis by RFLP, PCR across repetitive DNA elements, SNPs).

* DNA markers are short segments of DNA whose sequence and location are known. They represent landmarks along the chromosome. * Restriction fragment length polymorphisms (RFLPs) and PCR - Assesses variable DNA repeats that can differ among individuals. * Small nucleotide polymorphisms (SNPs) - Single base-pair differences among individuals. * With pedigree diagrams, several million of these variations help give geneticists the ability to identify and locate genes associated with a particular condition.

Genetic recombination in bacteria

* Genetic information from one bacterium is transferred to another * DNA from one bacteria recombines with the DNA from a second bacterium - Can replace a region in the chromosome, which will include all genes in that region. - The second bacterium does not transfer DNA to the first (not a reciprocal exchange)

In the fruit fly, Drosophila melanogaster, white eye (w) and miniature wing (m) are X-linked recessive mutant alleles that are linked. A fully mutant female crossed with a wild-type male produces wild-type females and fully mutant males (the F1 generation). What are the genotypes of these F1 flies?

*** - Female: ++/wm ; Male: wm/Y chromosome.

In the fruit fly, Drosophila melanogaster, dumpy wing (dp), black bodied (b) and cinnabar eyes (cn) are all recessive mutant alleles for genes located on the same chromosome (autosome). A female with dumpy wing that was HOMOZYGOUS for all three genes is mated to a black-bodied male with cinnabar eyes that is also HOMOZYGOUS for all three genes. Fully mutant males were then mated to phenotypically wild-type F1 females, and the following progeny (1000 total) were observed (phenotypes not shown are assumed to be wild-type: 170 dumpy wing, black body, cinnabar eyes 265 dumpy wing 180 wild-type 275 black body, cinnabar eyes 54 black body 4 dumpy wing, black body 46 dumpy wing, cinnabar eyes 6 cinnabar eyes

**This information relates to the first 4 questions below.**

E. coli strain A (met-, leu+, thi+) was mix with E. coli strain B (met+, leu-, thi-) in complete medium and incubated overnight. The following day cells from the mixture were plated on agar with complete medium as follows: 0.1 mL (milliliters) was diluted into 9.9 mL sterile liquid to produce dilution 1 (D1); 0.1 mL of D1 was diluted into 9.9 mL sterile liquid to produce D2; 0.1 mL of D2 was diluted into 9.9 mL sterile liquid to produce D3; 1 mL of D3 produced 30 colonies. Cells from the mixture plated on minimal medium without dilution produced 3 colonies. What was the frequency of recombination?

- 1/10

Dominant lethal allele

- 100% death if dominant, possibly won't even survive birth. - Still dies if heterozygous, but can possibly live long enough to produce offspring.

Many of the color varieties of summer squash are determined by several interacting loci: AA or Aa gives white, aaBB or aaBb gives yellow, and aabb produces green. Assume that two fully heterozygous plants are crossed. Which shows the phenotypes and frequencies of the offspring.

- 12 white: 3 yellow: 1 green

Flower color for a pea plant species is determined by interacting genes (not linked): alleles A and B together produce purple flowers. Homozygous recessive alleles for either gene produce white flowers. Assume a fully heterozygous plant is crossed with a fully homozygous plant. Of 400 progeny plants, how many are expected to have white flowers.

- 300

What is the distance between black body and dumpy wing?

- 36 mu

If a cross between two creeper chickens is to result in six offspring, how many chicks are expected to be creepers?

- 4 (2/3 = Pp and 1/3 = pp) 6 * (2/3) == 4 (In the question 6 offspring are produced by crossing 2 creepers (Pp); there is a 25% chance of (PP), however it is deadly and is not used in the calculations so you calculate using the squares, of the punnet square, that don't have (PP).

What is the likelihood that the above couple's first daughter will be color blind?

- 50%; because it refers to the probability of the females ONLY, so, on the punnet square, 1/2 of the girls will be color blind

A recessive gene for red-green color blindness is located on the X chromosome in humans. Assume that a woman with normal vision (her father is color blind) marries a color-blind male. What is the likelihood that this couple's first son will be color blind?

- 50%; because it refers to the probability of the males ONLY, so, on the punnet square, 1/2 of the boys will be color blind

Genetic Anticipation

- A phenomenon in which a genetic disease appears earlier and the size of the repeated segment increases with each succeeding generation - With each generation, the disease gets worse than it was in the previous generation.

In males, pattern baldness is dominant due to the B allele located on an autosomal gene. In females, pattern baldness (due to the SAME allele) is recessive. In females, which genotype will be associated with pattern baldness?

- BB

Which gene is in the middle?

- Black Body

For a synteny test, hybrid cell line A contains human chromosomes 1, 3 and 5, hybrid cell line B contains human chromosomes 1-4 and hybrid cell line C contains human chromosomes 1, 2 and 5. A gene product of interest is produced by cell lines A and C but not B. Which human chromosome is responsible for the expression of the gene product?

- Chromosome 5

Pea plant flower color in the question above is controlled by which mode of inheritance?

- Complementary gene interaction. (It requires both dominant forms of the genes involved.)

Summer squash color in the question above is controlled by which mode of inheritance?

- Dominant Epistasis

Which of the following genotypes would be found in a strain of bacteria that is a prototroph? A. lac-, leu-, gal-, thi-, thr- B. lac+, leu+, gal-, thi-, thr- C. lac-, leu-, gal+, thi+, thr+ D. lac-, leu+, gal+, thi-, thr- E. lac+, leu+, gal+, thi+, thr+

- E: lac+, leu+, gal+, thi+, thr+

Drosophila flies that are homozygous for the eyeless mutant allele can either lack eyes OR exhibit varying reductions in eye size. This variation in the intensity of the mutant phenotype is known as _________.

- Expressivity (The range of expression of the mutant phenotype. The degree to which a mutant phenotype appears in one offspring.)

T/F: Positive interference occurs when there is an enhancement of double crossover events (i.e. the number of double crossovers occurring in a given experiment is greater than the number of double crossovers that would be expected).

- False

T/F: in transferring the F factor to F- bacteria, the F+ bacteria lose the F factor and become F-.

- False

T/F: pattern baldness is an example of sex-limited inheritance.

- False

T/F: the three-point mapping procedure used to map three genes on a chromosome in Drosophila and plants can also be used to map three linked genes in humans.

- False *******

A cross between an organism with the genotype AaBb (two dominant phenotypes) and an organism that is homozygous recessive for both alleles produces 1003 progeny that are dominant for both phenotypes and 999 progeny that are recessive for both phenotypes. No progeny are produced that are dominant for just one of the two phenotypes. Which of the following statements is true?

- Genes a and b are on the same chromosome and close enough together that recombination never occurs between them - Genes a and b are members of the same linkage group

Because of the mechanism of sex determination, males of many species can be neither homozygous nor heterozygous for alleles on the sex chromosomes. Such males are said to be____

- Hemizygous.

Examples of Genetic Anticipation

- Huntington disease, myotonic dystrophy, fragile X - Due to the expansion of a trinucleotide repeat sequence in the DNA. * Delayed Onset of Phenotypic Expression * Tay-Sachs disease - Autosomal recessive Hexosaminidase A, lipid metabolism, baby normal for a few months, dies by age 3 * Huntington disease - Autosomal dominant - Trinucleotide repeat expansion in the HTT gene - The age of onset of the disease is usually between 30 and 50 * Duchene muscular dystrophy - X-linked recessive - Diagnosis at 3-5 years old

Name a human condition that is associated with a dominant lethal allele

- Huntington's Disease.

In Angelman syndrome, a small region is deleted from the maternal chromosome 15. Which term explains why genes in this region on the paternal chromosome 15 do not compensate for the deletion?

- Imprinting

Genetic imprinting

- Imprinting occurs when genes inherited from only one parent are expressed. Those same genes inherited from the other parent are not expressed due to DNA methylation. (one gene is turned one, while the other is turned off). - due to epigenetic modifications (e.g. DNA methylation) Prader-Willi versus Angelman syndromes.

Which characteristic applies to Huntington's disease?

- It is caused by a dominant lethal allele. - It is an example of a trinucleotide repeat expansion disease. - It is an example of genetic anticipation.

Recessive lethal allele

- Lethal when homozygous. (lives whole life) - Distinct mutant phenotype when heterozygous. - Recessive lethal, but is dominant with respect to the phenotype.

Conditional mutations.

- Mutations affected by temperature are called conditional or temperature-sensitive mutations. - They are useful in studying mutations that affect essential processes. (Ex: nutritional temperature effects)

In chickens, the P allele causes stunted legs in chickens (known as creepers). Chickens with the genotype pp have normal legs. Creepers are always heterozygous (Pp) and never homozygous (PP). How is this observation explained?

- P is a recessive lethal allele

Despite a genotype of AaBB, a summer squash plant has produced fruit with yellow rather than the expected white color. Which mechanism might explain this phenomenon?

- The locus containing the A allele underwent a translocation from a euchromatic (loose; expresses gene) to a heterochromatic (highly condensed; restricts gene expression) region

Penetrance

- The percentage of individuals that show at least some degree of expression of the mutant genotype in a population. - Penetrance is the frequency with which a mutant phenotype (any degree of the phenotype) appears in a population relative to its expected frequency.

Expressivity

- The range of expression of the mutant phenotype. - The degree to which a mutant phenotype appears in one offspring.

T/F: Differences in milk production among cows is an example of sex-limited inheritance.

- True

State or describe two distinctions between an F+ strain and an Hfr strain of a certain bacterial species.

- Unlike F+ strains, Hfr strains can be used for time mapping. - In Hfr strains, the F factor is not transferred to the recipient cell. - Unlike F+ cells, the F factor plasmid in Hfr cells becomes integrated into the chromosome and are not free along the cytoplasm.

Given that loci A and B in Drosophila are sex-linked and 20 map units apart, what phenotypic frequencies would you expect in male and female offspring resulting from an aabb female × AB/Y male? (Assume A and B are dominant to a and b, respectively.)

- all males = ab ; all females = AB

Assume that a man who carries a certain X-linked allele has children. Assuming normal meiosis and random combinations of gametes, the man would pass this allele to ________.

- all of his daughters.

What is somatic cell hybridization

- allows human genes to be assigned to their respective chromosomes. - involves fusion of two cells in culture to form a single hybrid cell, called a heterokaryon. - Upon continued culturing of the hybrid cell, chromosomes from one of the two parental species are gradually lost until only a few chromosomes of one species remain and most chromosomes are from the other species, creating what is termed a synkaryon

With respect to the three genes mentioned in the problem, what would be genotypes of the HOMOZYGOUS parents used in making the phenotypically wild-type F1 females?

- dp + +/dp + + and + b cn/+ b cn

How to calculate expected double crossover. (DCO(exp))

1) divide 2 the map units distances by 100 (to convert them from their percent/mu form to their decimal form.) 2) multiply them together. Ex: DCO(exp) = (0.201) * (0.291) = 0.058491 == 0.06

A white-eyed F1 female is then crossed with a wild type F1 male to produce the following F2 progeny: 200 wild-type females, 180 white-eyed males, 10 wild-type males and 10 males displaying white/red mottling across the ommatidia (the facets of their compound eye). What is the penetrance for this particular experiment.

180 (white-eyed males) + 10 (wild-type males) + 10 (males white/red mottling) = 200 180 + 10 (males white/red mottling) = 190 190 / 200 == 0.95 * 100 == 95% penetrance.

Assume that the a and b alleles in the above species of pea plant are mutant alleles. A plant with the genotype aaBb is self-fertilized to produce 400 progeny with the following phenotypes: 350 with white flowers, 40 with purple flower and 10 with white flowers but some slight purple mottling. What is the penetrance in this experiment?

350 + 10 == 360 --> 360/400 == 90%

What is a bacterial colony?

A bacterial colony consists of multiple microorganisms that are all from one mother cell, and they gather together and are genetically identical.

How to calculate penetrance.

Add up the number of progeny with the desired trait (even if the progeny displays a mix of traits), then divide by the total number of progeny and finally multiply by 100%.

In mice the allele for yellow coat color, AY, is dominant to agouti, A, but is also a recessive lethal allele. Assume two yellow mice will be mated and will produce six pups. How many pups are expected to be agouti?

Approximately 2 pups. (25%) - Recessive lethal alleles means you must homozygous (have 2 of the allele) to die. (Even though the yellow coat allele is dominant to agouti, it's requires must be homozygous for the allele for it to be lethal.)

Why might the restriction endonuclease fail to cut the middle site of the disease locus shown on the previous page?

Because mutations change the DNA sequence. The sequence would fail to be noticed by the restriction endonuclease.

How to calculate coefficient of coincidence

C = observed double crossover (DCO(obs)) / expected double crossover (DCO(exp)) Ex: C = 0.04 (DCO(obs)) / 0.06 (DCO(exp)) = 0.6667

Which of the following processes can lead to the independent assortment of the alleles of two genes that are linked?

Crossing over.

Certain types of breast cancer are associated with loss of BRCA1 expression, without any detection of mutations in the chromosomal region encompassing the BRCA1 gene. What epigenetic modification is suppressing BRCA1 expression in these instances?

DNA Hypomethylation

X linkage

Describes genes that are only found on the X chromosome. - Red/green color blindness - Hemophilia

Hemiozygous

Describes males because they only have one pair of genes encoded by their only X chromosome.

Interlocus

Distance between two genes on a chromosome

Dominant epistasis and how to evaluate this mode of inheritance

Dominant epistasis: dominant allele in one genetic locus masks the expression of the alleles in the second locus. (Ex: W (white) masks G (green) and g (yellow) in Wwgg/WWgg/WwGg/WWGg/WWGG making it white.)

Why might the restriction endonuclease fail to cut the middle site of the disease locus shown on the previous page?

Due to mutations that changed the DNA, the endonuclease may fail to recognize the correct area to cut.

Essential gene

Essential genes are those that are absolutely required for survival The absence of their protein product leads to a lethal phenotype (~1/3 of all genes are essential)

What term describes the variation in the phenotypes of different male flies above despite the fact that they all carry the recessive, mutant w allele?

Expressivity

How does conjugation work

F+ cells serve as DNA donors and F- cells are the recipients. F+ cells donate DNA to the F- cells through the F pilus.

What is the difference between F+ and an Hfr strain of bacteria

Hfr strains can also donate genetic information to the F- recipient but instead on converting it to F+, it remains F-. That's how I understand it

How is interference calculated?

Interference (I) = 1 - C Ex: 1 - 0.6667 == 0.3333 (borderline between Positive interference and no interference).

Non-essential gene

Nonessential genes are those not absolutely required for survival (some sets of genes have redundant functions)

Sex-limited inheritance

Occurs in cases where the expression of a specific phenotype is absolutely limited to one sex. (exclusively in one sex.)

Prader-Willi syndrome

Occurs when, at first, the father's gene is expressed (turned on), and supplies UBE3A, while the mother's gene is not expressed (turned off) and then the father's gene suddenly turns off too during development, in which UBE3A is no longer being supplied by the father's gene.

Angelman syndrome

Occurs when, at first, the mother's gene is expressed (turned on), and supplies UBE3A, while the father's gene is not expressed (turned off) and then the mother's gene suddenly turns off too during development, in which UBE3A is no longer being supplied by the mother's gene.

What is a plasmid

Plasmids are composed of a double-stranded closed circle of DNA - Exist in multiple copies in the cytoplasm. - May contain one or more genes. - Replicate independently of the bacterial chromosome.

Difference between positive and negative interference.

Positive: fewer double-crossover events occur than expected. Negative: more double-crossover

What is the difference between an auxotroph and a prototroph; how are they used for genetic analysis

Prototroph: can synthesize all essential organic compounds and therefore can be grown on minimal medium (usually have a (+) ) Auxotroph: Through mutation, an auxotroph has lost the ability to synthesize one or more essential compounds; it must be provided with them in the medium if it is to grow. (usually have a (-) )

How to calculate the distance in map units between two genes

SCO (Single crossover; crossover between 1st gene and middle gene (or middle gene and last gene)) + DCO (Double crossover) / total * 100% (% = mu) (ex: distance between s and v; SCO(82+79) + DCO(23 +17)/1000 * 100% == 20.1% == 20.1 mu)

Assume that an F1 plant is crossed with the parental plant exhibiting the dominant phenotype. For 400 progeny plants, state the expected phenotypes AND the expect numbers for each phenotype.

Tall/Broad leaves; 400

Complementary gene interaction

The complementary gene is a interaction of two dominant non inter-allelic gene in which each gene have its own effect but when come together to interact a new trait is developed and the Mendelian ratio 9:3:3:1 is changed to 9:7 due to complimentation of both genes.

The white-eye gene in Drosophila is recessive and X-linked. A red-eyed female (wild-type phenotype) crossed with a white-eyed male produces the following F1 progeny: 133 white-eyed females, 127 wild-type females, 129 white-eyed males and 137 wild-type males. What are the genotypes of the parents?

The genotype of the mother is w w+ and the genotype of the father is w Y. ( w+ = red eyes (dominant and the wildtype), w = white eyes (recessive), Y = the Y chromosome of the male)

Color varieties of summer squash are determined by interacting loci: AA or Aa gives white, aaBB or aaBb gives yellow, and aabb produces green. A cross between a plant with white squash and one with yellow squash produces the following progeny: 206 plants with white squash, 146 plants with yellow squash and 48 plants with green squash. What are the genotypes of the parental plants?

The genotype of the parental squash is: White (AaBb) and Yellow (aaBb).

. If a DNMT modifies the promoter of a specific gene, how would the level of expression of that gene be expected to change?

The level of expression would be expected to decrease.

Locus

The position of a gene on a genetic map; the specific place on a chromosome where a gene is located

Sex-influenced

The sex of an individual influence the expression of a phenotype that is not limited to one sex or the other. (occurs mainly in one sex) (can be autosomal genes that are influenced by hormone levels.)

Complete Linkage

There are only parental phenotypes, no recombinants/crossovers.

Incomplete Linkage

There are parental and recombinant phenotypes, however, the recombinant phenotypes occur in small percentages.

BRIEFLY describe the imprinting defect that leads to Beckwith-Wiedeman syndrome including the gene that is affected.

This is a parental overgrowth disorder with abdominal wall defects, enlarged organs, large birth weight, and predisposition to cancer due to abnormal methylation patterns which result from altered patterns of gene expression. The genes included in this disease are H19 and IGF2, located on chromosome 11. The H19 is maternally expressed, with the parental is imprinted and silence, while the IGF2 allele is not imprinted. The factors lead to the overgrowth mutation.

What is the difference between vertical and horizontal gene transfer

Vertical: The transfer of bacterial DNA to that of the same species. Horizontal: The transfer of bacterial DNA to that of a related but distinct species.

Position effects

translocation of a chromosome segment containing a gene of interest from a euchromatic region (lightly packed DNA) to a heterochromatic region (Tightly packed DNA) or vice versa