If a pea plant that is herterozygous for rounds, yellow peas is crossed with a pea plant that is homozygous for round peas but h
eterozygous for yellow peas, how many different phenotypes are their offspring expected to show?
1 answer:
Answer: There re two different types of phenotypes in the offsprings that is Round yellow and Round green.
Explanation-
As per the Mendelian inheritance, Round shape and yellow color of pea are dominant over wrinkled shape and green color of pea.
According to the question, genotypes of parents are- RrYy ( heterzygous for round and yellow) and RRYy ( homozygous for round and heterozygous for yellow).
The first parent can produce four different gametes that is RY, Ry, rY, ry whereas the second parent will produce only two types of gametes that is RY and Ry.
When they are crossed, they will produce two different phenotypes in the offsprings.
Refer punnett square.
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Given:
The heat content of the pizza is Q = 500 kcal
The volume of the cold water is V = 50 L
The mass of a liter of cold water is 1 kg.
Calculate the mass, M, of cold water.
M = (50 L)*(1 kg/L) = 50 kg
The specific heat of water is
c = 4.184 kJ/(kg-K)
Also,
1 kcal = 4184 J, therefore
Q = (500 kcal)*(4184 J/kcal) = 2.092 x 10⁶ J
Let ΔT = increase in temperature of the cold water, °C (same as K).
Then
Q = M*c*ΔT
or
ΔT = Q/(M*c)
Answer: The temperature increases by 10 °C (or 10 K)
The heat content of the pizza is Q = 500 kcal
The volume of the cold water is V = 50 L
The mass of a liter of cold water is 1 kg.
Calculate the mass, M, of cold water.
M = (50 L)*(1 kg/L) = 50 kg
The specific heat of water is
c = 4.184 kJ/(kg-K)
Also,
1 kcal = 4184 J, therefore
Q = (500 kcal)*(4184 J/kcal) = 2.092 x 10⁶ J
Let ΔT = increase in temperature of the cold water, °C (same as K).
Then
Q = M*c*ΔT
or
ΔT = Q/(M*c)
Answer: The temperature increases by 10 °C (or 10 K)