The fifth week of pregnancy.
Answer:
Option C, disruptive selection
Explanation:
Disruptive selection selects features and traits which are against the average population. The population would have extreme phenotypes but nothing in middle of that and hence could lead to deviation.
The higher percentage of male human population has deeper voice as dominant characteristics. The intermediate male population with a finer or less deep voice is small. As per the disruptive selection theory, the traits at the extremities are favored more over and above the mediocre traits and hence intermediate species (males with less deep or fine voice) will compete among themselves to produce species with ability on the extremes i.e deep voice or with higher level of testosterone.
Hence, option C is correct
Answer:
The constants: 14 hours of light per day; food one time per day 5 oz; wheel for exercise; cage size
The independent variable: the temperature
The dependent variables: length and weight
Data to be collected: weight of mice; length of mice
A flaw in the experiment: the sample size is to small to be random; it doesn't specify a control group
Explanation:
The constants are the controlled variables. The scientist controls these items because a variable that remains unchanged prevents it from having any effect on the outcome.
The independent variable is the condition manipulated by the scientist to see its effects.
The dependent variables are the ones being measured or tested for in an experiment.
Randomization means every subject gets an equal chance at being in the control group. There are to few subjects. A control group would consist of elements that present exactly the same characteristics of the experimental group, except for the variable applied to the latter.
Answer:
"Brachiopoda"
Drosal and ventral valves; Individuals resemble clams; May have no anus
"Bryozoa"
Possesses a chitinous zoecium; Anus opening near mouth; Colonies resemble moss
<span>Polydactyly in its most common form, is an autosomal dominant mutation. Let's represent the mutation as P and the normal state as +. Thus, if both parents are heterozygous (P+), both would have the mutation (and thus have six fingered). They have 3/4 chances of having a child with the extra digit, 1/4 normal. </span>