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Fofino [41]
3 years ago
12

The influence of how a mother’s nutrient intake may change gene expression to increase the infant’s development of obesity or ch

ronic diseases later in life is referred to as ________.
Biology
1 answer:
8_murik_8 [283]3 years ago
3 0

Answer:

The influence that maternal nutrition -the mother's intake of nutrients- has on the fetal gene expression, as well as its development and growth, is referred to as fetal programming, responsible for the predisposition to obesity or chronic diseases during its lifetime.

Explanation:

Both gene expression and fetal growth and development depend on three relevant factors:

  • <em>Proper placental function.</em>
  • <em>Maternal nutrition</em><em>.</em>
  • <em>How the fetus uses maternal nutrients. </em>

The above factors can eventually <u>induce changes in the metabolism of the developing fetus</u>, which manages to adapt, but with consequences on the normal developmental pathway, which can trigger negative metabolic disturbances.

Fetal programming - also referred to as fetal metabolism programming - includes a number of factors that influence postnatal metabolic behavior, including susceptibility to obesity or chronic disease.

Learn more:

Fetal alcohol syndrome brainly.com/question/1083413

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3 years ago
4. The Rh blood type response system is controlled by the D allele. The genotypes DD and Dd are Rh (Rh positive); dd is Rh- (Rh
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Answer:

A - Frequency of D allele = 0.35

Frequency of d allele = 0.65

B- Frequency of DD genotype = 0.1225

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Frequency of dd genotype = 0.425.

C - Total population = 400

Frequency of heterozygous population = 400*0.455 = 183.

EXPLANATION:

The equilibrium sum of all the allelic frequency of a gene is always 1 and sum of all the genotypic frequency of all the genotype is always 1 - This is according to Hardy-Weinberg.

So p+q =1

p2+ 2pq+q2 =1

where p = frequency of dominant allele

q is the frequency of the recessive allele.

from the question, it was given that there were  170 individual out of 400, which were Rh- negative,

So q2 = 170/400 = 0.425

q= 0.65

Also p+q =1

so p = 1-q

or p = 1-0.65

Hence p =0.35

Frequency of homozgupus for D allele = 0.35*0.35 = 0.1225

Frequency of heterozygous or Dd will be 2pq.

or 2*0.35*0.65 = 0.455

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