Answer:
cell wall, cell membrane, cytoplasm, vacuole, ER and nucleous
Explanation:
<u>Given:</u>
Present population of bison = 3652
Maximum growth rate = 0.28
<u>To find:</u>
The population of bison after one year.
<u>Solution:</u>
The future population can be calculated with the help of the following formula,
where 
is the future population, 
is the present population, r is the growth rate and n is the number of years.
On substituting the given values in the formula we get,
Therefore, the population size of bison after one year is 4675. The park does not have to control the population because the growth rate is on an average scale.
Answer:
F1 Generation -
Genotype - GgRr
Phenotype - Green and round
F2 generation -
Genotype Ratio –
GGRR: GGRr: GgRR: GgRr: GGrr: Ggrr: ggRR: ggRr: ggrr
1:2:2:4:1:2:1:2:1
Phenotypic ration
Green & round: Green & wrinkled: yellow & round:yellow : wrinkled
9: 3:3:1
Explanation:
G represents Green seed color
g represents yellow seed color
R represents round seed
r represents wrinkled seed
Here G is dominant over g and R is dominant over r
F1 generation -
Genotype of true breeding parents will be
GGRR and ggrr
The offspring produced will be -
GgRr
Genotype - GgRr
Phenotype - Green and round
F2 Generation -
Genotype Ratio –
GGRR: GGRr: GgRR: GgRr: GGrr: Ggrr: ggRR: ggRr: ggrr
1:2:2:4:1:2:1:2:1
Phenotypic ration
Green & round: Green & wrinkled: yellow & round:yellow : wrinkled
9: 3:3:1
Answer:
See the answer below
Explanation:
Let the disorder be represented by the allele a.
Since the disease is an autosomal recessive one, affected individuals will have the genotype aa and normal individuals will have the genotype Aa or AA.
Since the four adults are carriers, their genotypes would be Aa.
Aa x Aa
Progeny: AA 2Aa aa
Probability of being affected = 1/4
Probability of being a carrier = 1/2
Probability of not being affected = 3/4
(a) The chance that the child second child of Mary and Frank will have alkaptonuria = 1/2
(b) The chance that the third child of Sara and James will be free of the condition = 3/4
(c)
(d) If someone has no family history of the disorder, their genotype would be AA.
AA x aa
4 Aa
<em>The chance that a child with alkaptonuria will have an offspring with alkaptonuria if the child's mate has no family history </em>= 0
(e)
(f) <em>The chance that a child with alkaptonuria will have an offspring with alkaptonuria if the child's mate has no family history</em> = 0
