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T/F Dominance results from biochemical processes involving enzymes associated with the gene.
- ANSWER ✔✔ True
T/F If a heterozygote (Aa) makes some enzymes, but not enough to catalyze as many reactions
as a homozygote (AA) can, all genotypes would have the same phenotype (AA=Aa=aa)? -
ANSWER ✔✔ False
Unit of heritable information on chromosome and consisting of a sequence of DNA. - ANSWER
✔✔ Gene
A heritable feature, such as flower color. - ANSWER ✔✔ Character
A variant of a character, such as purple or white flowers. - ANSWER ✔✔ Trait
T/F Each trait carries two copies of a gene, one inherited from the mother and the other from the
father. - ANSWER ✔✔ True
Probability - ANSWER ✔✔ Describes our certainty about an outcome or event
Independent Events - ANSWER ✔✔ The outcome of one event does not affect the outcome of
the second event
Joint Probability - ANSWER ✔✔ The probability of two events occurring together
P(A and B) = P(A) · P(B) - ANSWER ✔✔ Joint Probability Calculation
P(A OR B) = P(A) + P(B) - P(A AND B) - ANSWER ✔✔ Probability that either event occurs
-Each parent in the monohybrid cross contributed one of two alleles to each offspring
-Every combination of alleles was equally likely - ANSWER ✔✔ Mendels postulated the
following rules of inheritance
Different phenotypes for a single trait - ANSWER ✔✔ Monohybrid cross
T/F A probability of 0.98 means that an event occurs rarely. - ANSWER ✔✔ False
, T/F When a parent is heterozygous, the probability of passing the dominant allele to its offspring
is 25%. - ANSWER ✔✔ False
T/F When both parents are heterozygous (Aa), the probability of an offspring being homozygous
recessive (aa) equals 0.25. - ANSWER ✔✔ True
T/F A Punnett square enables one to determine the frequency of observed genotypes without
using a mathematical formula to calculate the joint probability. - ANSWER ✔✔ True
T/F When both parents are heterozygous (Aa), the probability of an offspring being heterozygous
(Aa) equals 0.75. - ANSWER ✔✔ False
Why do experiments? - ANSWER ✔✔ 1. Replicate Observations
2. Control Conditions
3. Randomize Subjects
T/F If two heterozygotes mate to produce two offspring, the chance that both offspring are
homozygous equals 50%. - ANSWER ✔✔ False
The probability of the observed deviation or a greater deviation, when the model is correct -
ANSWER ✔✔ P Value
A statistical test used to determine the probability of obtaining observed proportions by chance,
under a specific hypothesis. - ANSWER ✔✔ Chi-square test
T/F Given a P value less than 0.01, our model of genetics probably fails to explain the observed
phenotypes. - ANSWER ✔✔ True
T/F A statistical test estimates the probability of obtaining a certain deviation from predictions
based on sampling error. - ANSWER ✔✔ True
T/F We cannot use Mendel's model alone to explain variation in continuous traits, such as height.
- ANSWER ✔✔ True
T/F Mendel's model applies only to discrete traits, which have only two or three phenotypes
(e.g., round or wrinkled peas; or red, pink, and white flowers). - ANSWER ✔✔ True
T/F Fisher's revised model of inheritance started the field of quantitative genetics, which focuses
on continuous traits. - ANSWER ✔✔ True
The resulting theory of evolution grew during a period from 1920-1940 is known as - ANSWER
✔✔ Modern Synthesis