Module 3 Flashcards
match the term with the correct answer choice:
- Mendel’s first law of inheritance
- Mendel’s second law of inheritance
- Monohybrid cross
- Dihybrid cross
- Test cross
- Sum rule
- Product rule
ANSWER CHOICES:
a. Probability of two independent events = product of their probabilities.
b. Cross involving two traits with two alleles each.
c. Cross involving one trait with two alleles.
d. Alleles of different genes assort independently.
e. Probability of either of two exclusive events = sum of their probabilities.
f. Two alleles for a gene separate during gamete formation.
g. Cross dominant phenotype with homozygous recessive to determine genotype.
Mendel’s first law of inheritance → f
Mendel’s second law of inheritance → d
Monohybrid cross → c
Dihybrid cross → b
Test cross → g
Sum rule → e
Product rule → a
Match the terms to the correct definition:
Binomial factor expansion
Linkage
Centimorgan
LOD analysis
Pedigree
Autosomal dominant
Autosomal recessive
Punnett square
a. Family tree tracing inheritance of traits.
b. Genes close together on the same chromosome that do not assort independently.
c. Statistical method to determine if genes are linked.
d. Unit measuring genetic distance; 1% recombination frequency.
e. Diagram used to predict offspring genotypes and phenotypes.
f. Trait appears in every generation; affects both sexes equally.
g. Trait often skips generations; affects both sexes equally.
h. Method to predict outcomes for events with two traits.
Binomial factor expansion → h
Linkage → b
Centimorgan → d
LOD analysis → c
Pedigree → a
Autosomal dominant → f
Autosomal recessive → g
Punnett square → e
- What does Mendel’s Law of Independent Assortment state?
alleles for different traits are passed down separately during gamete formation, as long as the genes are on different chromosomes.
How does Mendel’s Law of Segregation apply to gamete formation in a heterozygous pea flower (Pp)?
Each gamete gets one allele (P or p) as alleles separate during meiosis
List all possible gametes for the following genotypes assuming independent assortment:
a. AABbCc
b. AaBbCc
c. AABBCc
d. aabbCc
a. ABC, ABc, AbC, Abc
b. ABC, ABc, AbC, Abc, aBC, aBc, abC, abc
c. ABC, ABc
d. aBC, aBc, abC, abc
What are the phenotypes and proportions in a dihybrid cross of AaBb x AaBb? Show your work.
Phenotypes:
9/16 tall and green (A_B_),
3/16 tall and red (A_bb),
3/16 short and green (aaB_),
1/16 short and red (aabb).
What is a testcross, and how did Mendel use it?
crossing a dominant phenotype with a homozygous recessive to determine genotype.
Mendel used it to confirm F1 heterozygosity
Nancy ‘Corner Pocket’ LeSharke is dominant on the professional snooker circuit. She misses on average only once every 10 shots (success rate = 90%). What are the probabilities for
a. Nancy makes 9 balls straight, followed by a miss?
a. 9 straight makes, then a miss:
P=(0.9) ^ 9 ×(0.1)=0.3874
Mendel’s Law of ___________ __________ means that alleles of separate genes assort independently.
__________ cross (PpTt): Alleles for flower color and height sort independently.
Offspring ratio: _____ tall purple, _____tall white, ______ short purple, _____ short white.
Independent Assortment
Dihybrid
9;3;3;1
Nancy ‘Corner Pocket’ LeSharke is dominant on the professional snooker circuit. She misses on average only once every 10 shots (success rate = 90%). What are the probabilities for
b. Nancy makes 1 ball, then a miss, then 8 more makes?
b. 1 make, 1 miss, 8 more makes:
P=(0.9) ^9 ×(0.1)=0.3874
Nancy ‘Corner Pocket’ LeSharke is dominant on the professional snooker circuit. She misses on average only once every 10 shots (success rate = 90%). What are the probabilities for
c. Nancy makes 2 balls, then a miss, then 7 more makes?
c. 2 makes, 1 miss, 7 more makes:
P=(0.9) ^9 ×(0.1)=0.3874
Nancy ‘Corner Pocket’ LeSharke is dominant on the professional snooker circuit. She misses on average only once every 10 shots (success rate = 90%). What are the probabilities for
d. Nancy makes 9 balls and 1 miss, in any order?
d. 9 makes and 1 miss, any order:
Use binomial probability:
P=(1/10)×(0.9) ^9 ×(0.1) ^1 =10×0.3874=0.3874
Nancy ‘Corner Pocket’ LeSharke is dominant on the professional snooker circuit. She misses on average only once every 10 shots (success rate = 90%). What are the probabilities for
e. Nancy switches sports mid-career and joins the professional Unicycling Darts circuit?
e. Switches sports:
This is a hypothetical question. The probability cannot be calculated but depends on Nancy’s decision-making.
How can the product rule and sum rule be used to calculate the odds of obtaining an offspring with two dominant phenotypes in a dihybrid cross of heterozygous parents (𝑌𝑦𝑅𝑟)?
Product Rule: Multiply probabilities of independent events
P(Y andR ) =3/4×3/4=9/16.
Sum Rule: Add probabilities for mutually exclusive events (e.g., , 𝑌 or𝑦𝑦)
P(YellowORGreen)=3/4+1/4=1
Result: The probability of both dominant phenotypes is 9/16.
A pea plant is purple-flowered, tall, yellow-seeded and has inflated pods. All of these
traits are Mendelian, and the expressed phenotypes are all dominant. Describe two ways
in which you could determine the genotype of this plant, and what the genotypic and
phenotypic results would be if the plant was heterozygous at each locus.
homozygous dominant, heterozygous, homozygous recessive).
In a test cross with a homozygous recessive plant, a 1:1 dominant to recessive ratio in the offspring suggests the plant is heterozygous.
In a self-cross, a 3:1 dominant to recessive ratio and a 1:2:1 genotypic ratio shows it’s heterozygous.
What are the genotypes of the parent pea plants in a cross resulting in offspring with yellow and green seeds, round and wrinkled shapes, and purple and white flowers?
Parent 1 (Yellow, Round, Purple): YyRrWw
Parent 2 (Green, Wrinkled, Purple): yyrrWw
This cross produces offspring with a 9:3:3:1 phenotypic ratio.
Yellow, round, purple flowers: Y_R_W_
Yellow, round, white flowers: Y_R_ww
Yellow, wrinkled, purple flowers: Y_rrW_
Yellow, wrinkled, white flowers: Y_rrww
Green, round, purple flowers: yyR_W_
Green, round, white flowers: yyR_ww
Green, wrinkled, purple flowers: yyrrW_
Green, wrinkled, white flowers: yyrrww
One hundred sets of parental flowering snapdragons are crossed. In each case, both
parents have purple flowers (the dominant phenotype), but both are heterozygous for the
flower color gene (Pp). Each cross results in five offspring.
a. What is the expected number of sets of offspring that will result in all white-
flowered individuals?
b. What is the expected number of sets of offspring that will result in 4 purple and 1
white offspring, in any order?
c. What is the expected number of sets of offspring that will result in 2 purple and 3
white flowered offspring, in any order?
d. What are the odds a single cross in this experiment resulting in production of first
a purple flowered offspring, and then a white flowered offspring?
a. All white-flowered offspring:
Answer: Expected number = 0 (probability = 1/1024).
b. 4 purple and 1 white offspring (in any order):
Answer: Expected number = 40 (probability = 405/1024).
c. 2 purple and 3 white offspring (in any order):
Answer: Expected number = 9 (probability = 90/1024).
d. Odds of getting first a purple, then a white offspring:
Answer: Odds = 3:13 (probability = 3/16).
What’s the definition of probability based on previous occurrences of an event?
Probability is the likelihood of an event happening, calculated by comparing how often it’s happened before to the total possible outcomes.
A pea plant heterozygous for alleles at the seed color and flower color genes (YyPp) is
selfed and produces the following offspring:
150 yellow seeded, purple flowered offspring
40 yellow seeded, white flowered offspring
65 green seeded, purple flowered offspring
10 green seeded, white flowered offspring
Assuming the dominant alleles are for yellow seeds (Y) and purple flowers (P), do these
traits exhibit strict Mendelian inheritance patterns? Whether you answered yes or no, how
confident are you (statistically speaking) in your answer?
No. The observed offspring ratios don’t fit the expected 9:3:3:1 Mendelian ratio. It should be:
Yellow, purple: 9/16 × 265 ≈ 149
Yellow, white: 3/16 × 265 ≈ 50
Green, purple: 3/16 × 265 ≈ 50
Green, white: 1/16 × 265 ≈ 16
Brown:
Bb x bb always gives 1:1 ratio
odds of brown (dominant) = ½
wiry-haired:
Ff x Ff always gives 3:1 ratio
odds of wiry-haired (recessive) = ¼
small size:
Ll x ll always gives 1:1 ratio
odds of small size (recessive) = 1/2
Odds of brown AND wiry-haired AND small size:
½ x ¼ x ½ = 1/16
