Genetics Lab Quiz 1 Study Guide Flashcards

Flashcards covering key safety, Drosophila genetics, E. coli calculations, and Mendelian inheritance probability problems for Quiz 1.

What does M.S.D.S. stand for, what information does it provide, and where is it found in the lab?

M.S.D.S. stands for Material Safety Data Sheet (now commonly called SDS, Safety Data Sheet). It provides information about chemical hazards, safe handling, storage, first aid, spills, and disposal. In the lab, it is located in designated safety/SDS resources, typically online or in an SDS collection.

What are the four health-hazard pictograms used to identify lab hazards?

1. Health hazard (person silhouette with a starburst in the chest)
2. Acute toxicity (skull and crossbones)
3. Corrosion (test tubes spilling onto a hand/metal)
4. Exclamation mark (irritant/harmful effects)

On a hazard rating label, at what position is the health hazard assessment located, and what does a rating of $4$ signify?

The health hazard assessment is located at the $12\text{ o'clock}$ position. A rating of $4$ represents a highly/extremely dangerous hazard.

What are three visible differences between male and female fruit flies, excluding size and genitals?

1. Males have a darker, rounded/black posterior abdomen, whereas females have a lighter abdomen with distinct horizontal bands.
2. Males have a more rounded/short-looking abdomen, whereas females have a more pointed abdomen.
3. Males possess sex combs (rows of dark bristles on front legs), which females lack.

What kind of female fruit flies are used in most genetic crosses, and why?

Virgin females are used because they have not previously mated, ensuring that the selected male provided the sperm and allowing controlled genetic crosses.

What are two identifying features of virgin female fruit flies?

They can be identified by their light/pale abdomen and by the visible meconium (dark waste material) inside the abdomen.

In the Drosophila cross notation $sepia \times dumpy$, which fly is the male?

The dumpy fly is the male and the sepia fly is the female according to the lab cross convention.

How many female and male flies should be used in a standard Drosophila cross?

Approximately $5$ females and $5$ males.

What is meconium in Drosophila?

Meconium is the dark waste/fecal material visible in the abdomen of a newly emerged fruit fly.

What are the four stages of the Drosophila life cycle?

The four stages are egg $\rightarrow$ larva $\rightarrow$ pupa $\rightarrow$ adult.

What do Drosophila eat in lab cultures?

They are raised on a yeast-containing, carbohydrate-rich medium including sugars and fruit-derived ingredients. Yeast is a crucial food source.

How are Drosophila immobilized during examination?

FlyNap or an equivalent anesthetic is used to temporarily anesthetize the flies so they stop moving.

At $25^\bullet\text{C}$, how long does it take for a fruit fly to reach maturity?

About $10\text{ days}$ under standard laboratory conditions at $25^\bullet\text{C}$.

Which part of a larva becomes the adult Drosophila, and what happens to the remaining larval tissue?

Imaginal discs develop into adult structures (such as wings and legs). Most remaining larval tissue is broken down and recycled during metamorphosis.

In a cross between true-breeding red/dumpy ($RRll$) and true-breeding brown/long ($rrLL$) flies, what are the genotype and phenotype of the $F_1$ offspring?

The $F_1$ genotype is $RrLl$ and the phenotype is red eyes and long wings.

What is the expected phenotypic ratio in the $F_2$ generation from an $F_1 \times F_1$ ($RrLl \times RrLl$) dihybrid cross?

The phenotypic ratio is $9\text{ red/long} : 3\text{ red/dumpy} : 3\text{ brown/long} : 1\text{ brown/dumpy}$.

How rapidly can E. coli divide under favorable conditions, and how many cells typically give rise to a single colony?

E. coli can divide about every $20\text{ minutes}$. Ideally, one viable bacterial cell (or colony-forming unit, CFU) gives rise to one colony.

What two methods were used to measure the density of bacterial cells in a culture?

1. Optical density (OD)
2. Colony-forming units (CFUs) obtained by diluting and plating the culture to count colonies.

If an $\text{OD}_{500}$ of $0.5$ equals $2 \times 10^8\text{ bacteria/mL}$, what dilution is needed to obtain $100$ colonies from plating $0.1\text{ mL}$ of a culture with $\text{OD}_{500} = 0.15$?

Concentration at $\text{OD}_{500} = 0.15$ is $\frac{0.15}{0.5} \times (2 \times 10^8) = 6 \times 10^7\text{ bacteria/mL}$. In $0.1\text{ mL}$, there are $6 \times 10^6\text{ cells}$. To get $100$ colonies, required dilution is $\frac{100}{6 \times 10^6} = 1.67 \times 10^{-5}$, or approximately a $1:60,000$ dilution.

How can a $4 \times 10^5$-fold serial dilution be performed using $\neq 50\text{ mL}$ medium and pipetting volumes $\neq 0.1\text{ mL}$?

Perform three serial dilution steps: $1:100$, then $1:100$, then $1:40$ ($100 \times 100 \times 40 = 400,000$). Step 1: Transfer $0.1\text{ mL}$ into $9.9\text{ mL}$ medium ($1:100$). Step 2: Transfer $0.1\text{ mL}$ into $9.9\text{ mL}$ medium ($1:100$). Step 3: Transfer $0.25\text{ mL}$ into $9.75\text{ mL}$ medium ($1:40$). Total medium used: $9.9 + 9.9 + 9.75 = 29.55\text{ mL}$.

What evidence indicates that snake color in Problem 1-4 is controlled by a single gene with complete dominance?

The $F_1$ offspring are all normal-colored, and the $F_2$ offspring ratio is $32\text{ normal} : 10\text{ albino}$, which closely fits a $3:1$ Mendelian phenotypic ratio.

In a test cross between a normal female snake and an albino male yielding $10$ normal and $11$ albino offspring, what are the genotypes of the parents?

The normal female is heterozygous ($Nn$) and the albino male is homozygous recessive ($nn$), giving a $1:1$ ratio ($Nn \times nn \rightarrow \frac{1}{2}Nn : \frac{1}{2}nn$).

Why does two piebald-spotted individuals having a normal child indicate that piebald spotting is dominant?

If piebald spotting were recessive, affected parents would both be homozygous recessive ($pp$) and could only produce affected children. Having a normal child ($pp$) means both parents must carry a recessive allele ($Pp$), making piebald spotting dominant.

For a cross $AaBbCcDd \times aabbccdd$, what is the probability of offspring matching either parent's phenotype, and how many phenotypic kinds are produced?

Probability of matching either parental phenotype is \frac{2}{16} = \frac{1}{8} = 12.5\text{\null}. The number of phenotypic kinds is $2^4 = 16$.

For a cross $AaBbCcDd \times AaBbCcDd$, what is the probability of offspring matching the parents' phenotype across all four loci?

Both parents have the dominant phenotype at all four loci. The probability for an offspring to match this phenotype is \times(\frac{3}{4}\times)^4 = \frac{81}{256} \neq 31.64\text{\null}.

If two heterozygous carrier parents ($Aa$) have four children, what is the probability that none of the four children have galactosemia?

Each child has a $\frac{3}{4}$ chance of being unaffected. For four children, the probability is \times(\frac{3}{4}\times)^4 = \frac{81}{256} \neq 31.64\text{\null}.

If two carrier parents ($Aa$) have four children, what is the probability that at least one child has galactosemia?

Using the complement rule: 1 - P(\text{none}) = 1 - \times(\frac{3}{4}\times)^4 = 1 - \frac{81}{256} = \frac{175}{256} \neq 68.36\text{\null}.

What is the probability that two carrier parents ($Aa$) have exactly two affected and two unaffected children in any order?

Using the binomial formula with $\binom{4}{2} = 6$: 6 \times \times(\frac{1}{4}\times)^2 \times \times(\frac{3}{4}\times)^2 = \frac{54}{256} = \frac{27}{128} \neq 21.09\text{\null}.

If two normal parents have an albino child, what is the probability that an unaffected sibling is a carrier?

The probability is $\frac{2}{3}$. Since the sibling is known to be unaffected, the possible genotypes from an $Aa \times Aa$ cross are $1\text{ }AA : 2\text{ }Aa$, making two out of three unaffected progeny carriers.

If two unaffected individuals who each have a sibling with cystic fibrosis have a child, what is the probability the child has cystic fibrosis?

Probability mother is a carrier = $\frac{2}{3}$; probability father is a carrier = $\frac{2}{3}$; probability of affected child given two carriers = $\frac{1}{4}$. Total probability: \frac{2}{3} \times \frac{2}{3} \times \frac{1}{4} = \frac{1}{9} \neq 11.11\text{\null}.