Since it's been crossed with a homozygous wrinkled green, the offspring has a genotype for heterozygous round and yellow. As round and yellow are dominant traits, they're expressed in the phenotype. But when self pollinated in the f2 generation, the recessive ones will show as well
Hope it helps :')
Answer:
Activated
Explanation:
In the presence of lactose, and in the absence of glucose, lactose will bind to a protein called a "repressor," deactivating it. Through this, RNA polymerase has a free way to synthesize the mRNA that will give enzymes for lactose degradation.
I think I understand even without the picture. I'll add a picture of the Punnett Square filled in, but what you're crossing is
Tt x tt (heterozygous crossed with a homozygous recessive)
The ratio you get in the end is 2 heterozygous (Tt) and 2 homozygous (tt) offspring, so the ratio is 1:1.
So the percentage of offspring that are homozygous recessive is 50%.
Cerebral cortex
1. Emotion can invoke an autonomic response.
2. Sexual thoughts or image which can increase blood flow to the genitals.
Hypothalamus.
1. It activates the flight or fight response.
2. It is the major control of ANS.
3. It is the integrating center for thermoregulation.
Brainstem.
1. It is integrating centers for reflexes which control heart rate and blood pressure.
2. It is the integrating center for pupillary reflex
Spinal cord
1. It is the integrating center for urination, erection, defecation and ejaculation reflexes.
<span>F- allele for freckles
f- </span><span>allele without freckles
1) The man is heterozygote and has freckles, its indicating that the allele for freckles is dominant.
A cross between him and a woman who is also </span><span>heterozygote: Ff x Ff
it would result in the following probabilities:
- 1/4 - homozygote with freckles: FF
- 2/4 - </span><span>heterozygote with freckles: Ff
- 1/4- </span><span>homozygote without freckles:ff
Their son would have a probability of 75% of being born with freckles.
2) The cross resulted in this probabilities:
</span><span><span>- 1/4 - homozygote with freckles: FF
- 2/4 - </span><span>heterozygote with freckles: Ff
- 1/4- </span><span>homozygote without freckles:ff
So, the chance of being born heterozygote for this gene is 2/4, which is the same as half (50%).
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