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UNIT 4
NS270 NUTRITIONAL ASSESSMENT
AND MANAGEMENT
Amy Habeck, RD, MS, LDN
Unit 4 Learning Objectives:
Practice Calculations
1)
1)
Review- Nutrition and Diet Therapy
2)
a)
3)
Unit 4 project calculations
Chapter 7 – Assessment of the Hospitalized Patient
Answer your questions
Nutrition Screening

Screening of hospitalized patients
 Completed
within first 24-48 hours
 Identifies characteristics known to be associated with
nutrition problems
 Malnutrition
 Nutritional
risk
 What are some diagnoses that put a patient at increased
nutritional risk?
 Patients
at nutritional risk should have a nutrition
assessment
Diagnoses Associated with
Increased Nutritional Risk

Trauma: fracture, burn, closed head
injury, GSW, spinal cord injury, MVA

HIV/AIDS

V/D

Dysphagia

Anemia

Bowel resection

CVA or hemiparesis

Short bowel syndrome

GI bleed

Small bowel obstruction

Crohn’s disease

Hypoglycemia

Dumping syndrome

FTT

Pressure ulcers

Congenital heart disease

Organ transplant

COPD

DM

Anorexia

CAD

Cancer

pancreatitis
Case Study-Question 1
How can you estimate her
height?
Table 7.1 page 219
S=75.00 +(1.91 KH)-(0.17A)

Equation for white female >60 y.o.
= (substitute numbers and complete calculation)
72 year old female
History of osteoporosis with
compression fracture
KH = 16.5 inches
S=stature
KH=knee height (cm)
A= age
Case Study- Question 1
S=stature
KH=knee height (cm)
A= age
KH=16.5 in x 2.54cm/in=41.91cm
A= age=72
S=75.00 +(1.91x KH)-(0.17xA)
S=75.00+(1.91x41.91)-(0.17x72)
S=75.00+80-12.2
72 year old female
S=142.8 cm
History of osteoporosis with
compression fracture
KH = 16.5 inches
Case Study – Question 2

Calculate adjusted body weight (ABW)
 Equation
and example on page 225
 Table 7.5 : % of body weight contributed by body
parts


=% of body weight = lower leg + foot
= (substitute numbers and complete calculation)
Helen
Wt = 115
Amputation =Lower leg and foot
Case Study – Question 2





Calculate adjusted body weight (ABW)
Equation and example on page 225
Table 7.5 - % of body weight contributed by body parts
ABW= current wt/(100-% of amputation)x 100
= (substitute numbers and complete calculation)
Helen
Wt = 115
Amputation =Lower leg and foot = 5.3 + 1.8= 7.1%
Case Study – Question 2







Calculate adjusted body weight
Equation and example on page 225
Table 7.5 - % of body weight contributed by body parts
Adjusted wt = current wt/(100-% of amputation)x 100
Adjusted wt = 115/(100 - 7.1) x 100
Adjusted wt = 115/(92.9) x 100
Adjusted wt = 123.79 pounds
Helen
Wt = 115
Amputation =Lower leg and foot = 5.3 + 1.8= 7.1%
Calculate BMI based on adjusted
body weightQuestion 3
P176 Lee and Nieman, classification table 6.6
BMI based on adjusted body weight=
=ABW(pounds)/ht(in)/ht(in)x703=
= (substitute numbers and complete calculation)

Justin
Entire right leg amputation
Ht: 5’6” = 66”
ABW: 178 pounds
Calculate BMI based on adjusted
body weightQuestion 3
P176, classification table 6.6
Ht= 5’6”=66”
 BMI based on adjusted body weight=
=ABW(pounds)/ht(in)/ht(in)x703=
=178/(66)2x703=28.7

Justin
Entire right leg amputation
Ht: 5’6”
ABW: 178 pounds
Evaluating Desirable Body WeightQuestion 4

Hamwi equations
 Hamwi
equation is found on page 170-171 of Lee and
Nieman
 Men
 5’
= 106# + 6# for every additional inch +/- 10%
 Women
 5’=100#

+ 5# for every additional inch +/- 10%
Height-weight tables
 What
are some of the limitations of the height-weight
tables?
IBW-Number 4

What is her desirable or ideal body weight?
 Use
the Hamwi equation
 = (substitute numbers and complete calculation)
 Page 170-171
48 year old patient, Ms. Geneva
Female
Height: 5’6”
Actual body weight: 155#
Let’s Practice – Question 4




What is her desirable or ideal body weight?
Women: 5’=100 + (5x6) +/- 10%
IBW=130+/- 10% or 130x.9 to 130x1.1
IBW=117-143#
48 year old patient, Ms. Geneva
Female
Height: 5’6”
Actual body weight: 155#
Using Anthropometric Measures to
estimate weight-Number 5

Equation: page 224, table 7.3
Female: (MAC x 1.63)+(CCx1.43)-37.46
Male: (MACx2.31)+(CCx1.5)-50.10
Estimated weight =

= (substitute numbers and complete calculation)



Marjorie
MAC = 30 cm
CC = 34 cm
Using Anthropometric Measures to
estimate weight-Number 5






Equation: page 224, table 7.3
Female: (MAC x 1.63)+(CCx1.43)-37.46
Estimated weight = (MAC x 1.63)+(CCx1.43)-37.46
Estimated weight =(30 x 1.63)+(34x1.43)-37.46=
Estimated weight = 48.9+48.62-37.46=60kg
Estimated weight = 60kgx2.2pounds/kg=132 pounds
Marjorie
MAC = 30 cm
CC = 34 cm
Using Anthropometric Measures to
estimate weight-Number 6

Known: KH and MAC (both in cm)
Equation: page 225, table 7.4
Convert your answer to pounds

Est. weight = (KHx1.09)+(MAC x 3.14)-83.72

= (substitute numbers and complete calculation)


50 y.o. black male
KH = 42 cm
MAC = 30 cm
Using Anthropometric Measures to
estimate weight-Number 6








Known: KH and MAC (both in cm)
Equation: page 225, table 7.4
Convert your answer to pounds
50 y.o. black male
KH = 42 cm
MAC = 30 cm
Est. weight = (KHx1.09)+(MAC x 3.14)-83.72
Est. weight = (42x1.09)+(30 x 3.14)-83.72
Est. weight = (45.78)+(94.2)-83.72
Est. weight = 55.96 kg
55.96 kg x 2.2 pound/kg = 123 pounds
Calculating REE
Question 7

Calculate the REE for Ms. Geneva
 What
results do you get with each of the methods
below?
48 year old patient, Ms. Geneva
Female with right below-knee
Jeor
amputation
 Harris-Benedict
Height: 5’6”
Actual body weight: 155#
 WHO
 National Academy of Sciences
 Mifflin-St.
 How
do they compare?
 Which method do you like better and why?
 Apply an activity factor for average activity to each of
your results. List the answer separately.
Ms. Geneva- Mifflin-St. Jeor







48 year old female patient, s/p R BKA
Height: 5’6”=66inx2.54cm/in=167.6cm
Actual body weight: 155# x 1kg/2.2#=70.5kg
Female: REE = 10×weight + 6.25×height - 5×age - 161
REE= (10x70.5)+(6.25x167.6)-(5x48)-161
REE= (705)+(1047.5)-(240)-161
REE= 1351.5 kcal/day
Ms. Geneva-Harris Benedict








48 year old female patient, s/p R BKA
Height: 5’6”=66inx2.54cm/in=167.6cm
Actual body weight: 155#x1kg/2.2#=70.5kg
p 232, table 7.7
REE=655.1+9.6W+1.9S-4.7A
REE=655.1+(9.6x70.5)+(1.9x167.6)-(4.7x48)
REE=655.1+676.8+318.4-225.6
REE=1425 kcal/day
Ms. Geneva- WHO








48 year old female patient, s/p R BKA
Height: 5’6”=66inx2.54cm/in=167.6cm
Actual body weight: 155#x1kg/2.2#=70.5kg
p 232, table 7.7
REE=8.7W+829
REE=8.7x70.5+829
REE=613.4+829
REE=1442kcal/d
Ms. Geneva- NAS








48 year old female patient, s/p R BKA
Height: 5’6”=66inx2.54cm/inx1m/100cm=1.68m
Actual body weight: 155#x1kg/2.2#=70.5kg
p 233, table 7.8
REE=247-(2.67xA)+(401.5xht)+(8.6xwt)
REE=246-(2.67x48)+(401.5x1.68)+(8.6x70.5)
REE=245-128.2+674.5+606.3
REE=1398kcal/d
Compare

Mifflin-St. Jeor
 REE=1351

Harris Benedict
 REE=1425

kcal/d
NAS
 REE=1398

kcal/d
WHO
 REE=1442

kcal/d
kcal/d
How do you think these equations compare?
EER-Number 8

P 233-234, table 7.9
EER for males >19 years

EER= (substitute numbers and complete calculation)

29 y.o. male
PA=physical activity factor
Ht in m= 1.78m
wt in kg=90.5
PA=1.11 (low active)
age=45
EER-Number 8

P 233-234, table 7.9
EER for males >19 years
PA=physical activity factor

Ht in m= 1.78m, wt in kg=90.5, PA=1.11 (low active), age=45

EER=662-9.53(age)+PA x(15.91x wt+539.6 x ht)







EER=662-9.53(45y)+1.11 x(15.91x 90.5kg+539.6 x 1.78m)
EER=662-428.85+1.11 x( 1439.9+ 960.49)
EER=662-428.85+1.11 x( 2400.39)
EER=662-428.85+ 2664.4
EER=2897.55 kcal
Calculate TEE for Overweight
Adults- Number 9

Complete the calculations for Ms. Geneva using the
TEE equation in Table 7.10 on page 235.
48 year old female patient, s/p R BKA
Height: 5’6”=66inx2.54cm/in=167.6cm
Actual body weight: 155#x1kg/2.2#=70.5kg
Use an activity factor for active adults.

= (substitute numbers and complete calculation)




Calculate TEE for Overweight
Adults- Number 9









48 year old female patient, s/p R BKA
Height: 5’6”=66inx2.54cm/in=167.6cm
Actual body weight: 155#x1#/2.2kg=70.5kg
TEE=448-(7.95xage)+PA x(11.4x Wt + 619x ht)
TEE=448-(7.95x48)+1.27 x(11.4x70.5+619x1.68)
TEE=448-(7.95x48)+1.27 x(803.7+1039.9)
TEE=448-384.6+1.27 x 1843.6
TEE=448-384.6+2341.4
TEE=2405kcal/d
Ireton-Jones-number 10





Ireton-Jones recommended for overweight
individuals in critical condition
Activity factors (AF): box 7.3, page 233
Injury factors (IF): table 7.11, page 237
Obesity factor: 1=BMI>27, 0=BMI<27
IJEE: 629-11(age)+25(wt)-609(obesity factor)
Ireton-Jones Energy Expenditure
Number 10


Ireton-Jones Equation
IJEE = 629 - 11 (A) + 25 (W) - 609 (O)
A = age in years, W = weight in kg, O= obesity
 Obesity

: BMI >27 = 1, BMI ≤27 = 0
Kcals = IJEE x AF x IF
 Activity
factors (AF) - Box 7.3, p. 233
 Injury factors (IF) -Table 7.11, p. 237
 IJEE = (substitute numbers and complete calculation)
50 year old female, confined to bed after
minor surgery
Ht: 5’4”
Actual body weight: 176#
BMI = 30.2
Ireton-Jones Energy Expenditure
Number 10





Ireton-Jones Equation
IJEE = 629 - 11 (A) + 25 (W) - 609 (O)
IJEE=629 –(11x 50)+(25x80)-609(1)
IJEE=629 – 550+2000-609
IJEE=1470 kcal/day
50 year old female, confined to bed after
minor surgery
Ht: 5’4”
Actual body weight: 176#
BMI = 30.2
Ireton-Jones Energy Expenditure
Number 10






Ireton-Jones Equation
IJEE = 629 - 11 (A) + 25 (W) - 609 (O)
IJEE=629 –(11x 50)+(25x80)-609(1)
IJEE=629 – 550+2000-609
IJEE=1470 kcal/day
Kcals = IJEE x AF x IF
 Activity
factors (AF) – confined to bed=1.2
 Injury factors (IF) –minor surgery=1.0-1.1


Kcals = 1470 x 1.2 x 1.1
Kcals= 1940 kcal/day
50 year old female, confined to bed after
minor surgery
Ht: 5’4”
Actual body weight: 176#
Estimating Protein Needs- Number
11


Estimate Ms. Geneva’s protein needs for an
individual who has undergone major surgery.
P 239, table 7.13
48 year old patient, Ms. Geneva
Female
Height: 5’6”
Actual body weight: 155#
Estimating Protein Needs Number
11

Estimate Ms. Geneva’s protein needs for an
individual who has undergone major surgery.
 Moderate
stress level: 1.2-1.8 gm/kg
Estimating Protein Needs Number
11

Estimate Ms. Geneva’s protein needs for an
individual who has undergone major surgery.
 Moderate



stress level: 1.2-1.8 gm/kg
155 # = 70.45 kg
70.45 x 1.2 = 84.54 gm protein/d
70.45 x 1.8 = 126.81 gm protein/d
Questions About Assessment of
Hospitalized Patients?
Farewell



Thank you for your kind attention and
participation!
Email any time - [email protected]
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
@ProfAmyH