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algol13
3 years ago
12

Dylan is on a strict vegan diet. He doesn’t consume meat, fish, eggs, or dairy products. Which nutrient is difficult to obtain n

aturally from a vegan diet
Biology
1 answer:
stepladder [879]3 years ago
5 0

Answer:

According to the given situation, the diet of Dylan lacks Vitamin B12. It is an essential nutrient that is not found in any plant source of food. It is mainly found in animal sources of food like fish, meat, beef, milk and dairy products. It is an essential nutrient that is involved in the development of red blood cells, and proper functioning of brain.

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A daughter has hemophilia a recessive sex-linked what are the possible genotypes of her parents
marishachu [46]

Answer:

If a female child has hemophilia it is possible that the mother is a carrier of the hemophilia gene and the father has hemophilia.

Explanation:

  • A daughter gets X chromosome from both her parents.
  • It is generally seen in males.
  • Hemophilia is generally recessive in females.They act as carriers of hemophilia. This occurs because they have a X chromosome that dominates the hemophilia affected gene that they inherit from any parent.
  • But, if both the parents have faulty genes ,i.e the mother is the carrier of the gene and the father is hemophiliac, then the chances are the daughter has hemophilia too.
8 0
3 years ago
In animals, the inability to make the pigment melanin results in albinism, a recessive autosomal condition. Two unaffected paren
Fittoniya [83]

Answer: 1/16, or approximately 6.25% (see explanation below)

Explanation:

Answering this question requires two steps.

First, we need to figure out the probability that this couple will have a child with albinism in the first place. We know the following:

- Both parents are unaffected.

- The couple has already had one affected child.

- Albinism follows an autosomal recessive inheritance pattern.

Let ( M = normal gene ) and ( m = mutated gene ). Since the condition is recessive, the affected child can be assumed to have a “mm” genotype. Barring the possibility of a de novo mutation (which are assumed to be rare), the affected child must have inherited one ”m” allele from each parent. Since both of them are unaffected, however, we can assume that they are both carriers (genotype “Mm”). In conclusion, 1/4 of their offspring (25%) <em>for any given pregnancy</em> may be expected to have albinism. See the resulting Punnett square:

<u>       |      M    |      m   </u>

<u>M    |    MM   |    Mm  </u>

<u>m    |    Mm   |    mm  </u>

Note that the question asks about the probability that not one but two consecutive births result in affected children. Since it can be assumed that both events are independent (meaning: the outcome of a pregnancy does not influence the outcome of following ones), we may apply the rule of multiplication for probabilities. The final answer is therefore 1/4 * 1/4 = 1/16.

7 0
4 years ago
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