plants cannot sustain themselves because they need dirt for fertilizer the sunlight to grow in water for hydration. they also need homeostasis for a heathy balance
Answer:
½ c raw red bell pepper (190 mg per cup)
1 medium orange (96 mg per cup)
½ cup cooked broccoli (81 mg per cup, uncooked)
¾ cup tomato juice (44 mg per cup, 33 mg in 3/4 cup)
Explanation:
Even if the reference measurements is 1 cup the proportion of vitamin C is maintained when half a cup of the measure is used.
As I do not know the size of the orange I am considering that half an orange is equal to half a cup.
Brococoli is cooked and Vitamin C is vulnerable to the cooking process and breaks down easily. So its vitamin C content will vary depending on how it was cooked and how long it lasted.
ANSWER: MITRAL VALVE STENOSIS
EXPLANATION:
The child have the risk of having MITRAL VALVE STENOSIS. It is also referred to as mitral stenosis.
Mitral valve stenosis occurs as results of the mitral valve opening narrowing. Which effect to less blood flowing through it.
The mitral valve is located between two chambers (the atrium and the ventricle) on the left side of your heart.
However, Mitral valve stenosis can lead to different health issues, including blood clots, difficulty breathing, fatigue, and heart failure.
Mitral valve stenosis is specifically caused by rheumatic fever (a childhood disease). This rheumatic fever occurs has a result of the body's immune response to an infection associated with the streptococcal bacteria.
Acute rheumatic fever affects the joints and the heart greatly. It causes joints inflammation temporarily and in severe case causes chronic disability.
Nevertheless, this cardiac complication have treatment and it is based on whether the affected individuals shows symptoms. Medications like blood thinners or anticoagulants (to reduce the risk of blood clots), diuretics, antiarrhythmics (to cure abnormal heart rhythms), beta-blockers (to slow your heart) etc, are being administered based on the level of the complication.
Admitting that the "a" is a capital A for normal pigmentation and "d" is a capital D for dimpled chin, meaning that these are the dominant traits, the fraction expected to be albino with a non-dimpled chin is of 1/16.
When two heterozygous are crossed and two characteristics are being analysed, the offspring quantity that will possess the two recessive traits can be represented by 1/16. This is easily confirmed when a Punnett square is made. Considering that both parents were heterozygous, on both sides of the crossing in the Punnett square, you would have the following alleles' combination: AD, Ad, aD, and ad. The offspring that would be homozygous recessive (aadd) would correspond to only 1/16.
